| Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
Exercise 4.4
Question 1
Fill in the blanks
(i) sin2 θ cosec2 θ = ……..(ii) 1 + tan2 θ = ……
(iii) Reciprocal sin θ. cot θ = ……
(iv) 1–.......= cos2θ
(v) $\tan A=\frac{\cdots \cdots \cdot}{\cos A}$
(vi) $\ldots \ldots=\frac{\cos A}{\sin A}$
(vii) cos θ is reciprocal of .........
(viii) Reciprocal of sin θ is.........
(ix) Value of sin θ in terms of cos θ is
(x) Value of cos θ in terms of sin θ is
Sol :
(i) Given: sin2 θ cosec2 θ
$\Rightarrow \sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\csc \theta}\right]$
= 1
= 1
(ii) Given: 1 + tan2 θ
$=1+\left(\frac{\sin \theta}{\cos \theta}\right)^{2}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}$[∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec2 θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
$=1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}$[∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec2 θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
(iii) Given : sin θ cot θ
Firstly, we simplify the given trigonometry
Firstly, we simplify the given trigonometry
$=\sin \theta \times \frac{\cos \theta}{\sin \theta}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
= cos θ
Now, the reciprocal of cos θ is
$=\frac{1}{\cos \theta}$
=sec θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
= cos θ
Now, the reciprocal of cos θ is
$=\frac{1}{\cos \theta}$
=sec θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
(iv) Given: 1 – x = cos2θ
Subtracting 1 to both the sides, we get
1 – x –1 = cos2θ – 1
⇒ –x = – sin2θ
⇒ x =sin2 θ
(v) $\tan A=\frac{\sin A}{\cos A}$
(vi) $\cot A=\frac{\cos A}{\sin A}$
(vii) $\cos \theta=\frac{1}{\sec \theta}$
(viii) $\sin \theta=\frac{1}{\cos \theta}$
(ix) We know that
cos2 θ + sin2 θ = 1
⇒ sin2 θ = 1 – cos2 θ
⇒ sin θ = √(1 – cos2 θ)
(x) We know that
cos2 θ + sin2 θ = 1
⇒ cos 2 θ = 1 – sin2 θ
⇒ cos θ = √(1 – sin2 θ)
Question 2
If sin θ= p and cos θ = q, what is the relation between p and q ?
We know that,
cos2 θ + sin2 θ = 1 …(i)
Given : sin θ = p and cos θ = q
Putting the values of sin θ and cos θ in eq. (i), we get
(q)2 + (p)2 =1
⇒ p2 + q2 =1
Question 3
If cos A = x, express sin A in terms of x
We know that
cos2 θ + sin2 θ = 1
⇒ sin2 θ = 1 – cos2 θ
⇒ sin θ = √(1 – cos2 θ)
And Given that cos θ = x
⇒ sin θ = √(1– x2)
Question 4
If x cos θ = 1 and y sin θ = 1 find the value of tan θ.
Sol :Given x cosθ = 1 and y sinθ = 1
$\Rightarrow \cos \theta=\frac{1}{x}$ and $\sin \theta=\frac{1}{\mathrm{y}}$
Now, we know that
Now, we know that
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
Putting the value of sin θ and cos θ, we get
$\tan \theta=\frac{\frac{1}{y}}{\frac{1}{x}}$
$\Rightarrow \tan \theta=\frac{x}{y}$
Sol :
We know that
cos2 θ + sin2 θ = 1
⇒ cos2 40° + sin2 40° = 1
⇒ sin2 40° = 1 – cos2 40°
⇒ sin 40° = √(1 – cos2 40°)
And Given that cos 40° = p
⇒ sin 40° = √(1– p2)
We know that
cos2 θ + sin2 θ = 1
⇒ cos2 77° + sin2 77° = 1
⇒ cos 2 77° = 1 – sin2 77°
⇒ cos 77° = √(1 – sin2 77°)
And Given that sin 77° = x
⇒ cos 77° = √(1 – x2 )
We know that
cos2 θ + sin2 θ = 1
⇒ cos2 55° + sin2 55° = 1
⇒ sin2 55° = 1 – cos2 55°
⇒ sin 55° = √(1 – cos2 55°)
And Given that cos 55° = x2
⇒ sin 55° = √{1– (x2)2}
⇒ sin 55° = √(1 – x4)
Sol :
We know that
cos2 θ + sin2 θ = 1
⇒ cos2 50° + sin2 50° = 1
⇒ cos 2 50° = 1 – sin2 50°
⇒ cos 50° = √(1 – sin2 50°)
And Given that sin 50° = a
⇒ cos 50° = √(1 – a2 )
Sol :
Given x cos A = 1 and tan A =y
$\Rightarrow \mathrm{x}=\frac{1}{\cos \mathrm{A}}$ and $\frac{\sin A}{\cos A}=y$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
To find: x2 – y2
Putting the value of sin θ and cos θ, we get
$\tan \theta=\frac{\frac{1}{y}}{\frac{1}{x}}$
$\Rightarrow \tan \theta=\frac{x}{y}$
Question 5
If cos40o = p, then write the value of sin 40o in terms of p.
We know that
cos2 θ + sin2 θ = 1
⇒ cos2 40° + sin2 40° = 1
⇒ sin2 40° = 1 – cos2 40°
⇒ sin 40° = √(1 – cos2 40°)
And Given that cos 40° = p
⇒ sin 40° = √(1– p2)
Question 6
If sin 77° = x, then write the value of cos 77o in terms of x.
Sol :We know that
cos2 θ + sin2 θ = 1
⇒ cos2 77° + sin2 77° = 1
⇒ cos 2 77° = 1 – sin2 77°
⇒ cos 77° = √(1 – sin2 77°)
And Given that sin 77° = x
⇒ cos 77° = √(1 – x2 )
Question 7
If cos55° = x2, then write the value of sin 55o in terms of x.
Sol :We know that
cos2 θ + sin2 θ = 1
⇒ cos2 55° + sin2 55° = 1
⇒ sin2 55° = 1 – cos2 55°
⇒ sin 55° = √(1 – cos2 55°)
And Given that cos 55° = x2
⇒ sin 55° = √{1– (x2)2}
⇒ sin 55° = √(1 – x4)
Question 8
If, sin 50° = α then write the value of cos 50° in terms of α.
We know that
cos2 θ + sin2 θ = 1
⇒ cos2 50° + sin2 50° = 1
⇒ cos 2 50° = 1 – sin2 50°
⇒ cos 50° = √(1 – sin2 50°)
And Given that sin 50° = a
⇒ cos 50° = √(1 – a2 )
Question 9
If x cos A = 1 and tan A = y, then what is the value of x2 – y2.
Given x cos A = 1 and tan A =y
$\Rightarrow \mathrm{x}=\frac{1}{\cos \mathrm{A}}$ and $\frac{\sin A}{\cos A}=y$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
To find: x2 – y2
Putting the values of x and y , we get
$\left(\frac{1}{\cos A}\right)^{2}-\left(\frac{\sin A}{\cos A}\right)^{2}$
$=\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A}{\cos ^{2} A}$ [∵ cos2 θ + sin2 θ = 1]
= 1
Sol :
Taking LHS = (1 – sinθ)(1+ sinθ)
Using identity , (a + b) (a – b) = (a2 – b2) , we get
= (1)2 – (sinθ)2
= 1 – sin2 θ
= cos2 θ [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Sol :
Taking LHS =(1 – cosθ)(1+ cosθ)
Using identity , (a + b) (a – b) = (a2 – b2) , we get
= (1)2 – (cosθ)2
= 1 – cos2 θ
= sin2 θ [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Sol :
Taking LHS $\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{(1)^{2}-(\cos \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$ [Using identity , (a + b) (a – b) = (a2 – b2)]
$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
= tan2 θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= RHS
Hence Proved
Sol :
Taking LHS $=\frac{1}{\sec \theta+\tan \theta}$
Multiplying and divide by the conjugate of sec θ + tan θ
$=\frac{1}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta}$ [Using identity , (a + b) (a – b) = (a2 – b2)]
= sec θ – tan θ [∵ 1+ tan2 θ = sec2 θ ]
= RHS
Hence Proved
Sol :
Taking LHS = sin θ cot θ
$=\sin \theta \times \frac{\cos \theta}{\sin \theta}\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
= cos θ
=RHS
Hence Proved
Sol :
Sol :
Taking LHS = cos2 A (tan2 A+1)
= cos2 θ (sec2 θ) [∵ 1+ tan2 θ = sec2 θ ]
$\left(\frac{1}{\cos A}\right)^{2}-\left(\frac{\sin A}{\cos A}\right)^{2}$
$=\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A}{\cos ^{2} A}$ [∵ cos2 θ + sin2 θ = 1]
= 1
Question 10
Prove the followings identities:
(1 – sin θ)(1 + sin θ) = cso2 θSol :
Taking LHS = (1 – sinθ)(1+ sinθ)
Using identity , (a + b) (a – b) = (a2 – b2) , we get
= (1)2 – (sinθ)2
= 1 – sin2 θ
= cos2 θ [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Question 11
Prove the followings identities:
(1 + cos θ)(1 – cos θ) = sin2θSol :
Taking LHS =(1 – cosθ)(1+ cosθ)
Using identity , (a + b) (a – b) = (a2 – b2) , we get
= (1)2 – (cosθ)2
= 1 – cos2 θ
= sin2 θ [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Question 12
Prove the followings identities:
$\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}=\tan ^{2} \theta$Taking LHS $\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{(1)^{2}-(\cos \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$ [Using identity , (a + b) (a – b) = (a2 – b2)]
$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
= tan2 θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= RHS
Hence Proved
Question 13
Prove the followings identities:
$\frac{1}{\sec \theta+\tan \theta}=\sec \theta-\tan \theta$Taking LHS $=\frac{1}{\sec \theta+\tan \theta}$
Multiplying and divide by the conjugate of sec θ + tan θ
$=\frac{1}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta}$ [Using identity , (a + b) (a – b) = (a2 – b2)]
= sec θ – tan θ [∵ 1+ tan2 θ = sec2 θ ]
= RHS
Hence Proved
Question 14 A
Prove the following identities :
sinθ. cotθ = cos θSol :
Taking LHS = sin θ cot θ
$=\sin \theta \times \frac{\cos \theta}{\sin \theta}\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
= cos θ
=RHS
Hence Proved
Question 14 B
Prove the following identities :
sin2 θ(1+ cot2 θ) = 1Sol :
Taking LHS = sin2 θ(1+ cot2 θ)
= sin2 θ (cosec2 θ) [∵ cot2 θ +1= cosec2 θ ]
= sin2 θ (cosec2 θ) [∵ cot2 θ +1= cosec2 θ ]
$=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\operatorname{cscec} \theta}\right]$
= 1
=RHS
Hence Proved
= 1
=RHS
Hence Proved
Question 14 C
Prove the following identities :
cos2 A (tan2 A+1) = 1Sol :
Taking LHS = cos2 A (tan2 A+1)
= cos2 θ (sec2 θ) [∵ 1+ tan2 θ = sec2 θ ]
$=\cos ^{2} \theta \times \frac{1}{\cos ^{2} \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
= 1
=RHS
Hence Proved
Sol :
Taking LHS = tan4 θ + tan2 θ
= (tan2 θ)2 + tan2 θ
= ( sec2 θ – 1)2 + (sec2 θ – 1) [∵ 1+ tan2 θ = sec2 θ ]
= sec4 θ + 1 – 2 sec2 θ + sec2 θ – 1 [∵ (a – b)2 = (a2 + b2 – 2ab)]
= sec4 θ – sec2 θ
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\left(1+\tan ^{2} \theta\right) \sin ^{2} \theta}{\tan \theta}$
$=\frac{\left(\sec ^{2} \theta\right) \sin ^{2} \theta}{\frac{\sin \theta}{\cos \theta}}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{1 \times \sin \theta \times \cos \theta}{\cos ^{2} \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin \theta}{\cos \theta}$
= tan θ
=RHS
Hence Proved
Sol :
Taking LHS$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+1$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec2θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
= 1
=RHS
Hence Proved
Question 14 D
Prove the following identities :
tan4θ + tan2θ = sec4θ – sec2θSol :
Taking LHS = tan4 θ + tan2 θ
= (tan2 θ)2 + tan2 θ
= ( sec2 θ – 1)2 + (sec2 θ – 1) [∵ 1+ tan2 θ = sec2 θ ]
= sec4 θ + 1 – 2 sec2 θ + sec2 θ – 1 [∵ (a – b)2 = (a2 + b2 – 2ab)]
= sec4 θ – sec2 θ
=RHS
Hence Proved
Question 14 E
Prove the following identities :
$\frac{\left(1+\tan ^{2} \theta\right) \sin ^{2} \theta}{\tan \theta}=\tan \theta$Sol :
Taking LHS $=\frac{\left(1+\tan ^{2} \theta\right) \sin ^{2} \theta}{\tan \theta}$
$=\frac{\left(\sec ^{2} \theta\right) \sin ^{2} \theta}{\frac{\sin \theta}{\cos \theta}}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{1 \times \sin \theta \times \cos \theta}{\cos ^{2} \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin \theta}{\cos \theta}$
= tan θ
=RHS
Hence Proved
Question 14 F
Prove the following identities :
$\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+1=\frac{\tan ^{2} \theta}{\sin ^{2} \theta}$Sol :
Taking LHS$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+1$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec2θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
Now, RHS $=\frac{\tan ^{2} \theta}{\sin ^{2} \theta}$
$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\sin ^{2} \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta \times \cos ^{2} \theta}$
$=\frac{1}{\cos ^{2} \theta}$
= sec2θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
∴ LHS = RHS
Hence Proved
Sol :
Taking LHS $=\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{3}{\cos ^{2} \theta}-\frac{4 \sin ^{2} \theta}{\cos ^{2} \theta}$
= 3 sec2 θ – 4 tan2 θ
We know that,
1+ tan2 θ = sec2 θ
= 3(1+ tan2 θ) – 4 tan2 θ
= 3 + 3 tan2 θ – 4 tan2 θ
= 3 – tan2 θ
=RHS
Hence Proved
Sol :
Taking LHS = (1+ tan2 θ) cos θ sin θ
= (sec2 θ) cos θ sin θ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta \times \cos ^{2} \theta}$
$=\frac{1}{\cos ^{2} \theta}$
= sec2θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
∴ LHS = RHS
Hence Proved
Question 14 G
Prove the following identities :
$\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}=3-\tan ^{2} \theta$Sol :
Taking LHS $=\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{3}{\cos ^{2} \theta}-\frac{4 \sin ^{2} \theta}{\cos ^{2} \theta}$
= 3 sec2 θ – 4 tan2 θ
We know that,
1+ tan2 θ = sec2 θ
= 3(1+ tan2 θ) – 4 tan2 θ
= 3 + 3 tan2 θ – 4 tan2 θ
= 3 – tan2 θ
=RHS
Hence Proved
Question 14 H
Prove the following identities :
(1+ tan2 θ) cos θ. sin θ = tan θSol :
Taking LHS = (1+ tan2 θ) cos θ sin θ
= (sec2 θ) cos θ sin θ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{1}{\cos ^{2} \theta} \times \cos \theta \times \sin \theta$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin \theta}{\cos \theta}$
= tan θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS = sin2 θ – cos2 φ
=( 1 – cos2 θ) – (1 – sin2 φ) [∵ cos2 θ + sin2 θ = 1] & [∵ cos2 φ + sin2 φ = 1]
= 1 – cos2 θ – 1 + sin2 φ
= sin2 φ – cos2 θ
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}$
$=\frac{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1}$
$=\frac{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\sin ^{2} \theta}}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$
= tan2 θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS = (1 – cosθ)(1+ cosθ)(1+ cot2 θ)
Using identity , (a + b) (a – b) = (a2 – b2) in first two terms , we get
= (1)2 – (cosθ)2 (cosec2 θ) [∵ cot2 θ +1= cosec2 θ]
= (1 – cos2 θ) (cosec2 θ)
= (sin2 θ) (cosec2 θ) [∵ cos2 θ + sin2 θ = 1]
$=\frac{\sin \theta}{\cos \theta}$
= tan θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved
Question 14 I
Prove the following identities :
sin2θ – cos2 ϕ = sin2ϕ – cos2θSol :
Taking LHS = sin2 θ – cos2 φ
=( 1 – cos2 θ) – (1 – sin2 φ) [∵ cos2 θ + sin2 θ = 1] & [∵ cos2 φ + sin2 φ = 1]
= 1 – cos2 θ – 1 + sin2 φ
= sin2 φ – cos2 θ
=RHS
Hence Proved
Question 14 J
Prove the following identities :
$\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta$Taking LHS $=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}$
$=\frac{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1}$
$=\frac{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\sin ^{2} \theta}}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$
= tan2 θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved
Question 15 A
Prove the following identities :
(1 – cosθ)(1+ cosθ)(1+ cot2 θ) = 1Sol :
Taking LHS = (1 – cosθ)(1+ cosθ)(1+ cot2 θ)
Using identity , (a + b) (a – b) = (a2 – b2) in first two terms , we get
= (1)2 – (cosθ)2 (cosec2 θ) [∵ cot2 θ +1= cosec2 θ]
= (1 – cos2 θ) (cosec2 θ)
= (sin2 θ) (cosec2 θ) [∵ cos2 θ + sin2 θ = 1]
$=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\csc \theta}\right]$
=1
= RHS
Hence Proved
Sol :
Taking LHS $=\frac{(1+\sin \theta)^{2}(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}$
$=\frac{1+\sin ^{2} \theta+2 \sin \theta+1+\sin ^{2} \theta-2 \sin \theta}{2 \cos ^{2} \theta}$
[∵ (a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)]
$=\frac{2+2 \sin ^{2} \theta}{2 \cos ^{2} \theta}$
$=\frac{2\left(1+\sin ^{2} \theta\right)}{2\left(1-\sin ^{2} \theta\right)}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)}$
=1
= RHS
Hence Proved
Question 15 B
Prove the following identities :
$\frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}$Sol :
Taking LHS $=\frac{(1+\sin \theta)^{2}(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}$
$=\frac{1+\sin ^{2} \theta+2 \sin \theta+1+\sin ^{2} \theta-2 \sin \theta}{2 \cos ^{2} \theta}$
[∵ (a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)]
$=\frac{2+2 \sin ^{2} \theta}{2 \cos ^{2} \theta}$
$=\frac{2\left(1+\sin ^{2} \theta\right)}{2\left(1-\sin ^{2} \theta\right)}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}$
=RHS
Hence Proved
Question 15 C
Prove the following identities :
$\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)}=\frac{1+\sin \theta}{1+\cos \theta}$Sol :
Taking LHS $=\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)}$
Multiplying and divide by the conjugate of (1 – sinθ), we get
$=\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)} \times \frac{(1+\sin \theta)}{(1+\sin \theta)}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta\left[(1)^{2}-(\sin \theta)^{2}\right]}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta \times \cos ^{2} \theta}$
$=\frac{(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)} \times \frac{(1+\sin \theta)}{(1+\sin \theta)}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta\left[(1)^{2}-(\sin \theta)^{2}\right]}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta \times \cos ^{2} \theta}$
$=\frac{(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta}$
Now, multiply and divide by conjugate of 1 – cos θ, we get
$=\frac{(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta} \times \frac{(1+\cos \theta)}{(1+\cos \theta)}$
$=\frac{\left(1^{2}-\cos ^{2} \theta\right)(1+\sin \theta)}{\sin ^{2} \theta(1+\cos \theta)}$
$=\frac{\sin ^{2} \theta(1+\sin \theta)}{\sin ^{2} \theta(1+\cos \theta)}$
$=\frac{1+\sin \theta}{1+\cos \theta}$
=RHS
Hence Proved
Sol :
Taking LHS = (sin θ – cos θ)2
Using the identity,(a – b)2 = (a2 + b2 – 2ab)
= sin2 θ + cos2 θ – 2sin θ cos θ
= 1 – 2sin θ cos θ [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Sol :
Taking LHS = (sin θ + cos θ)2 + (sin θ – cos θ)2
Using the identity,(a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)
= sin2 θ + cos2 θ + 2sin θ cos θ + sin2 θ + cos2 θ – 2sin θ cos θ
= 1 +1 [∵ cos2 θ + sin2 θ = 1]
= 2
=RHS
Hence Proved
Sol :
Taking LHS = (asin θ + bcos θ)2 + (acos θ – bsin θ )2
Using the identity,(a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)
= a2 sin2 θ + b2 cos2 θ + 2 ab sin θ cos θ + a2 cos2 θ + b2 sin2 θ – 2 ab sin θ cos θ
= a2 sin2 θ+ a2 cos2 θ + b2 sin2 θ + b2 cos2 θ
= a2 (sin2 θ + cos2 θ) + b2 (sin2 θ + cos2 θ)
= a2 + b2 [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Sol :
Taking LHS = cos4 A + sin4 A + 2 sin2 A cos2 A
Using the identity,(a + b)2 = (a2 + b2 + 2ab)
Here, a = cos2 A and b = sin2 A
= ( cos2 A + sin2 A) [∵ cos2 θ + sin2 θ = 1]
= 1
Sol :
Given:
$\begin{matrix}\sin ^{4} A-\cos ^{4} A&=2 \sin ^{2} A-1&=1-2 \cos ^{2} A&=\sin ^{2} A-\cos ^{2} A\\ \text{I}&\text{II}&\text{III}&\text{IV}\end{matrix}$
Taking I term
= sin4 A – cos4 A → I term
= (sin2 A)2 – (cos2 A)2
= (sin2 A – cos2 A)(sin2 A+ cos2 A )
[∵ (a2 – b2) = (a + b) (a – b)]
= (sin2 A – cos2 A)(1) [∵ cos2 θ + sin2 θ = 1]
= (sin2 A – cos2 A) …(i) → IV term
From Eq. (i)
= {sin2 A – (1 – sin2 A)} [∵ cos2 θ + sin2 θ = 1]
= sin2 A – 1 + sin2 A
= 2 sin2 A – 1 → II term
Again, From Eq. (i)
= {(1 – cos2 A) – cos2 A } [∵ cos2 θ + sin2 θ = 1]
=1 – 2 cos2 A → III term
Hence, I = II = III = IV
Hence Proved
Sol :
Given:
$=\frac{(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta} \times \frac{(1+\cos \theta)}{(1+\cos \theta)}$
$=\frac{\left(1^{2}-\cos ^{2} \theta\right)(1+\sin \theta)}{\sin ^{2} \theta(1+\cos \theta)}$
$=\frac{\sin ^{2} \theta(1+\sin \theta)}{\sin ^{2} \theta(1+\cos \theta)}$
$=\frac{1+\sin \theta}{1+\cos \theta}$
=RHS
Hence Proved
Question 15 D
Prove the following identities :
(sin θ – cos θ)2 = 1 – 2 sinθ . cos θSol :
Taking LHS = (sin θ – cos θ)2
Using the identity,(a – b)2 = (a2 + b2 – 2ab)
= sin2 θ + cos2 θ – 2sin θ cos θ
= 1 – 2sin θ cos θ [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Question 15 E
Prove the following identities :
(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2Sol :
Taking LHS = (sin θ + cos θ)2 + (sin θ – cos θ)2
Using the identity,(a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)
= sin2 θ + cos2 θ + 2sin θ cos θ + sin2 θ + cos2 θ – 2sin θ cos θ
= 1 +1 [∵ cos2 θ + sin2 θ = 1]
= 2
=RHS
Hence Proved
Question 15 F
Prove the following identities :
(asin θ + bcos θ)2 + (acos θ – bsin θ)2 = a2 + b2Sol :
Taking LHS = (asin θ + bcos θ)2 + (acos θ – bsin θ )2
Using the identity,(a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)
= a2 sin2 θ + b2 cos2 θ + 2 ab sin θ cos θ + a2 cos2 θ + b2 sin2 θ – 2 ab sin θ cos θ
= a2 sin2 θ+ a2 cos2 θ + b2 sin2 θ + b2 cos2 θ
= a2 (sin2 θ + cos2 θ) + b2 (sin2 θ + cos2 θ)
= a2 + b2 [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Question 15 G
Prove the following identities :
cos4 A + sin4 A + 2 sin2 A. cos2 A = 1Sol :
Taking LHS = cos4 A + sin4 A + 2 sin2 A cos2 A
Using the identity,(a + b)2 = (a2 + b2 + 2ab)
Here, a = cos2 A and b = sin2 A
= ( cos2 A + sin2 A) [∵ cos2 θ + sin2 θ = 1]
= 1
Question 15 H
Prove the following identities :
sin4 A – cos4 A = 2 sin2 A – 1 = 1 – 2 cos2 A = sin2 A – cos2 ASol :
Given:
$\begin{matrix}\sin ^{4} A-\cos ^{4} A&=2 \sin ^{2} A-1&=1-2 \cos ^{2} A&=\sin ^{2} A-\cos ^{2} A\\ \text{I}&\text{II}&\text{III}&\text{IV}\end{matrix}$
Taking I term
= sin4 A – cos4 A → I term
= (sin2 A)2 – (cos2 A)2
= (sin2 A – cos2 A)(sin2 A+ cos2 A )
[∵ (a2 – b2) = (a + b) (a – b)]
= (sin2 A – cos2 A)(1) [∵ cos2 θ + sin2 θ = 1]
= (sin2 A – cos2 A) …(i) → IV term
From Eq. (i)
= {sin2 A – (1 – sin2 A)} [∵ cos2 θ + sin2 θ = 1]
= sin2 A – 1 + sin2 A
= 2 sin2 A – 1 → II term
Again, From Eq. (i)
= {(1 – cos2 A) – cos2 A } [∵ cos2 θ + sin2 θ = 1]
=1 – 2 cos2 A → III term
Hence, I = II = III = IV
Hence Proved
Question 15 I
Prove the following identities :
cos4 θ – sin4 θ = cos2 θ – sin2 θ = 2 cos2 θ – 1Sol :
Given:
$\begin{matrix}\cos ^{4} \theta-\sin ^{4} \theta&=\cos ^{2} \theta-\sin ^{2} \theta&=2 \cos ^{2} \theta-1\\ \text{I}&\text{II}&\text{III}\end{matrix}$
Taking I term
= cos4 θ – sin4 θ → I term
= (cos2 θ)2 – (sin2 θ)2
= (cos2 θ – sin2 θ)(cos2 θ+ sin2 θ )
[∵ (a2 – b2) = (a + b) (a – b)]
= (cos2 θ – sin2 θ) (1) [∵ cos2 θ + sin2 θ = 1]
= (cos2 θ – sin2 θ) …(i) → II term
From Eq. (i)
= {cos2 θ – (1 – cos2 θ)} [∵ cos2 θ + sin2 θ = 1]
= 2 cos2 θ – 1 → III term
Hence, I = II = III
Hence Proved
Sol :
Taking LHS = 2 cos2 θ – cos4 θ + sin4 θ
= 2 cos2 θ – (cos4 θ – sin4 θ)
= 2 cos2 θ – [(cos2 θ)2 – (sin2 θ)2]
Using identity, (a2 – b2) = (a + b) (a – b)
= 2 cos2 θ – [(cos2 θ – sin2 θ)(cos2 θ+ sin2 θ )]
= 2 cos2 θ – [(cos2 θ – sin2 θ)(1)] [∵ cos2 θ + sin2 θ = 1]
=2 cos2 θ – cos2 θ + sin2 θ
= cos2 θ + sin2 θ
= 1 [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Sol :
Taking LHS = 1 – 2 cos2 θ + cos4 θ
We know that,
cos2 θ + sin2 θ = 1
= 1– 2 cos2 θ + (cos2 θ)2
= 1 – 2 cos2 θ + (1 – sin2 θ)2
= 1 – 2 cos2 θ +1 + sin4 θ – 2sin2θ
= 2 – 2(cos2 θ + sin2θ) + sin4 θ
= 2 – 2(1) + sin4 θ
= sin4 θ
=RHS
Hence Proved
Sol :
Taking LHS = 1 – 2 sin2 θ + sin4 θ
We know that,
cos2 θ + sin2 θ = 1
= 1– 2 sin2 θ + (sin2 θ)2
= 1 – 2 sin2 θ + (1 – cos2 θ)2
= 1 – 2 sin2 θ +1 + cos4 θ – 2cos2θ
= 2 – 2(cos2 θ + sin2θ) + cos4 θ
= 2 – 2(1) + cos4 θ
= cos4 θ
=RHS
Hence Proved
Sol :
Taking LHS = sec2 θ + cosec2 θ
Taking I term
= cos4 θ – sin4 θ → I term
= (cos2 θ)2 – (sin2 θ)2
= (cos2 θ – sin2 θ)(cos2 θ+ sin2 θ )
[∵ (a2 – b2) = (a + b) (a – b)]
= (cos2 θ – sin2 θ) (1) [∵ cos2 θ + sin2 θ = 1]
= (cos2 θ – sin2 θ) …(i) → II term
From Eq. (i)
= {cos2 θ – (1 – cos2 θ)} [∵ cos2 θ + sin2 θ = 1]
= 2 cos2 θ – 1 → III term
Hence, I = II = III
Hence Proved
Question 15 J
Prove the following identities :
2 cos2 θ – cos4 θ + sin4 θ = 1Sol :
Taking LHS = 2 cos2 θ – cos4 θ + sin4 θ
= 2 cos2 θ – (cos4 θ – sin4 θ)
= 2 cos2 θ – [(cos2 θ)2 – (sin2 θ)2]
Using identity, (a2 – b2) = (a + b) (a – b)
= 2 cos2 θ – [(cos2 θ – sin2 θ)(cos2 θ+ sin2 θ )]
= 2 cos2 θ – [(cos2 θ – sin2 θ)(1)] [∵ cos2 θ + sin2 θ = 1]
=2 cos2 θ – cos2 θ + sin2 θ
= cos2 θ + sin2 θ
= 1 [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Question 15 K
Prove the following identities :
1 – 2 cos2 θ + cos4 θ = sin4θSol :
Taking LHS = 1 – 2 cos2 θ + cos4 θ
We know that,
cos2 θ + sin2 θ = 1
= 1– 2 cos2 θ + (cos2 θ)2
= 1 – 2 cos2 θ + (1 – sin2 θ)2
= 1 – 2 cos2 θ +1 + sin4 θ – 2sin2θ
= 2 – 2(cos2 θ + sin2θ) + sin4 θ
= 2 – 2(1) + sin4 θ
= sin4 θ
=RHS
Hence Proved
Question 15 L
Prove the following identities :
1 – 2 sin2 θ + sin4 θ = cos4θSol :
Taking LHS = 1 – 2 sin2 θ + sin4 θ
We know that,
cos2 θ + sin2 θ = 1
= 1– 2 sin2 θ + (sin2 θ)2
= 1 – 2 sin2 θ + (1 – cos2 θ)2
= 1 – 2 sin2 θ +1 + cos4 θ – 2cos2θ
= 2 – 2(cos2 θ + sin2θ) + cos4 θ
= 2 – 2(1) + cos4 θ
= cos4 θ
=RHS
Hence Proved
Question 16 A
Prove that the following identities :
sec2θ + cosec2θ = sec2θ.cosec2θSol :
Taking LHS = sec2 θ + cosec2 θ
$=\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}$ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta \sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos ^{2} \theta \sin ^{2} \theta}$
= sec2 θ × cosec2 θ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta}$
$=\frac{1}{\sin \theta}$ [∵ cos2 θ + sin2 θ = 1]
= cosec θ
=RHS
Hence Proved
Sol :
Taking LHS = cot θ + tan θ
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta \sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos ^{2} \theta \sin ^{2} \theta}$
= sec2 θ × cosec2 θ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved
Question 16 B
Prove that the following identities :
$\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta=\operatorname{cosec} \theta$Sol :
Taking LHS $=\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta}$
$=\frac{1}{\sin \theta}$ [∵ cos2 θ + sin2 θ = 1]
= cosec θ
=RHS
Hence Proved
Question 16 C
Prove that the following identities :
cotθ + tan θ = cosec θ . sec θSol :
Taking LHS = cot θ + tan θ
$=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}$
$\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$ and $\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos \theta \sin \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos \theta \sin \theta}$
= cosec θ sec θ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1-\sin \theta}{1+\sin \theta}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos \theta \sin \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1}{\cos \theta \sin \theta}$
= cosec θ sec θ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved
Question 17
Prove that the following identities :
$\frac{1-\sin \theta}{1+\sin \theta}=\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}$Sol :
Taking LHS $=\frac{1-\sin \theta}{1+\sin \theta}$
Multiplying and divide by the conjugate of 1 + sin θ , we get
$=\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta}$
$=\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$
$=\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta}$
$=\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}$
=RHS
Hence Proved
Question 18
Prove that the following identities :
$\frac{1-\cos \theta}{1+\cos \theta}=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$
Multiplying and divide by the conjugate of 1 + cos θ , we get
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
=RHS
Hence Proved
Question 19
Prove that the following identities :
$\frac{1-\cos \theta}{1+\cos \theta}=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$
Multiplying and divide by the conjugate of 1 + cos θ , we get
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\cos \theta}{1+\sin \theta}$
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
=RHS
Hence Proved
Question 20
Prove that the following identities :
$\frac{\cos \theta}{1+\sin \theta}=\frac{1-\sin \theta}{\cos \theta}$Sol :
Taking LHS $=\frac{\cos \theta}{1+\sin \theta}$
Multiplying and divide by the conjugate of 1 + sin θ , we get
$=\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{\cos \theta(1-\sin \theta)}{(1)^{2}-(\sin \theta)^{2}}$ [∵ (a + b) (a – b) = (a2 – b2)]
$=\frac{\cos \theta(1-\sin \theta)}{1-\sin ^{2} \theta}$
$=\frac{\cos \theta(1-\sin \theta)}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1-\sin \theta}{\cos \theta}$
=RHS
Hence Proved
Sol :
Taking LHS
= sin8 θ – cos8 θ
= (sin4 θ)2 – (cos4 θ)2
= (sin4 θ – cos4 θ)(sin4 θ+ cos4 θ )
[∵ (a2 – b2) = (a + b) (a – b)]
= {(sin2 θ)2 – (cos2 θ)2}{(sin2 θ)2 + (cos2 θ)2}
= (sin2 θ + cos2 θ) (sin2 θ – cos2 θ) [(sin2 θ + cos2 θ) – 2 sin2 θ cos2 θ]
[ ∵ (a2 + b2) = (a +b)2 – 2ab]
= (1)[ sin2 θ –cos2 θ][(1) – 2 sin2 θ cos2 θ]
= (sin2 θ – cos2 θ)(1 – 2 sin2 θ cos2 θ)
=RHS
Hence Proved
Sol :
Taking LHS
= 2(sin6 θ – cos6 θ) – 3(sin4 θ+ cos4 θ ) + (sin2 θ + cos2 θ)
= 2[(sin2 θ)3 – (cos2 θ)3 ] – 3[(sin2 θ)2 + (cos2 θ)2 ]+1 [∵ cos2 θ + sin2 θ = 1]
Now, we use these identities, (a3 – b3)= (a + b)3 – 3ab(a+b) and (a2 + b2) = (a +b)2 – 2ab]
= 2[(sin2 θ + cos2 θ)3 – 3sin2θ cos2θ (sin2 θ+ cos2 θ)] –3[(sin2 θ + cos2 θ) – 2 sin2 θ cos2 θ]+ 1
=2[(1) – 3sin2θ cos2θ (1)] – 3[(1) – 2 sin2 θ cos2 θ] + 1 [∵ cos2 θ + sin2 θ = 1]
=2(1 – 3 sin2 θ cos2 θ )– 3 + 6sin2 θ cos2 θ+ 1
= 2– 6sin2θ cos2θ – 2 + 6sin2 θ cos2 θ
=0
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$
$=\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{\cos \theta(1-\sin \theta)}{(1)^{2}-(\sin \theta)^{2}}$ [∵ (a + b) (a – b) = (a2 – b2)]
$=\frac{\cos \theta(1-\sin \theta)}{1-\sin ^{2} \theta}$
$=\frac{\cos \theta(1-\sin \theta)}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1-\sin \theta}{\cos \theta}$
=RHS
Hence Proved
Question 21
Prove that the following identities :
(sin8θ – cos8θ) = (sin2θ – cos2θ)(1 – 2sin2θ .cos2θ)Sol :
Taking LHS
= sin8 θ – cos8 θ
= (sin4 θ)2 – (cos4 θ)2
= (sin4 θ – cos4 θ)(sin4 θ+ cos4 θ )
[∵ (a2 – b2) = (a + b) (a – b)]
= {(sin2 θ)2 – (cos2 θ)2}{(sin2 θ)2 + (cos2 θ)2}
= (sin2 θ + cos2 θ) (sin2 θ – cos2 θ) [(sin2 θ + cos2 θ) – 2 sin2 θ cos2 θ]
[ ∵ (a2 + b2) = (a +b)2 – 2ab]
= (1)[ sin2 θ –cos2 θ][(1) – 2 sin2 θ cos2 θ]
= (sin2 θ – cos2 θ)(1 – 2 sin2 θ cos2 θ)
=RHS
Hence Proved
Question 22
Prove that the following identities :
2(sin6 θ – cos6 θ) – 3(sin4 θ + cos4 θ ) + (sin2 θ + cos2 θ)Sol :
Taking LHS
= 2(sin6 θ – cos6 θ) – 3(sin4 θ+ cos4 θ ) + (sin2 θ + cos2 θ)
= 2[(sin2 θ)3 – (cos2 θ)3 ] – 3[(sin2 θ)2 + (cos2 θ)2 ]+1 [∵ cos2 θ + sin2 θ = 1]
Now, we use these identities, (a3 – b3)= (a + b)3 – 3ab(a+b) and (a2 + b2) = (a +b)2 – 2ab]
= 2[(sin2 θ + cos2 θ)3 – 3sin2θ cos2θ (sin2 θ+ cos2 θ)] –3[(sin2 θ + cos2 θ) – 2 sin2 θ cos2 θ]+ 1
=2[(1) – 3sin2θ cos2θ (1)] – 3[(1) – 2 sin2 θ cos2 θ] + 1 [∵ cos2 θ + sin2 θ = 1]
=2(1 – 3 sin2 θ cos2 θ )– 3 + 6sin2 θ cos2 θ+ 1
= 2– 6sin2θ cos2θ – 2 + 6sin2 θ cos2 θ
=0
=RHS
Hence Proved
Question 23
Prove the following identities
$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$Sol :
Taking LHS $=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$
$=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}$
$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}}$
$=\frac{\cos ^{2} A}{\cos A-\sin A}+\frac{\sin ^{2} A}{\sin A-\cos A}$
$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos A-\sin A}$
$=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}}$
$=\frac{\cos ^{2} A}{\cos A-\sin A}+\frac{\sin ^{2} A}{\sin A-\cos A}$
$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos A-\sin A}$
Using the identity, (a2 – b2)= (a + b) (a – b)
$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{(\cos A-\sin A)}$
= sin A +cos A
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$
$=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{(1+\cos \theta)(\sin \theta)}$
$=\frac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{(1+\cos \theta)(\sin \theta)}$
$=\frac{2+2 \cos \theta}{(1+\operatorname{ses} \theta)(\sin \theta)}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2(1+\cos \theta)}{(1+\cos \theta)(\sin \theta)}$
$=\frac{2}{\sin \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}$
$=\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}$
$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{(\cos A-\sin A)}$
= sin A +cos A
=RHS
Hence Proved
Question 24
Prove the following identities
$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=\frac{2}{\sin \theta}$Sol :
Taking LHS $=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$
$=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{(1+\cos \theta)(\sin \theta)}$
$=\frac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{(1+\cos \theta)(\sin \theta)}$
$=\frac{2+2 \cos \theta}{(1+\operatorname{ses} \theta)(\sin \theta)}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2(1+\cos \theta)}{(1+\cos \theta)(\sin \theta)}$
$=\frac{2}{\sin \theta}$
=RHS
Hence Proved
Question 25
Prove the following identities
$\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=2 \sec ^{2} \theta$Sol :
Taking LHS $=\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}$
$=\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}$
Using the identity, (a2 – b2)= (a + b) (a – b)
$=\frac{2}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2}{1-\sin ^{2} \theta}$
$=\frac{2}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
= 2 sec2 θ
=RHS
Hence Proved
$\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta$
Sol :
Taking LHS $=\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}$
$=\frac{(1+\sin \theta)^{2}+\cos ^{2} \theta}{(1+\sin \theta)(\cos \theta)}$
$=\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{(\cos \theta)(1+\sin \theta)}$
$=\frac{2+2 \sin \theta}{(\cos \theta)(1+\sin \theta)}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2(1+\sin \theta)}{(\cos \theta)(1+\sin \theta)}$
$=\frac{2}{\cos \theta}$
=2 sec θ
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)+\cos \theta(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{\cos \theta+\cos \theta \sin \theta+\cos \theta-\cos \theta \sin \theta}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{2 \cos \theta}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{2}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2}{1-\sin ^{2} \theta}$
$=\frac{2}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
= 2 sec2 θ
=RHS
Hence Proved
Question 26
Prove the following identities$\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta$
Sol :
Taking LHS $=\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}$
$=\frac{(1+\sin \theta)^{2}+\cos ^{2} \theta}{(1+\sin \theta)(\cos \theta)}$
$=\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{(\cos \theta)(1+\sin \theta)}$
$=\frac{2+2 \sin \theta}{(\cos \theta)(1+\sin \theta)}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2(1+\sin \theta)}{(\cos \theta)(1+\sin \theta)}$
$=\frac{2}{\cos \theta}$
=2 sec θ
=RHS
Hence Proved
Question 27
Prove the following identities
$\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=\frac{2}{\cos \theta}$Sol :
Taking LHS $=\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)+\cos \theta(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{\cos \theta+\cos \theta \sin \theta+\cos \theta-\cos \theta \sin \theta}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{2 \cos \theta}{(1-\sin \theta)(1+\sin \theta)}$
Using the identity, (a2 – b2)= (a + b) (a – b)
$=\frac{2 \cos \theta}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2 \cos \theta}{1-\sin ^{2} \theta}$
$=\frac{2 \cos \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2 \cos \theta}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2 \cos \theta}{1-\sin ^{2} \theta}$
$=\frac{2 \cos \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2}{\cos \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1}{1+\cos \theta}+\frac{1}{1-\cos \theta}$
$=\frac{1-\cos \theta+1+\cos \theta}{(1+\cos \theta)(1-\cos \theta)}$
=RHS
Hence Proved
Question 28
Prove the following identities
$\frac{1}{1+\cos \theta}+\frac{1}{1-\cos \theta}=\frac{2}{\sin ^{2} \theta}$Sol :
Taking LHS $=\frac{1}{1+\cos \theta}+\frac{1}{1-\cos \theta}$
$=\frac{1-\cos \theta+1+\cos \theta}{(1+\cos \theta)(1-\cos \theta)}$
Using the identity, (a2 – b2)= (a + b) (a – b)
$=\frac{2}{(1)^{2}-(\cos \theta)^{2}}$
$=\frac{2}{1-\cos ^{2} \theta}$
$=\frac{2}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta-1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}$
$=\frac{2}{(1)^{2}-(\cos \theta)^{2}}$
$=\frac{2}{1-\cos ^{2} \theta}$
$=\frac{2}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Question 29
Prove the following identities
$\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}=\frac{2 \tan \theta}{\cos \theta}$Sol :
Taking LHS $=\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta-1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}$
Using the identity, (a2 – b2)= (a + b) (a – b)
$=\frac{2 \sin \theta}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2 \sin \theta}{1-\sin ^{2} \theta}$
$=\frac{2 \sin \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{2 \tan \theta}{\cos \theta}$
$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS = cot2 θ – cos2 θ
$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved
Question 30
Prove the following identities
cot2θ – cos2θ = cot2θ . cos2θSol :
Taking LHS = cot2 θ – cos2 θ
$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-\cos ^{2} \theta$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta}$
$=\frac{\cos ^{2} \theta\left(1-\sin ^{2} \theta\right)}{\sin ^{2} \theta}$
[∵ cos2 θ + sin2 θ = 1]
$=\frac{\cos ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta}$
= cot2 θ cos2 θ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS = tan2 φ – sin2 φ – tan2 φ sin2 φ
$=\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi}-\sin ^{2} \varphi-\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi} \sin ^{2} \varphi$
$=\frac{\sin ^{2} \varphi-\sin ^{2} \varphi \cos ^{2} \varphi-\sin ^{4} \varphi}{\cos ^{2} \varphi}$
$=\frac{\sin ^{2} \varphi\left(1-\cos ^{2} \varphi-\sin ^{2} \varphi\right)}{\cos ^{2} \varphi}$
$=\frac{\sin ^{2} \varphi\left\{1-\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right)\right]}{\cos ^{2} \varphi}$ [∵ cos2 φ + sin2 φ = 1]
$=\frac{\sin ^{2} \varphi\{1-1\}}{\cos ^{2} \varphi}$
= 0
=RHS
Hence Proved
Sol :
Taking LHS = tan2 φ + cot2 φ + 2
$=\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi}+\frac{\cos ^{2} \varphi}{\sin ^{2} \varphi}+2$
$=\frac{\sin ^{4} \varphi+\cos ^{4} \varphi+2 \sin ^{2} \varphi \cos ^{2} \varphi}{\cos ^{2} \varphi \sin ^{2} \varphi}$ [∵ (a + b)2 = (a2 + b2 + 2ab)]
$=\frac{\left(\sin ^{2} \varphi+\cos ^{2} \varphi\right)^{2}}{\cos ^{2} \varphi \sin ^{2} \varphi}$ [∵ cos2 φ + sin2 φ = 1]
$=\frac{1}{\cos ^{2} \varphi \sin ^{2} \varphi}$
= sec2 φ cosec2 φ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right.$ and $\left.\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$ [∵ cot2 θ – cosec2 θ = 1]
$=\frac{(\cot \theta+\operatorname{cosec} \theta)-\{(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)\}}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1-\operatorname{cosec} \theta+\cot \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$
= cot θ + cosec θ
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta}$
$=\frac{\cos ^{2} \theta\left(1-\sin ^{2} \theta\right)}{\sin ^{2} \theta}$
[∵ cos2 θ + sin2 θ = 1]
$=\frac{\cos ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta}$
= cot2 θ cos2 θ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
=RHS
Hence Proved
Question 31
Prove the following identities
tan2 φ – sin2 φ – tan2 φ . sin2 φ = 0Sol :
Taking LHS = tan2 φ – sin2 φ – tan2 φ sin2 φ
$=\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi}-\sin ^{2} \varphi-\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi} \sin ^{2} \varphi$
$=\frac{\sin ^{2} \varphi-\sin ^{2} \varphi \cos ^{2} \varphi-\sin ^{4} \varphi}{\cos ^{2} \varphi}$
$=\frac{\sin ^{2} \varphi\left(1-\cos ^{2} \varphi-\sin ^{2} \varphi\right)}{\cos ^{2} \varphi}$
$=\frac{\sin ^{2} \varphi\left\{1-\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right)\right]}{\cos ^{2} \varphi}$ [∵ cos2 φ + sin2 φ = 1]
$=\frac{\sin ^{2} \varphi\{1-1\}}{\cos ^{2} \varphi}$
= 0
=RHS
Hence Proved
Question 32
Prove the following identities
tan2 φ + cot2 φ + 2 = sec2ϕ. cosec2ϕSol :
Taking LHS = tan2 φ + cot2 φ + 2
$=\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi}+\frac{\cos ^{2} \varphi}{\sin ^{2} \varphi}+2$
$=\frac{\sin ^{4} \varphi+\cos ^{4} \varphi+2 \sin ^{2} \varphi \cos ^{2} \varphi}{\cos ^{2} \varphi \sin ^{2} \varphi}$ [∵ (a + b)2 = (a2 + b2 + 2ab)]
$=\frac{\left(\sin ^{2} \varphi+\cos ^{2} \varphi\right)^{2}}{\cos ^{2} \varphi \sin ^{2} \varphi}$ [∵ cos2 φ + sin2 φ = 1]
$=\frac{1}{\cos ^{2} \varphi \sin ^{2} \varphi}$
= sec2 φ cosec2 φ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right.$ and $\left.\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved
Question 33
Prove the following identities
$\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}$Sol :
Taking LHS $=\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$ [∵ cot2 θ – cosec2 θ = 1]
$=\frac{(\cot \theta+\operatorname{cosec} \theta)-\{(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)\}}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1-\operatorname{cosec} \theta+\cot \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$
= cot θ + cosec θ
$=\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right.$ and $\left.\sin \theta=\frac{1}{\csc \theta}\right]$
$=\frac{1+\cos \theta}{\sin \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$
$=\frac{1+\cos \theta}{\sin \theta}$
=RHS
Hence Proved
Question 34
Prove the following identities
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$Sol :
Taking LHS $=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$
$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right.$ and $\left.\tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{\frac{\sin \theta}{\sin \theta-\cos \theta}}{\sin \theta}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$=\frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta-\sin \theta}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta\{-(\sin \theta-\cos \theta)\}}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{(\cos \theta \sin \theta)(\sin \theta-\cos \theta)}$
$=\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{(\cos \theta \sin \theta)(\sin \theta-\cos \theta)}$ [∵ (a3 – b3)= (a – b)(a2 +b2 +ab)]
$=\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta \sin \theta}+\frac{\sin \theta \cos \theta}{\cos \theta \sin \theta}$
= tan θ cot θ + 1
=RHS
Hence Proved
Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$
$=\frac{\frac{\sin \theta}{\sin \theta-\cos \theta}}{\sin \theta}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$=\frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta-\sin \theta}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta\{-(\sin \theta-\cos \theta)\}}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{(\cos \theta \sin \theta)(\sin \theta-\cos \theta)}$
$=\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{(\cos \theta \sin \theta)(\sin \theta-\cos \theta)}$ [∵ (a3 – b3)= (a – b)(a2 +b2 +ab)]
$=\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta \sin \theta}+\frac{\sin \theta \cos \theta}{\cos \theta \sin \theta}$
= tan θ cot θ + 1
=RHS
Hence Proved
Question 35
Prove the following identities
$\frac{1-\cos \theta}{1+\cos \theta}=(\cot \theta-\operatorname{cosec} \theta)^{2}$Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$
Multiplying and divide by the conjugate of 1 + cos θ , we get
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
$=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$
= (cosec θ – cot θ)2
= { – (cot θ – cosec θ)}2
= (cot θ – cosec θ)2
=RHS
Hence Proved
Sol :
Taking LHS $=\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$ [multiplying and divide by conjugate of 1– cosθ]
$=\sqrt{\frac{(1+\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\left(1-\cos ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1+\cos \theta}{\sin \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$ [multiplying and divide by conjugate of 1– cosθ]
$=\sqrt{\frac{(1+\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\left(1-\cos ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1+\cos \theta}{\sin \theta}$
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)2 and (a + b) (a – b) = (a2 – b2)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
$=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$
= (cosec θ – cot θ)2
= { – (cot θ – cosec θ)}2
= (cot θ – cosec θ)2
=RHS
Hence Proved
Question 36
Prove the following identities
$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{1+\cos \theta}{\sin \theta}$Sol :
Taking LHS $=\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$ [multiplying and divide by conjugate of 1– cosθ]
$=\sqrt{\frac{(1+\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\left(1-\cos ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1+\cos \theta}{\sin \theta}$
=RHS
Hence Proved
Question 37
Prove the following identities
$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{\sin \theta}{1-\cos \theta}$Sol :
Taking LHS $=\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$ [multiplying and divide by conjugate of 1– cosθ]
$=\sqrt{\frac{(1+\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\left(1-\cos ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{1+\cos \theta}{\sin \theta}$
Multiply and divide by conjugate of 1+ cosθ, we get
$=\frac{1+\cos \theta}{\sin \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{1-\cos ^{2} \theta}{\sin \theta \times(1-\cos \theta)}$
$=\frac{\sin ^{2} \theta}{\sin \theta \times(1-\cos \theta)}$
$=\frac{\sin \theta}{1-\cos \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}$ [multiplying and divide by conjugate of 1+sinθ]
$=\sqrt{\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}$
$=\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$
$=\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
=$=\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
= sec θ – tan θ
$=\frac{1+\cos \theta}{\sin \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{1-\cos ^{2} \theta}{\sin \theta \times(1-\cos \theta)}$
$=\frac{\sin ^{2} \theta}{\sin \theta \times(1-\cos \theta)}$
$=\frac{\sin \theta}{1-\cos \theta}$
=RHS
Hence Proved
Question 38
Prove the following identities
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$Sol :
Taking LHS $=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}$ [multiplying and divide by conjugate of 1+sinθ]
$=\sqrt{\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}$
$=\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$
$=\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
=$=\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
= sec θ – tan θ
$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cos \theta=\frac{1}{\sec \theta}\right]$
=RHS
Hence Proved
Sol :
Taking LHS $=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}$ [multiplying and divide by of 1+ sin θ]
$=\sqrt{\frac{(1)^{2}-(\sin \theta)^{2}}{(1+\sin \theta)^{2}}}$
$=\sqrt{\frac{1-\sin ^{2} \theta}{(1+\sin \theta)^{2}}}$
$=\sqrt{\frac{\cos ^{2} \theta}{(1+\sin \theta)^{2}}}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{\cos \theta}{1+\sin \theta}$
=RHS
Hence Proved
Sol :
Taking LHS $=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
[multiplying and divide by conjugate of 1– sinθ in 1st term and 1+sinθ in 2nd term]
$=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}+\sqrt{\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
=$\frac{1+\sin \theta}{\cos \theta}+\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1+\sin \theta+1-\sin \theta}{\cos \theta}$
$=\frac{2}{\cos \theta}$
= 2 sec θ
=RHS
Hence Proved
Sol :
Given : secθ+tanθ=m and sec θ–tanθ=n
To Prove : √mn = 1
Taking LHS = √mn
Putting the value of m and n, we get
$=\sqrt{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$
Using the identity, (a + b) (a – b) = (a2 – b2)
$=\sqrt{\sec ^{2} \theta-\tan ^{2} \theta}$
=√(1) [∵ 1+ tan2 θ = sec2 θ]
=±1
=RHS
Hence Proved
Given: cos θ+sinθ=1
On squaring both the sides, we get
(cos θ +sin θ)2 =(1)2
⇒ cos2 θ + sin2 θ + 2sinθ cos θ = 1
⇒ cos2 θ + sin2 θ = cos2 θ + sin2 θ – 2sinθ cos θ
[∵ cos2 θ + sin2 θ = 1]
⇒ cos2 θ + sin2 θ = (cosθ – sinθ)2
[∵ (a – b)2 = (a2 + b2 – 2ab)]
⇒ 1 = (cos θ – sin θ)2
⇒ (cos θ – sin θ) = ±1
Hence Proved
Given : sin θ + sin2 θ = 1
⇒ sin θ = 1 – sin2 θ
Taking LHS = cos2θ+ cos4θ
= cos2 θ + (cos2 θ)2
= (1– sin2 θ) + (1– sin2 θ)2 …(i)
Putting sin θ = 1 – sin2 θ in Eq. (i), we get
= sin θ + (sin θ)2
= sin θ + sin2 θ
= 1 [Given: sin θ + sin2 θ = 1]
=RHS
Hence Proved
Sol :
To show : $\sin \theta=\frac{x^{2}-1}{x^{2}+1}$
Taking RHS $=\frac{x^{2}-1}{x^{2}+1}$
Given tanθ+secθ=x
$=\frac{(\tan \theta+\sec \theta)^{2}-1}{(\tan \theta+\sec \theta)^{2}+1}$
$=\frac{\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta-1}{\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta+1}$
$=\frac{\tan ^{2} \theta+\tan ^{2} \theta+2 \tan \theta \sec \theta}{\sec ^{2} \theta-1+\sec ^{2} \theta+2 \tan \theta \sec \theta+1}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{2 \tan ^{2} \theta+2 \tan \theta \sec \theta}{2 \sec ^{2} \theta+2 \tan \theta \sec \theta}$
$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}}{\frac{1}{\cos ^{2} \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta} \text { and } \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin ^{2} \theta+\sin \theta}{1+\sin \theta}$
$=\frac{\sin \theta(\sin \theta+1)}{1+\sin \theta}$
=sin θ
=LHS
Hence Proved
Sol :
Given: sin θ + cos θ = p and sec θ + cosec θ = q
To show q(p2 – 1) = 2p
Taking LHS = q(p2 – 1)
Putting the value of sin θ + cos θ = p and sec θ + cosec θ = q, we get
=(sec θ + cosec θ){( sin θ + cos θ )2 – 1)
=(sec θ + cosec θ){(sin2 θ + cos2 θ + 2sin θ cosθ) – 1)}
[∵ (a + b)2 = (a2 + b2 + 2ab)]
=(sec θ + cosec θ)(1+2sin θ cos θ – 1)
=(sec θ + cosec θ)(2sin θcosθ)
$=\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right) \times(2 \sin \theta \cos \theta)$ $\left[\because \cos \theta=\frac{1}{\sec \theta} \text { and } \sin \theta=\frac{1}{\cos \theta}\right]$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta} \times 2 \sin \theta \cos \theta$
= 2(sin θ +cos θ)
= 2p [ given sin θ + cos θ = p]
=RHS
Hence Proved
Sol :
Given: x cosθ = a and y = a tanθ
⇒$\mathrm{x}=\frac{\mathrm{a}}{\cos \theta}$ and $y=a \tan \theta$
To Prove : x2–y2=a2
Taking LHS = x2–y2
Putting the values of x and y, we get
$=\left(\frac{a}{\cos \theta}\right)^{2}-(a \tan \theta)^{2}$
$=\frac{a^{2}}{\cos ^{2} \theta}-a^{2} \tan ^{2} \theta$
$=\frac{a^{2}}{\cos ^{2} \theta}-a^{2} \frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{a^{2}-a^{2} \sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{a^{2}\left(1-\sin ^{2} \theta\right)}{\cos ^{2} \theta}$
$=\frac{a^{2} \cos ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
= a2
= RHS
Hence Proved
Taking LHS = x2 + y2 + z2
Putting the values of x, y and z , we get
=(r cos α sin β)2 + (r sin α sin β)2 + (r cos α)2
= r2 cos2α sin2β + r2 sin2α sin2β + r2 cos2α
Taking common r2 sin2 α , we get
= r2 sin2α (cos2β + sin2 β) + r2cos2 α
= r2 sin2α + r2 cos2 α [∵ cos2 β + sin2 β = 1]
=r2 ( sin2 α + cos2 α)
= r2 [∵ cos2 α + sin2 α = 1]
=RHS
Hence Proved
(ii) $\sin \theta=\frac{1-x^{2}}{1+x^{2}}$
Sol :
(i) Given sec θ – tan θ = x
Taking RHS $=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$
Putting the value of x, we get
$=\frac{2(\sec \theta-\tan \theta)}{1+(\sec \theta-\tan \theta)^{2}}$
$=\frac{2(\sec \theta-\tan \theta)}{1+\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}$
$=\frac{2(\sec \theta-\tan \theta)}{\sec ^{2} \theta+\sec ^{2} \theta-2 \sec \theta \tan \theta}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{2(\sec \theta-\tan \theta)}{2 \sec \theta(\sec \theta-\tan \theta)}$
$=\frac{1}{\sec \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
= cos θ
=RHS
Hence Proved
(ii) Given sec θ – tan θ = x
Taking RHS $=\frac{1-x^{2}}{1+x^{2}}$
Putting the value of x, we get
$=\frac{1-(\sec \theta-\tan \theta)^{2}}{1+(\sec \theta-\tan \theta)^{2}}$
$=\frac{1-\sec ^{2} \theta-\tan ^{2} \theta+2 \sec \theta \tan \theta}{1+\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}$
$=\frac{\left.-\tan ^{2} \theta-\tan ^{2} \theta+2 \sec \theta \tan \theta\right)}{\sec ^{2} \theta+\sec ^{2} \theta-2 \sec \theta \tan \theta}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{2 \tan \theta(\sec \theta-\tan \theta)}{2 \sec \theta(\sec \theta-\tan \theta)}$
=$\frac{\sin \theta}{\cos \theta} \times \cos \theta$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= sin θ
=RHS
Hence Proved
Let
(a cos θ + b sin θ)2 + (a sin θ – b cos θ)2 = a2cos2θ + b2 sin2θ + 2ab cos θ sin θ + a2sin2θ
+ b2 cos2θ – 2ab cos θ sin θ
⇒ c2 + (a sin θ – b cos θ)2 = a2 (cos2 θ + sin2 θ) + b2 (cos2 θ + sin2 θ)
⇒ c2 + (a sin θ – b cos θ)2 = a2 + b2
⇒ (a sin θ – b cos θ)2 = a2 + b2 – c2
⇒ (a sin θ – b cos θ) = ±√ (a2 + b2 – c2)
Sol :
Given: 1+sin2θ =3 sin θ cos θ
Divide by cos2 θ to both the sides, we get
⇒$\frac{1}{\cos ^{2} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{3 \sin \theta \cos \theta}{\cos ^{2} \theta}$
⇒ sec2 θ + tan2 θ = 3 tan θ
⇒ 1+ tan2 θ+ tan2 θ = 3tan θ
⇒ 2 tan2 θ –3tan θ +1 = 0
Let tanθ = x
⇒ 2x2 – 3x +1 = 0
⇒ 2x2 – 2x – x +1 = 0
⇒ 2x ( x – 1) – 1(x – 1) = 0
⇒ (2x – 1)(x – 1) = 0
Putting each of the factor = 0, we get
⇒ x = 1 or $\dfrac{1}{2}$
And above, we let tan θ = x
⇒$\tan \theta=1$ or $\frac{1}{2}$
Hence Proved
Taking RHS =x2 + y2
Putting the values of x and y, we get
(a cos θ – b sin θ)2 + (a sin θ + b cos θ)2
= a2cos2θ + b2 sin2θ – 2ab cos θ sin θ + a2sin2θ + b2 cos2θ + 2ab cos θ sin θ
= a2 (cos2 θ + sin2 θ) + b2 (cos2 θ + sin2 θ)
= a2 + b2 [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Sol :
Taking LHS =x2 – y2
Putting the values of x and y, we get
(a sec θ + b tan θ)2 – (a tan θ + b sec θ)2
= a2 sec2θ + b2 tan2θ + 2ab sec θ tan θ – a2tan2θ – b2 sec2θ – 2ab sec θ tan θ
= a2 (sec2θ – tan2θ) – b2 (sec2θ – tan2θ)
= a2 – b2 [∵ 1+ tan2 θ = sec2 θ]
=RHS
Hence Proved
$\tan \theta=\frac{\mathrm{a}^{2}-\mathrm{b}^{2}}{2 \mathrm{ab}}$
Sol :
Taking (a2 – b2) sin θ + 2ab cos θ = a2 + b2
We know that
$\sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$ and $\cos \theta=\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$
Then, substituting the above values in the given equation, we get
$=\mathrm{a}^{2}-\mathrm{b}^{2} \frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}+2 \mathrm{ab} \frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\mathrm{a}^{2}+\mathrm{b}^{2}$
Now, substituting,$t=\tan \frac{\theta}{2}$ , we have
$a^{2}-b^{2} \frac{2 t}{1+t^{2}}+2 a b \frac{1-t^{2}}{1+t^{2}}=a^{2}+b^{2}$
⇒ ( a2 – b2)2t – 2ab(1 – t2) = (a2 + b2)(1+t2)
Simplify, we get
(a2 + 2ab + b2)t2 – 2(a2 – b2)t + (a2 –2ab +b2)=0
⇒ (a+b)2 t2 – 2(a2 – b2)t + (a – b)2 = 0
⇒ (a+b)2 t2 –2 (a – b)(a+b)t + (a – b)2 =0
⇒ [(a+b)t – (a – b)]2 = 0 [∵ (a – b)2 = (a2 + b2 – 2ab)]
⇒ [(a+b)t – (a – b)] = 0
⇒ (a+b)t = (a – b)
⇒$ \mathrm{t}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}$
⇒$ \tan \frac{\theta}{2}=\frac{a-b}{a+b}$
We know that, $\frac{2 t}{1-t^{2}}=\tan \theta$ where $\mathrm{t}=\tan \frac{\theta}{2}$
$\Rightarrow \tan \theta=\frac{2\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)}{1-\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)^{2}}$
$\Rightarrow \tan \theta=\frac{2(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})}{(\mathrm{a}+\mathrm{b})^{2}-(\mathrm{a}-\mathrm{b})^{2}}$
$\Rightarrow \tan \theta=\frac{a^{2}-b^{2}}{2 a b}$
Hence Proved
=RHS
Hence Proved
Question 39
Prove the following identities
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{\cos \theta}{1+\sin \theta}$Sol :
Taking LHS $=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}$ [multiplying and divide by of 1+ sin θ]
$=\sqrt{\frac{(1)^{2}-(\sin \theta)^{2}}{(1+\sin \theta)^{2}}}$
$=\sqrt{\frac{1-\sin ^{2} \theta}{(1+\sin \theta)^{2}}}$
$=\sqrt{\frac{\cos ^{2} \theta}{(1+\sin \theta)^{2}}}$ [∵ cos2 θ + sin2 θ = 1]
$=\frac{\cos \theta}{1+\sin \theta}$
=RHS
Hence Proved
Question 40
Prove the following identities
$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta$Sol :
Taking LHS $=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
[multiplying and divide by conjugate of 1– sinθ in 1st term and 1+sinθ in 2nd term]
$=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}+\sqrt{\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}$ [∵ cos2 θ + sin2 θ = 1]
=$\frac{1+\sin \theta}{\cos \theta}+\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1+\sin \theta+1-\sin \theta}{\cos \theta}$
$=\frac{2}{\cos \theta}$
= 2 sec θ
=RHS
Hence Proved
Question 41
If secθ + tanθ=m and sec θ – tanθ = n, then prove that $\sqrt{m n}=1$
Given : secθ+tanθ=m and sec θ–tanθ=n
To Prove : √mn = 1
Taking LHS = √mn
Putting the value of m and n, we get
$=\sqrt{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$
Using the identity, (a + b) (a – b) = (a2 – b2)
$=\sqrt{\sec ^{2} \theta-\tan ^{2} \theta}$
=√(1) [∵ 1+ tan2 θ = sec2 θ]
=±1
=RHS
Hence Proved
Question 42
If cosθ + sinθ = 1, then prove that cosθ – sin θ = ± 1.
Sol :Given: cos θ+sinθ=1
On squaring both the sides, we get
(cos θ +sin θ)2 =(1)2
⇒ cos2 θ + sin2 θ + 2sinθ cos θ = 1
⇒ cos2 θ + sin2 θ = cos2 θ + sin2 θ – 2sinθ cos θ
[∵ cos2 θ + sin2 θ = 1]
⇒ cos2 θ + sin2 θ = (cosθ – sinθ)2
[∵ (a – b)2 = (a2 + b2 – 2ab)]
⇒ 1 = (cos θ – sin θ)2
⇒ (cos θ – sin θ) = ±1
Hence Proved
Question 43
If sinθ + sin2θ = 1, then prove that cos2θ +1 cos4θ = 1
Sol :Given : sin θ + sin2 θ = 1
⇒ sin θ = 1 – sin2 θ
Taking LHS = cos2θ+ cos4θ
= cos2 θ + (cos2 θ)2
= (1– sin2 θ) + (1– sin2 θ)2 …(i)
Putting sin θ = 1 – sin2 θ in Eq. (i), we get
= sin θ + (sin θ)2
= sin θ + sin2 θ
= 1 [Given: sin θ + sin2 θ = 1]
=RHS
Hence Proved
Question 44
If tanθ + secθ = x, show that sinθ =$\frac{x^{2}-1}{x^{2}+1}$
To show : $\sin \theta=\frac{x^{2}-1}{x^{2}+1}$
Taking RHS $=\frac{x^{2}-1}{x^{2}+1}$
Given tanθ+secθ=x
$=\frac{(\tan \theta+\sec \theta)^{2}-1}{(\tan \theta+\sec \theta)^{2}+1}$
$=\frac{\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta-1}{\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta+1}$
$=\frac{\tan ^{2} \theta+\tan ^{2} \theta+2 \tan \theta \sec \theta}{\sec ^{2} \theta-1+\sec ^{2} \theta+2 \tan \theta \sec \theta+1}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{2 \tan ^{2} \theta+2 \tan \theta \sec \theta}{2 \sec ^{2} \theta+2 \tan \theta \sec \theta}$
$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}}{\frac{1}{\cos ^{2} \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta} \text { and } \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin ^{2} \theta+\sin \theta}{1+\sin \theta}$
$=\frac{\sin \theta(\sin \theta+1)}{1+\sin \theta}$
=sin θ
=LHS
Hence Proved
Question 45
If sinθ + cosθ = p and secθ + cosecθ = q, then show q(p2–1) =2p
Given: sin θ + cos θ = p and sec θ + cosec θ = q
To show q(p2 – 1) = 2p
Taking LHS = q(p2 – 1)
Putting the value of sin θ + cos θ = p and sec θ + cosec θ = q, we get
=(sec θ + cosec θ){( sin θ + cos θ )2 – 1)
=(sec θ + cosec θ){(sin2 θ + cos2 θ + 2sin θ cosθ) – 1)}
[∵ (a + b)2 = (a2 + b2 + 2ab)]
=(sec θ + cosec θ)(1+2sin θ cos θ – 1)
=(sec θ + cosec θ)(2sin θcosθ)
$=\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right) \times(2 \sin \theta \cos \theta)$ $\left[\because \cos \theta=\frac{1}{\sec \theta} \text { and } \sin \theta=\frac{1}{\cos \theta}\right]$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta} \times 2 \sin \theta \cos \theta$
= 2(sin θ +cos θ)
= 2p [ given sin θ + cos θ = p]
=RHS
Hence Proved
Question 46
If x cosθ =a and y = a tanθ, then prove that x2–y2=a2
Given: x cosθ = a and y = a tanθ
⇒$\mathrm{x}=\frac{\mathrm{a}}{\cos \theta}$ and $y=a \tan \theta$
To Prove : x2–y2=a2
Taking LHS = x2–y2
Putting the values of x and y, we get
$=\left(\frac{a}{\cos \theta}\right)^{2}-(a \tan \theta)^{2}$
$=\frac{a^{2}}{\cos ^{2} \theta}-a^{2} \tan ^{2} \theta$
$=\frac{a^{2}}{\cos ^{2} \theta}-a^{2} \frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{a^{2}-a^{2} \sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{a^{2}\left(1-\sin ^{2} \theta\right)}{\cos ^{2} \theta}$
$=\frac{a^{2} \cos ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos2 θ + sin2 θ = 1]
= a2
= RHS
Hence Proved
Question 47
If x= r cos α sin β, y = r sin α sin β and z = r cos α then prove that x2 + y2 + z2 = r2.
Sol :Taking LHS = x2 + y2 + z2
Putting the values of x, y and z , we get
=(r cos α sin β)2 + (r sin α sin β)2 + (r cos α)2
= r2 cos2α sin2β + r2 sin2α sin2β + r2 cos2α
Taking common r2 sin2 α , we get
= r2 sin2α (cos2β + sin2 β) + r2cos2 α
= r2 sin2α + r2 cos2 α [∵ cos2 β + sin2 β = 1]
=r2 ( sin2 α + cos2 α)
= r2 [∵ cos2 α + sin2 α = 1]
=RHS
Hence Proved
Question 48
If secθ – tanθ = x, then prove that
(i)$\cos \theta=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$(ii) $\sin \theta=\frac{1-x^{2}}{1+x^{2}}$
Sol :
(i) Given sec θ – tan θ = x
Taking RHS $=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$
Putting the value of x, we get
$=\frac{2(\sec \theta-\tan \theta)}{1+(\sec \theta-\tan \theta)^{2}}$
$=\frac{2(\sec \theta-\tan \theta)}{1+\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}$
$=\frac{2(\sec \theta-\tan \theta)}{\sec ^{2} \theta+\sec ^{2} \theta-2 \sec \theta \tan \theta}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{2(\sec \theta-\tan \theta)}{2 \sec \theta(\sec \theta-\tan \theta)}$
$=\frac{1}{\sec \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
= cos θ
=RHS
Hence Proved
(ii) Given sec θ – tan θ = x
Taking RHS $=\frac{1-x^{2}}{1+x^{2}}$
Putting the value of x, we get
$=\frac{1-(\sec \theta-\tan \theta)^{2}}{1+(\sec \theta-\tan \theta)^{2}}$
$=\frac{1-\sec ^{2} \theta-\tan ^{2} \theta+2 \sec \theta \tan \theta}{1+\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}$
$=\frac{\left.-\tan ^{2} \theta-\tan ^{2} \theta+2 \sec \theta \tan \theta\right)}{\sec ^{2} \theta+\sec ^{2} \theta-2 \sec \theta \tan \theta}$ [∵ 1+ tan2 θ = sec2 θ]
$=\frac{2 \tan \theta(\sec \theta-\tan \theta)}{2 \sec \theta(\sec \theta-\tan \theta)}$
=$\frac{\sin \theta}{\cos \theta} \times \cos \theta$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= sin θ
=RHS
Hence Proved
Question 49
If a cos θ + b sinθ = c, then prove that a sinθ – b cos θ = ± $\sqrt{a^{2}+b^{2}-c^{2}}$
Sol :Let
(a cos θ + b sin θ)2 + (a sin θ – b cos θ)2 = a2cos2θ + b2 sin2θ + 2ab cos θ sin θ + a2sin2θ
+ b2 cos2θ – 2ab cos θ sin θ
⇒ c2 + (a sin θ – b cos θ)2 = a2 (cos2 θ + sin2 θ) + b2 (cos2 θ + sin2 θ)
⇒ c2 + (a sin θ – b cos θ)2 = a2 + b2
⇒ (a sin θ – b cos θ)2 = a2 + b2 – c2
⇒ (a sin θ – b cos θ) = ±√ (a2 + b2 – c2)
Question 50
If 1+sin2θ = 3 sinθ . cosθ, then prove that tan θ = 1 or 1/2.
Given: 1+sin2θ =3 sin θ cos θ
Divide by cos2 θ to both the sides, we get
⇒$\frac{1}{\cos ^{2} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{3 \sin \theta \cos \theta}{\cos ^{2} \theta}$
⇒ sec2 θ + tan2 θ = 3 tan θ
⇒ 1+ tan2 θ+ tan2 θ = 3tan θ
⇒ 2 tan2 θ –3tan θ +1 = 0
Let tanθ = x
⇒ 2x2 – 3x +1 = 0
⇒ 2x2 – 2x – x +1 = 0
⇒ 2x ( x – 1) – 1(x – 1) = 0
⇒ (2x – 1)(x – 1) = 0
Putting each of the factor = 0, we get
⇒ x = 1 or $\dfrac{1}{2}$
And above, we let tan θ = x
⇒$\tan \theta=1$ or $\frac{1}{2}$
Hence Proved
Question 51
If a cos θ – b sinθ = x and a sinθ + b cosθ = y that a2 + b2 = x2 + y2.
Sol :Taking RHS =x2 + y2
Putting the values of x and y, we get
(a cos θ – b sin θ)2 + (a sin θ + b cos θ)2
= a2cos2θ + b2 sin2θ – 2ab cos θ sin θ + a2sin2θ + b2 cos2θ + 2ab cos θ sin θ
= a2 (cos2 θ + sin2 θ) + b2 (cos2 θ + sin2 θ)
= a2 + b2 [∵ cos2 θ + sin2 θ = 1]
=RHS
Hence Proved
Question 52
If x = a sec θ + b tan θ a and y = a tan θ + b sec θ, then prove that x2 – y2 = a2 – b2.
Taking LHS =x2 – y2
Putting the values of x and y, we get
(a sec θ + b tan θ)2 – (a tan θ + b sec θ)2
= a2 sec2θ + b2 tan2θ + 2ab sec θ tan θ – a2tan2θ – b2 sec2θ – 2ab sec θ tan θ
= a2 (sec2θ – tan2θ) – b2 (sec2θ – tan2θ)
= a2 – b2 [∵ 1+ tan2 θ = sec2 θ]
=RHS
Hence Proved
Question 53
If (a2 – b2) sin θ + 2ab cosθ = a2 + b2, then prove that .
Sol :
Taking (a2 – b2) sin θ + 2ab cos θ = a2 + b2
We know that
$\sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$ and $\cos \theta=\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$
Then, substituting the above values in the given equation, we get
$=\mathrm{a}^{2}-\mathrm{b}^{2} \frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}+2 \mathrm{ab} \frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\mathrm{a}^{2}+\mathrm{b}^{2}$
Now, substituting,$t=\tan \frac{\theta}{2}$ , we have
$a^{2}-b^{2} \frac{2 t}{1+t^{2}}+2 a b \frac{1-t^{2}}{1+t^{2}}=a^{2}+b^{2}$
⇒ ( a2 – b2)2t – 2ab(1 – t2) = (a2 + b2)(1+t2)
Simplify, we get
(a2 + 2ab + b2)t2 – 2(a2 – b2)t + (a2 –2ab +b2)=0
⇒ (a+b)2 t2 – 2(a2 – b2)t + (a – b)2 = 0
⇒ (a+b)2 t2 –2 (a – b)(a+b)t + (a – b)2 =0
⇒ [(a+b)t – (a – b)]2 = 0 [∵ (a – b)2 = (a2 + b2 – 2ab)]
⇒ [(a+b)t – (a – b)] = 0
⇒ (a+b)t = (a – b)
⇒$ \mathrm{t}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}$
⇒$ \tan \frac{\theta}{2}=\frac{a-b}{a+b}$
We know that, $\frac{2 t}{1-t^{2}}=\tan \theta$ where $\mathrm{t}=\tan \frac{\theta}{2}$
$\Rightarrow \tan \theta=\frac{2\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)}{1-\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)^{2}}$
$\Rightarrow \tan \theta=\frac{2(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})}{(\mathrm{a}+\mathrm{b})^{2}-(\mathrm{a}-\mathrm{b})^{2}}$
$\Rightarrow \tan \theta=\frac{a^{2}-b^{2}}{2 a b}$
Hence Proved
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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