KC Sinha Mathematics Solution Class 11 Chapter 5 त्रिकोणमितीय फलन और उनके आलेख (Trigonometric function and their graphs) Exercise 5.1

Exercise 5.1

Question 1

निम्नलिखित तादात्म्यों को सिद्ध करें
(Prove the following identities)

(i) cos θ. tanθ=sin θ
Sol :
(i)
L.H.S$=cos \theta . tan \theta$
$\require{cancel} =\cancel{cos \theta} \times \dfrac{sin \theta}{\cancel{cos \theta}}$
$=sin \theta$


(ii) $\tan \theta=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$
Sol :
(ii)
R.H.S$=\dfrac{sin \theta}{\sqrt{1-sin^{2}\theta}}$
$=\dfrac{sin \theta}{cos \theta}=tan \theta$


(iii) sec²θ-cos²β=sin²β+tan²θ
Sol :
(iii)
L.H.S$=sec^{2}\theta - cos^{2} \beta$
$=(1+tan^{2} \theta)-(1-sin^{2} \beta)$
$=1+tan^{2} \theta-1+sin^{2} \beta$
$=sin^{2} \beta+tan^{2} \theta$


(iv) sin²A+cos²A.sin²B=sin²B+cos²B.sin²A
Sol :
(iv)
L.H.S$=sin^{2} A+cos^{2}A.sin^{2}B$
$=sin^{2}A+(1-sin^{2}A)(1-cos^{2}B)$
$=sin^{2} A+1-cos^{2}B-sin^{2}A+sin^{2}A.cos^{2}B$
$=sin^{2}B+cos^{2}B.sin^{2}A$

Question 2

सिद्ध करें कि 

(Prove that:)

(i) sec⁴θ-tan⁴θ=sec²θ+tan²θ
Sol :
L.H.S
=sec⁴θ-tan⁴θ
=(sec²θ)²-(tan²θ)²

=(sec²θ-tan²θ)(sec²θ+tan²θ)

[∵ sec²θ-tan²θ=1]

=1×(sec²θ+tan²θ)

=sec²θ+tan²θ

(ii) cosec²θ-cot⁴θ=cosec²θ+cot²θ

Sol :

(iii) (1+tanθ)²-sec²θ=2tanθ

Sol :

L.H.S

=(1+tanθ)²-sec²θ

=1²+tan²θ+2.1.tanθ-sec²θ

=1+tan²θ+2tanθ-sec²θ

[∵ 1+tan²θ=sec²θ]

=sec²θ+2tanθ-sec²θ

=2tanθ

(iv) (sinθ+cosθ)(1-sinθ.cosθ)=sin³θ+cos³θ

Sol :

=sin³θ+cos³θ

=(sinθ+cosθ)[sin²θ-sinθ.cosθ+cos²θ]

[∵sin²θ+cos²θ=1]

=(sinθ+cosθ)[1-sinθ.cosθ]

(v) sin⁴θ+cos⁴θ=1-2sin²θcos²θ

Sol :

[a²+b²=(a+b)²-2ab]

L.H.S
=sin⁴θ+cos⁴θ

=(sin2θ)2+(cos2θ)2

=(sin²θ+cos²θ)²-2sin²θcos²θ
(∵sin²θ+cos²θ=1)

=1²-2sin²θcos²θ

=1-2sin²θcos²θ

(vi) (cosecθ-sinθ)(secθ-cosθ)(tanθ+cotθ)=1
Sol :
L.H.S
(cosecθ-sinθ)(secθ-cosθ)(tanθ+cotθ)

$=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$

$=\left(\frac{1-\sin ^{2} \theta}{\sin \theta}\right)\left(\frac{\left(1-\cos ^{2} \theta\right)}{\cos \theta}\right)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}\right)$

$=\left(\frac{\cos ^{2} \theta}{\sin \theta}\right)\left(\frac{\left(\sin^{2} \theta\right)}{\cos \theta}\right)\left(\frac{1}{\cos \theta \sin \theta}\right)$

=1

(vii) (secθ+tanθ-1)(secθ-tanθ+1)=2tanθ

Sol :

L.H.S

=(secθ+tanθ-1)(secθ-tanθ+1)

[(a+b)(a-b)=a²-b²]

=[secθ+(tanθ-1)][secθ-(tanθ-1)]

=sec²θ-(tanθ-1)²
[(a-b)²=a²+b²-2ab]

=sec²θ-(tan²θ-1²-2tanθ.1)

=sec²θ-tan²θ-1+2tanθ

=1-1+2tanθ

=2tanθ

(viii) (1-sinθ-cosθ)²=2(1-sinθ)(1-cosθ)

Sol :
[(a-b-c)²=a²+b²+c²-2ab+2bc-2ca]

L.H.S
=(1-sinθ-cosθ)²

=1²+sin²θ+cos²θ-2.1.sinθ+2sinθcosθ-2cosθ.1

=1+1-2sinθ+2sinθcosθ-2cosθ

=2-2cosθ-2sinθ+2sinθcosθ

=2[1-cosθ-sinθ+sinθcosθ]

=2[1(1-cosθ)-sinθ(1-cosθ)]

=2(1-cosθ)(1-sinθ)

(ix) (1+sinθ+cosθ)²=2(1+sinθ)(1+cosθ)

Sol :

(x) $\cos ^{2} \theta \cdot \cos ^{2} \beta-\sin ^{2} \theta \cdot \sin ^{2} \beta=\cos ^{2} \theta-\sin ^{2} \beta$

Sol :

L.H.S

$\cos ^{2} \theta \cdot \cos ^{2} \beta-\sin ^{2} \theta \cdot \sin ^{2} \beta$

$=\left(1-\sin ^{2} \theta\right)\left(1-\sin ^{2} \beta\right)-\sin ^{2} \theta \sin ^{2} \beta$

$=1-\sin ^{2} \beta-\sin ^{2} \theta+\sin ^{2} \theta \sin ^{2} \beta-\sin ^{2} \theta \sin ^{2} \beta$

$=\cos ^{2} \theta-\sin ^{2} \beta$

Question 3

सिद्ध करें कि
Prove that :

(i) tanθ+cotθ=secθ.cosecθ
Sol :
L.H.S
=tanθ+cotθ

$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

=$\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}$

(∵cos²θ+sin²θ=1)

$=\frac{1}{\cos \theta \sin \theta}$

=secθcosecθ

(ii) $\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=2 \sec A$

Sol :
L.H.S
=$\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}$

$=\frac{\cos \cdot A(1+\sin A)+\cos A(1-\sin A)}{(1-\sin A)(1+\sin A)}$

$=\frac{\cos \theta+\cos A\sin A+\cos A-\cos A \sin A}{1^{2}-\sin ^{2} A}$

$=\frac{2 \cos A}{\cos ^{2} A}=2 \sec A$

(iii) $\frac{\cos ^{2} \alpha-\cos ^{2} \beta}{\cos ^{2} \alpha \cdot \cos ^{2} \beta}=\tan ^{2} \beta-\tan ^{2} \alpha$
Sol :
L.H.S
=$\frac{\cos ^{2} \alpha-\cos ^{2} \beta}{\cos ^{2} \alpha \cdot \cos ^{2} \beta}$

$=\frac{\cos^{2} \alpha}{\cos ^{2} \alpha \cos ^{2} \beta}-\frac{\cos ^{2} \beta}{\cos ^{2} \alpha \cos ^{2} \beta}$

$=\sec ^{2} \beta-\sec ^{2} \alpha$

$=\left(1+\tan ^{2} \beta\right)-\left(1+\tan ^{2} \alpha\right)$

$=1+\tan ^{2} \beta-1-\tan ^{2} \alpha=\tan ^{2} \beta-\tan ^{2} \alpha$

(iv) $\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}=2 \sec \theta \cdot \tan \theta$

Sol :

L.H.S

$\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}$

$=\frac{(1+\sin \theta)-(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$

$=\frac{1+\sin \theta-1+\sin \theta}{1^{2}-\sin ^{2} \theta}$

$=\frac{2 \sin \theta}{\cos ^{2} \theta}$

$=\frac{2 \sin \theta}{\cos \theta \cdot \cos \theta}$

=2tanθsecθ

(v) $\frac{\sec \theta}{1-\operatorname{cosec} \theta}+\frac{\operatorname{cosec} \theta}{\sec \theta}=\sec \theta \cdot \operatorname{cosec} \theta$

Sol :

(vi) $\frac{\cos A-\sin A}{\cos A+\sin A}=\frac{1-\tan A}{1+\tan A}$

Sol :

L.H.S

$\frac{\cos A-\sin A}{\cos A+\sin A}$

अंश(numeraot) तथा हर(denominator) मे cosA से भाग(divide) देने पर

$=\frac{\frac{\cos A}{\operatorname{cos} A}-\frac{\sin A}{\cos A}}{\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}}$

$=\frac{1-\tan A}{1+\tan A}$

(vii) $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1+\sin \theta}{\cos \theta}$

Sol :

L.H.S

$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$

अंश(numeraot) तथा हर(denominator) मे cosA से भाग(divide) देने पर

$=\frac{\frac{\sin \theta}{\cos \theta}-\frac{\cos \theta}{\cos \theta}+\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta }{\cos \theta}-\frac{1}{\cos \theta}}$

$=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta}$

$=\frac{\sec \theta+\tan \theta-1}{1-\sec \theta+\tan \theta}$

$\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$=\frac{(\sec \theta+\tan \theta)-\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{1-\sec \theta+\tan \theta}$

$\left[a^{2}-b^{2}=(a-b)(a+b)\right]$

$=\frac{(\sec \theta+\tan \theta)-(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}{1-\sec \theta+\tan \theta}$

$=\dfrac{(\sec \theta+\tan \theta)[1-\sec \theta-\tan \theta]}{1-\sec \theta+\tan \theta}$

$=\dfrac{(\sec \theta+\tan \theta)(1-\operatorname{sec} \theta+\tan \theta)}{1-\sec \theta +\tan \theta}$

=secθ+tanθ

$=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}$

$=\frac{1+\sin \theta}{\cos \theta}$

R.H.S

Question 4

सिद्ध करें कि
Prove that:

(i) $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=\sec \theta+\tan \theta$
Sol :
L.H.S
$=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$

$=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}$

$=\sqrt{\frac{(1+\sin \theta)^{2}}{1^2-\sin^2 \theta}}$

$=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}$

$=\frac{1+\sin \theta}{\cos \theta}$

$=\frac{1}{\cos \theta }+\frac{\sin \theta}{\cos \theta}$

=secθ+tanθ



(ii) $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\operatorname{cosec} \theta+\cot \theta$
Sol :


(iii) $\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\operatorname{cosec} \theta-\cot \theta$

Sol :

Question 5

सिद्ध करें कि
Prove that :

(i) $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\sin \theta+\cos \theta$

Sol :
L.H.S
$=\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$

$=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}$

$=\dfrac{\cos \theta}{\frac{\cos \theta-\sin \theta}{\cos \theta}}+\frac{\sin \theta}{\dfrac{\sin \theta-\cos \theta}{\sin \theta}}$

$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}+\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}$

$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta-\sin \theta}$

$=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}$

=cosθ+sinθ

(ii) $1+\frac{2 \tan ^{2} \theta}{\cos ^{2} \theta}=\tan ^{4} \theta+\sec ^{4} \theta$

Sol :
L.H.S
$=1+\frac{2 \tan ^{2} \theta}{\cos ^{2} \theta}$

$=1+2\left(\sec ^{2} \theta-1\right) \sec ^{2} \theta$

$=1+2 \sec ^{4} \theta-2 \sec ^{2} \theta$

$=\sec ^{4} \theta+1-2 \sec ^{2} \theta+\sec ^{4} \theta$

$=\left(\sec ^{2} \theta\right)^{2}+1^{2}-2 \cdot \sec ^{2} \theta \cdot 1+\sec ^{4} \theta$

$=\left(\sec ^{2} \theta-1\right)^{2}+\sec ^{4} \theta$

$=\left(\tan ^{2} \theta\right)^{2}+\sec ^{4} \theta$

$=\tan ^{4} \theta+\sec ^{4} \theta$

(iii) $\cos ^{2} A-\sin ^{2} A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$

Sol :
L.H.S
$=\cos ^{2} A-\sin ^{2} A$

=$\frac{\cos ^{2} A-\sin ^{2} A}{\cos^{2} A +\sin ^{2} A}$

अंश तथा हर मे $\cos ^{2} A$ से भाग देने पर,

$=\frac{\frac{\cos^2 A}{\cos^2 A}-\frac{\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\operatorname{cos}^{2} A}+\frac{\sin ^{2} A}{\cos ^{2} A}}$

$=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$

(iv) $\frac{1-\cos ^{4} \beta}{\sin ^{4} \beta}=2 \operatorname{cosec}^{2} \beta-1$

Sol :

L.H.S

$=\frac{1-\cos ^{4} \beta}{\sin ^{4} \beta}$

$=\frac{1^{2}-\left(\cos ^{2} \beta\right)^{2}}{\sin ^{4} \beta}$

$=\frac{\left(1-\cos ^{2} \beta\right)\left(1+\cos ^{2} \beta\right)}{\sin ^{2} \beta}$

$=\frac{\sin ^{2} \beta\left(1+\cos ^{2} \beta\right)}{\sin ^{4} \beta}$

$=\frac{1}{\sin ^{2} \beta}+\frac{\cos ^{2} \beta}{\sin ^{2} \beta}$

$=\operatorname{cosec}^{2} \beta+\cos ^{2} \beta$

$=\operatorname{cosec}^{2} \beta+\operatorname{cosec}^{2} \beta-1$

$=2 \operatorname{cosec}^{2} \beta-1$

(v) $\tan ^{2} \theta-\tan ^{2} \beta=\frac{\sin ^{2} \theta-\sin ^{2} \beta}{\cos ^{2} \theta \cdot \cos ^{2} \beta}$

Sol :

R.H.S

$=\frac{\sin ^{2} \theta-\sin ^{2} \beta}{\cos ^{2} \theta \cdot \cos ^{2} \beta}$

$=\frac{\left(1-\cos ^{2} \theta\right)-\left(1-\cos ^{2} \beta\right)}{\cos ^{2} \theta \cos ^{2} \beta}$

$=\frac{1-\cos ^{2} \theta-1+\cos ^{2} \beta}{\cos^2 \theta \cos ^{2} \beta}$

$=\frac{\cos ^{2} \beta-\cos ^{2} \theta}{\cos ^{2} \theta \cos ^{2} \beta}$

$=\frac{\cos^2 \beta}{\cos ^{2} \theta \cos ^{2} \beta}-\frac{\cos^2 \theta}{\cos ^{2} \theta \cos ^{2} \beta}$

$=\sec ^{2} \theta-\sec ^{2} \beta$

$=\left(1+\tan ^{2} \theta\right)-\left(1+\tan ^{2} \beta\right)$

$=1+\tan ^{2} \theta-1-\tan ^{2} \beta$

$=\tan ^{2} \theta-\tan ^{2} \beta$

(vi) $\frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}=\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}$

Sol :

L.H.S

$=\frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}$

$=\frac{\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta}$

$=\frac{\frac{1}{\cos ^{2} \theta}}{\frac{1}{\sin ^{2} 2 \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\tan ^{2} \theta$

R.H.S

$=\left(\frac{1-\tan \theta}{1-\cos \theta}\right)^{2}$

$=\left(\frac{1-\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}\right)^{2}$

$=\left(\frac{\frac{\cos \theta-\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}\right)^{2}$

$=\left(\frac{\cos \theta-\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta}\right)^{2}$

$=\left[\tan \theta\left(\frac{\cos \theta-\sin \theta}{\operatorname{sin} \theta-\cos \theta}\right)\right]^{2}$

$=\tan ^{2} \theta \frac{(\cos \theta -\sin \theta)^{2}}{\left(\sin \theta-\cos \theta\right)^{2}}$

L.H.S=R.H.S

$=\tan ^{2} \theta \frac{(\sin \theta-\cos \theta)^{2}}{\left(\sin \theta-\cos \theta\right)^{2}}$

$=\tan ^{2} \theta$

(vii) $\frac{\cos A+\cos B}{\sin A+\sin B}+\frac{\sin A-\sin B}{\cos A-\cos B}=0$

Sol :

L.H.S

$\frac{\cos A+\cos B}{\sin A+\sin B}+\frac{\sin A-\sin B}{\cos A-\cos B}$

$=\frac{(\cos A+\cos A)(\cos A-\cos B)+(\sin A+\sin B)(\sin A-\sin B)}{(\sin A+\sin B)(\cos A-\cos B)}$

$=\frac{\cos ^{2} A-\cos^{2} B+\sin^2 A-\sin^2 B}{(\sin A+\sin B)(\cos A-\cos B)}$

$=\frac{\left(\cos ^{2} A+\sin ^{2} A\right)-\left(\cos ^{2} B+\sin ^{2} B\right)}{(\sin A+\sin B)(\cos A-\cos B)}$

$=\frac{1-1}{(\sin A+\sin B)(\cos A-\cos B)}$

=0

(viii) $\frac{\tan \theta}{\sec \theta-1}-\frac{\sin \theta}{1+\cos \theta}=2 \cot \theta$

Sol :
L.H.S
$=\frac{\tan \theta}{\sec \theta-1}-\frac{\sin \theta}{1+\cos \theta}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}-1}-\frac{\sin \theta}{1+\cos \theta}$

$=\frac{\sin \theta}{1-\cos \theta}-\frac{\sin \theta}{1+\cos \theta}$

$=\frac{\sin \theta(1+\cos \theta)-\sin \theta(1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}$

$=\frac{\sin \theta+\sin \theta\cos \theta-\sin \theta+\sin \theta \cos \theta}{1^{2}-\cos ^{2} \theta}$

$=\frac{2+\sin\theta \cos \theta}{\sin ^{2} \theta}$

=2cotθ

(ix) $(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$

Sol :

L.H.S

$(\operatorname{cosec} \theta-\cot \theta)^{2}$

$=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$

$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^2$

$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$

$=\frac{(1-\cos \theta)^{2}}{1^{2}-\cos ^{2} \theta}$

$=\frac{(1-\cos \theta)^{2}}{(1-\cos \theta)(1+\cos \theta)}$

$=\frac{1-\cos \theta}{1+\cos \theta}$

(x) $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \cdot \operatorname{cosec} \theta$

Sol :

L.H.S

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$

$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)}$

$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$

$=\frac{\sin ^{3} \theta-\cos^3 \theta}{\cos \theta \sin \theta(\sin \theta-\cos \theta)}$

$=\frac{\left.(\sin \theta-\cos \theta)\left[(\sin \theta)^{2}+\sin \theta\cos\theta+\cos \theta\right)^{2}\right]}{\cos \theta \sin \theta(\sin \theta-\cos \theta)}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta}{\cos \theta \sin \theta}$

$=\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}$

$=\frac{1}{\cos \theta \sin \theta}+\frac{\sin \theta \cos \theta}{\cos \theta \sin \theta}$

=secθcosecθ+1

(xi) $\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}=2 \operatorname{cosec} \theta$

Sol :

L.H.S

=$\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$

=$\frac{\tan \theta(\sec \theta+1)+\tan \theta(\sec \theta-1)}{(\sec \theta-1)(\sec \theta+1)}$

$=\frac{\tan \theta \sec \theta+\tan \theta+\tan \theta \sec \theta-\tan \theta}{\sec ^{2} \theta-1^{2}}$

$=\frac{2 \tan \theta \sec \theta}{\tan ^{2} \theta}$

$=\frac{2 \times \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}=2 \operatorname{cosec} \theta$

(xii) $\frac{\sec \theta+\tan \theta}{\operatorname{cosec} \theta+\cot \theta}=\frac{\operatorname{cosec} \theta-\cot \theta}{\sec \theta-\tan \theta}$

Sol :

L.H.S

=$\frac{\sec \theta+\tan \theta}{\operatorname{cosec} \theta+\cot \theta}$

अंश तथा हर दोनो का परिमेय-करण करने पर,

$=\frac{\sec \theta+\tan \theta}{\operatorname{cosec} \theta+\cos \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta} \times \frac{\text{cosec } \theta-\cot \theta}{\operatorname{cosec} \theta-\cot \theta}$

$=\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\operatorname{cosec}^{2} \theta-\cos ^{2} \theta} \times \frac{\operatorname{cosec} \theta-\cot \theta}{\sec \theta-\tan \theta}$

$=\frac{1}{1} \times \frac{\operatorname{cosec} \theta-\operatorname{cot} \theta}{\sec \theta-\tan \theta}$

$=\frac{\operatorname{cosec} \theta-\cot \theta}{\sec \theta-\tan \theta}$

(xiii) $\frac{2 \sin \theta \cdot \tan \theta(1-\tan \theta)+2 \sin \theta \cdot \sec ^{2} \theta}{(1+\tan \theta)^{2}}=\frac{2 \sin \theta}{1+\tan \theta}$

Sol :
L.H.S
=$\frac{2 \sin \theta \tan \theta(1-\tan \theta)+2 \sin \theta \sec ^{2} \theta}{(1+\tan \theta)^{2}}$

=$\frac{2 \sin \theta\left[\tan \theta(1-\tan \theta)+\sec ^{2} \theta\right]}{(1+\tan \theta)^{2}}$

$=\frac{2 \sin \theta\left[\tan \theta-\tan^2 \theta+1+\tan^{2} \theta\right]}{(1+\tan \theta)^{2}}$

$=2 \sin \theta \frac{(1+\tan \theta)}{(1+\tan \theta)^{2}}$

$=\frac{2 \sin \theta}{1+\tan \theta}$

(xiv) $\sin A+\cos A=\frac{\cos A}{1-\tan A}+\frac{\tan A \cdot \sin A}{\tan A-1}$

Sol :

R.H.S

$=\frac{\cos A}{1-\tan A}+\frac{\tan A \cdot \sin A}{\tan A-1}$

$=\frac{\cos \theta}{1-\tan A}-\frac{\tan A \sin A}{1-\tan A}$

$=\frac{\cos \theta-\tan A \sin A}{1-\tan A}$

$=\frac{\cos A-\frac{\sin A}{\cos A} \cdot \sin A}{1-\frac{\sin A}{\cos A}}$

$=\frac{\frac{\cos ^{2} A-\sin ^{2} A}{\cos}}{\frac{\cos A-\sin A}{\cos A}}$

$=\frac{(\cos A \sin A)(\cos A+\sin A]}{\cos A \sin A}$

=sinA+cosA

(xv) $(\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta)^{2}+(\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta)^{2}=1$
Sol :
L.H.S
=$(\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta)^{2}+(\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta)^{2}$

[sin(A+B)=sinAcosB+cosAsinB

cos(A+B)=cosAcosB-sinAsinB]

$=\sin ^{2}(\alpha+\beta)+\cos ^{2}(\alpha+\beta)$

=1

(xvi) $\frac{1}{\operatorname{cosec} \theta-\cot \theta}-\frac{1}{\sin \theta}=\frac{1}{\sin \theta}-\frac{1}{\operatorname{cosec} \theta+\cot \theta}$

Sol :

L.H.S

=$\frac{1}{\operatorname{cosec} \theta-\cot \theta}-\frac{1}{\sin \theta}$

$=\frac{1}{\operatorname{cosec} \theta-\cot\theta} \times \frac{\operatorname{cosec} \theta+\cot \theta}{\text{cosec } \theta +\cot \theta}-\operatorname{cosec} \theta$

$=\frac{\operatorname{cosec} \theta+\cos \theta}{\operatorname{cosec}^{2} \theta-\operatorname{cot}^{2} \theta}-\operatorname{cosec} \theta$

=cosecθ+cotθ-cosecθ 

$\left[\because \operatorname{cosec}^{2} \theta-\cot^2 \theta=1\right]$

=cotθ

R.H.S
$=\frac{1}{\sin \theta}-\frac{1}{\operatorname{cosec} \theta+\cot \theta}$

$=\operatorname{cosec} \theta-\frac{1}{\operatorname{cosec} \theta+\cot \theta} \times \frac{\operatorname{cosec} \theta-\cot \theta}{\text{cosec } \theta-\cot \theta}$

$=\operatorname{cosec} \theta-\frac{\operatorname{cosec} \theta-\cot \theta}{\operatorname{cosec}^{2} \theta-\cot^{2} \theta}$

=cosecθ-(cosecθ-cotθ)

=cosecθ-(cosecθ-cotθ)

=cosecθ-cosecθ+cotθ

=cotθ

(xvii) $\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}=\frac{1+\cos A}{\sin A}$

Sol :

L.H.S

=$\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}$

$[\because\operatorname{cosec}^{2} A-\cot ^{2} A=1]$

$=\frac{\operatorname{cosec} A+\cos A-\left(\operatorname{cosec}^{2} A-\cos ^{2} A\right)}{1-\operatorname{cosec} A+\cot A}$

$=\frac{(\text{cosec }A+\cot A)-(\operatorname{cosec} A-\cot A) \operatorname{cosec} A+\cot A)}{1-\text{cosec A}+\cot A}$

=$\dfrac{(\operatorname{cosec A} +\cot A)-(\operatorname{cosec} \theta-\cot A)( \operatorname{cosec} A+\cot A}{1-\text{cosec A}+\cot A}$

$=\frac{(\operatorname{cosec} A+\cos A)[1-(\operatorname{cosec} A-\cot A)]}{1-\operatorname{cosec} A+\cot A}$

$=\frac{(\operatorname{cosec} A+\cos A)(1-\operatorname{cosec} A+\cot A)}{1-\operatorname{cosec} A+\cot A}$

$=\frac{1}{\sin A}+\frac{\cos A}{\sin A}$

$=\frac{1+\cos A}{\sin A}$

(xviii) (1+cotA-coescA)(1+tanA+secA)=2

Sol :

L.H.S

=(1+cotA-cosecA)(1+tanA+secA)

=1+tanA+secA+cotA+tanAcotA+secA.cotA-cosecA-cosecAtanA-cosecAsecA

=1+1+tanA+cotA+secA+$\dfrac{1}{\cos A} \times \dfrac{\cos A}{\sin A}$-cosecA$-\frac{1}{\sin A} \times \frac{\sin A}{\cos A}$-cosecAsecA

=2+$\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$+secA+cosecA-cosecA

=2+$\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$+secA+cosecA-cosecA-secA-cosecAsecA

=2+$\frac{\sin^2 A+\cos ^{2} A}{\cos A \sin A}$-cosecAsecA

=2+secAcosecA-cosecAsecA

=2

(xix) $\cos ^{8} \theta-\sin ^{8} \theta=\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\left(1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta\right)$
Sol :
L.H.S

=$\cos ^{8} \theta-\sin ^{8} \theta$

[$a^{2}-b^{2}=(a-b)(a+b)$

$a^{2}+b^{2}=(a+b)^{2}-2 a b$]

$=\left[\left(\cos ^{2} \theta\right)^{2}\right]^{2}-\left[\left(\sin ^{2} \theta\right)^{2}\right]^{2}$

=$\left[\left(\cos ^{2} \theta\right)^{2}-\left(\sin ^{2} \theta\right)^{2}\right]\left[\left(\cos ^{2} \theta\right)^{2}+\left(\sin ^{2} \theta\right)^{2}\right]$

$=\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left[\left(\cos ^{2} \theta+\sin^{2} \theta\right)^{2}-2 \cos ^{2} \theta_{2}\sin^2\theta\right]$

$=\left(\cos ^{2} \theta-\sin^{2} \theta\right)\left(1-2 \sin ^{2} \theta \cos ^{2} \theta\right)$

(xx) $\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}+\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=2 \operatorname{cosec} \theta$

Sol :

L.H.S

=$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}+\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$

$=\frac{\sqrt{(1-\cos \theta)^{2}}+\sqrt{(1+\cos \theta)^{2}}}{\sqrt{(1+\cos \theta)(1-\cos \theta)}}$

$=\frac{1-\cos \theta+1+\tan \theta}{\sqrt{1^{2}-\cos ^{2} \theta}}$

$=\frac{2}{\sin \theta}=2 \cdot \operatorname{cosec} \theta$

Question 6

निम्नलिखित में पांच त्रिकोणमितीय फलनों का मान ज्ञात कीजिए।
[Find the value f other five trigonometry functions in the function]

(i) $\cos x=-\frac{3}{5}, x$ तृतीय चतुर्थाश में है। [x is a third quadrant]
Sol :

Let B=3k , H=5k

$P=\sqrt{H^{2}-B^{2}}$

$=\sqrt{(5 k)^{2}-(3 k)^{2}}$

$=\sqrt{25 k^{2}-9 k^{2}}$

$=\sqrt{16 k^{2}}$

=4k


$\cos x=-\frac{3}{5}=\frac{B}{H}$

$\sin x=\frac{-4k}{5k}=-\frac{4}{5}$ $\left(\because \sin \theta=\frac{P}{H}\right)$

$\tan x=\frac{4 k}{3 k}=\frac{4}{3}$ $\left(\tan \theta=\frac{P}{B}\right)$

$\operatorname{cosec} x=\frac{-5 k}{4 k}=\frac{-5}{4}$ $\left(\because \operatorname{casec} \theta=\frac{H}{P}\right)$

$\sec x=\frac{-5 k}{3 k}=-\frac{5}{3}$ $\left(\sec \theta=\frac{H}{B}\right)$

$\cot x=\frac{3 k}{4 k}=\frac{3}{4}$ $\left(\cot \theta=\frac{B}{P}\right)$

(ii) $sinx=\frac{3}{5}$, x द्वितीय चतुर्थाश में है। [x is in 2nd quadrant]

Sol :

(iii) $secx=\frac{13}{5}$, x चतुर्थ चतुर्थाश में है। [x is in 4nd quadrant]

Sol :

(iv) $\cot x=\frac{-5}{12}$ , x द्वितीय चतुर्थाश में है। [x is in 2nd quadrant]

Sol :

(v) $\cotx=\frac{3}{4}$, x तृतीय चतुर्थाश में है। [x is in 3rd quadrant]

Sol :

Question 7

(i) यदि (If) $\cos \theta=-\frac{3}{5}$ तथा (and) $\pi<\theta<\frac{3 \pi}{2}$ तो ज्ञात करें (then find )

$\frac{\sec \theta-\tan \theta}{\operatorname{cosec} \theta+\cot \theta}$
Sol :

$P=\sqrt{H^{2}-B^{2}}$

$=\sqrt{5^{2}-3^2}$

=4

$\cos \theta=-\frac{3}{5}=\frac{B}{H}$

$\sec \theta=\frac{H}{B}=\frac{-5}{3}$

$\tan \theta=\frac{P}{B}=\frac{4}{3}$

$\operatorname{cosec} \theta=\frac{H}{P}=\frac{-5}{4}$

$\cos \theta=\frac{B}{P}=\frac{3}{4}$

$\frac{\sec \theta-\tan \theta}{\text{cosec } \theta +\cot \theta}$

$=\frac{-\frac{5}{3}-\frac{4}{3}}{-\frac{5}{4}+\frac{3}{4}}$

$=\frac{-\frac{5-4}{3}}{-\frac{5+3}{4}}$

$=\frac{-\frac{9}{3}}{\frac{-2}{4}}$

=6


(ii) यदि $\cos \theta =\frac{3}{5}$ तथा θ चौथे चतुर्थाश मे हो तो cosecθ+cotθ का मान निकाले ।

[If $\cos \theta =\frac{3}{5}$ and θ lies in the fourth quadrant find the value of cosecθ+cotθ]

Sol :
Diagram
$P=\sqrt{H^{2}-B^{2}}$

$=\sqrt{5^{2}-3^{2}}$

$=\sqrt{25-9}=\sqrt{16}$

=4

$\cos \theta=\frac{3}{5}$

$\operatorname{cosec} \theta=\frac{H}{P}=\frac{-5}{4}$

$\cot \theta=\frac{B}{P}=-\frac{3}{4}$

$\operatorname{cosec} \theta+\cot \theta=-\frac{5}{4}-\frac{3}{4}$

$=\frac{-8}{4}$

=-2

(iii)  यदि (If) $\tan \theta=\frac{-4}{3},\frac{3\pi}{2}<\theta<2\pi$ तो $9\sec^2 \theta-4cot \theta$ का मान निकाले। (Find the value of $9\sec^2 \theta-4cot \theta$)

Sol :

(iv) यदि (If) $\sin \alpha=\frac{3}{5}, \tan \beta=\frac{1}{2}$ and $\frac{\pi}{2}<\alpha<\pi<\beta<\frac{3 \pi}{2}$ show that $8 \tan \alpha-\sqrt{5} \sec \beta=-\frac{7}{2}$

Sol :

Diagram

$B=\sqrt{5^{2}-3^{2}}$

$=\sqrt{25-9}=\sqrt{16}$

=4

Diagram

$H=\sqrt{1^{2}+2^{2}}$

$=\sqrt{1+4}=\sqrt{5}$

$\sin \alpha=\frac{3}{5}$

$\tan \alpha=\frac{P}{B}=-\frac{3}{4}$

$\tan \beta=\frac{1}{2}$

$\sec \beta=\frac{H}{B}=-\frac{\sqrt{5}}{2}$

L.H.S
$8 \tan \alpha-\sqrt{5} \sec \beta$

$=8\left(-\frac{3}{4}\right)-\sqrt{5} \times\left(-\frac{\sqrt{5}}{2}\right)$

$=-6+\frac{5}{2}$

$=\frac{-12+5}{2}=\frac{-7}{2}$

Question 8

साबित करे कि $\sin \theta =x+\frac{p}{x}$,x के वास्तविक मानो के लिए तभी सम्भव है जब $p\leq \frac{1}{4}$

Sol :

$\sin \theta=x+\frac{P}{x}$

[$\sin^{2} \theta \leq 1$]

Squaring both the sides

दोनो तरफ वर्ग करने पर

$\sin^{2} \theta=\left(x+\frac{P}{x}\right)^2$


$=x^{2}+\left(\frac{P}{x}\right)^{2}+2 \cdot x \cdot \frac{p}{x}$

$=\left(x-\frac{p}{x}\right)^{2}+2 \cdot x \cdot \frac{p}{x}+2 p$

$=\left(x-\frac{p}{2}\right)^{2}+4 p$
∴$\left(x-\frac{p}{x}\right)^{2} \geqslant 0$

∴$P \leq \frac{1}{4}$

Question 9

सिद्ध करे कि $\cos ^{2} \theta+\sec ^{2} \theta$ का मान 2 से कम नहीं हो सकता है ।

Sol :

$\cos ^{2} \theta+\sec ^{2} \theta=1-\sin ^{2} \theta+1+\tan ^{2} \theta$

$=2+\tan ^{2} \theta-\sin ^{2} \theta$

$=2+\frac{\sin^2{\theta}}{\cos ^{2} \theta}-\sin ^{2} \theta$

$=2+\sin ^{2} \theta\left(\frac{1}{\cos ^{2} \theta}-1\right)$

$=2+\sin ^{2} \theta\left(\sec ^{2} \theta-1\right)$

$=2+\sin ^{2} \theta \cdot \tan ^{2} \theta$

$\sin ^{2} \theta \tan ^{2} \theta$≥0

$\cos ^{2} \theta+\sec ^{2} \theta \geqslant 2$


Question 10

साबित करें कि $\sin^2 \theta =\frac{(x+y)^2}{4xy}$ , x और y के वास्तविक मानो के लिए सम्भव है जब x=y तथा x≠0

Sol :

$\sin ^{-2} \theta=\frac{(x+y)^{2}}{4 x y}$

$\because \sin ^{2} \theta \leq 1$

$\frac{(x+y)^{2}}{4x y} \leq 1$

$(x+y)^{2} \leq 4 x y$

$(x+y)^{2}-4 x y \leq 0$

$x^{2}+y^{2}+2 x y-4 x y \leq 0$

$x^{2}+y^{2}-2 x y \leq 0$

$(x-y)^{2} \leq 0$

we know that

$(x-y)^{2} \geq 0$

$\therefore \quad(x-y)^{2}=0$

x-y=0

x=y

For real value of $\sin ^{2} \theta$
($\sin ^{2} \theta$ के वास्तविक मान के लिए)

$4 x y \neq 0$

$x \neq 0, y \neq 0$

∴ x=y and x≠0

Question 11

क्या निम्नलिखित समीकरण स्तय है ?

(i) $\sec \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$
Sol :
Diagram

$P=\sqrt{\left(x^{2}-y^{2}\right)^{2}-\left(x^{2}+y^{2}\right)^{2}}$

$P=\sqrt{x^{4}+y^{4}-2 x^{2} y^{2}-x^{4}-y^{4}-2x^2y^2}$

$=\sqrt{-4 x^{2} y^{2}} \notin R$

$\sec \theta=\frac{x^{2}+ y^{2}}{x^{2}+y^{2}}=\frac{H}{B}$

False

(ii) $\cos \theta=\frac{x^{2}+1}{x}$

Sol :

(iii) $\sin \theta=\frac{2 x y}{x^{2}+y^{2}}$

Sol :

Question 12

निम्नलिखित के मान निकाले

Find the values of the following

(i) sin 1830°

Sol :

=sin(20×90°+30°)

=sin30°

$=\frac{1}{2}$

(ii) sin 765°

Sol :

=sin(8×90°+45°)

=sin45°

$=\frac{1}{\sqrt{2}}$

(iii) cos(-1710)°

Sol :

[cos(-θ)=cosθ]

=cos1710°

=cos(19×90°+0)

=sin0°

=0

(iv) $\sin \left(-\frac{11 \pi}{3}\right)$

Sol :

[sin(-θ)=-sinθ]

$=-\sin \left(\frac{11}{3} \pi\right)$

$=-\sin \left(4 \pi-\frac{\pi}{3}\right)$

$=-\left[-\operatorname{sin} \frac{\pi}{3}\right]$

$=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$

(v) $\sin \frac{31 \pi}{3}$

Sol :

(vi) $\cot \left(-\frac{15 \pi}{4}\right)$

Sol :

$=-\cot\left(\frac{15 \pi}{4}\right)$

$=-\operatorname{cot}\left(4 \pi-\frac{\pi}{4}\right)$

$=-\left(-\cot \frac{\pi}{4}\right)$

$=\operatorname{cot} \frac{\pi}{4}=1$

Question 13

निम्नलिखित के मान निकाले

Find the value of the following :

(i) $\sin ^{2} 135^{\circ}+\cos ^{2} 120^{\circ}-\sin ^{2} 120^{\circ}+\tan ^{2} 150^{\circ}$

Sol :

=$\sin ^{2} 135^{\circ}+\cos ^{2} 120^{\circ}-\sin ^{2} 120^{\circ}+\tan ^{2} 150^{\circ}$

$=\sin ^{2}(180-45)+\cos ^{2}(180-60)-\sin ^{2}(180-60)+\tan ^{2}(180-30)$

$=\sin ^{2} 45+\left[-\cos 60^{\circ}\right]^{2}-\sin ^{2} 60+(-\tan 30)^{2}$

$=\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(-\frac{1}{\sqrt{3}}\right)^{2}$

$=\frac{1}{2}+\frac{1}{4}-\frac{3}{4}+\frac{1}{3}$

$=\frac{6+3-9+4}{12}$

$=\frac{4}{12}=\frac{1}{3}$


(ii) $2 \cos ^{2} 135^{\circ}+\sin 150^{\circ}+\frac{1}{2} \cos 180^{\circ}+\tan 135^{\circ}$
Sol :
$=2 \cos ^{2}(180-45)+\sin (180-30)+\frac{1}{2} \cos 180+\tan \left(180^{\circ}-45\right)$

$=2(\cos 45)^{2}+\sin 30+\frac{1}{2} \cos 180+(-\tan 45)$

$=2\left(-\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}+\frac{1}{2} \times(-1)+(-1)$

$=2 \times \frac{1}{2}+\frac{1}{2}-\frac{1}{2}-1$

=1-1
=0

(iii) $\tan ^{2} 45^{\circ}-4 \sin ^{2} 60^{\circ}+2 \cos ^{2} 45^{\circ}+\sec ^{2} 180^{\circ}$

Sol :

(iv) $\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}$

Sol :
$=\cot ^{2} \frac{\pi}{6}+\operatorname{cosec}\left(\pi-\frac{\pi}{6}\right)+3 \tan ^{2} \frac{\pi}{6}$

$=(\sqrt{3})^{2}+\operatorname{cosec} \frac{\pi}{6}+3\left(\frac{1}{\sqrt{3}}\right)^{2}$

$=3+2+3\left(\frac{1}{3}\right)$

=6

(v) $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
Sol :

$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$

$=\frac{1}{4}+\frac{1}{4}-1$

$=\frac{1+1-4}{4}=\frac{-2}{4}$

$=-\frac{1}{2}$

(vi) $3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}$
Sol :
$=3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \left(\pi-\frac{\pi}{6}\right)\cot\frac{\pi}{4}$

$=3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{\pi}{6} \cdot \cot \frac{\pi}{4}$

$=3 \times \frac{1}{2} \times 2-4 \times \frac{1}{2} \times 1$

=3-2

=1

Question 14

साबित करें कि

Prove that

(i) $\sin ^{2} 200^{\circ}+\sin ^{2} 250^{\circ}+\sin ^{2} 280^{\circ}+\sin ^{2} 370^{\circ}=2$

Sol :

L.H.S

$\sin ^{2} 200+\sin ^{2} 250+\sin ^{2} 280+\operatorname{sin}^{2} 370$

$=\sin ^{2}(2 \times 90+20)+\sin ^{2}(3 \times 90-20)+\sin ^{2}(3 \times 90+10)+\sin^{2}(4\times 90+10)$

$=\sin ^{2} 20^{\circ}+\cos ^{2} 20^{\circ}+\cos ^{2} 10^{\circ}+\sin ^{2} 10$

=1+1

=2

(ii) $\cos 570^{\circ} \cdot \sin 510^{\circ}+\sin \left(-330^{\circ}\right) \cdot \cos 390^{\circ}=.0$

Sol :

L.H.S

$\sin (-\theta)=-\sin \theta$

$\cos 570 \sin 510+\sin (-330) \cdot \cos 390$

=-cos30sin30-(-sin30)cos30

=-cos30sin30+sin30cos30

=0

Question 15

निम्नलिखित के मान ज्ञात करें 

[Find the values of the following]

(i) $\cos ^{2} \frac{\pi}{16}+\cos ^{2} \frac{3 \pi}{16}+\cos ^{2} \frac{5 \pi}{16}+\cos ^{2} \frac{7 \pi}{16}$

Sol :

$=\cos ^{2} \frac{\pi}{16}+\cos ^{2} \frac{3 \pi}{16}+\cos ^{2}\left(\frac{\pi}{2}-\frac{3 \pi}{16}\right)+\cos ^{2}\left(\frac{\pi}{2}-\frac{\pi}{16}\right)$

$=\cos^{2} \frac{\pi}{16}+\cos ^{2} \frac{3 \pi}{16}+\sin ^{2} \frac{3 \pi}{16}+\sin ^{2} \frac{\pi}{16}$

=1+1

=2

(ii) $\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}$

Sol :

$=\sin^{2} \frac{\pi}{8} +\sin ^{2} \frac{3 \pi}{8}+\sin ^{2}\left(\frac{\pi}{2}+\frac{\pi}{8}\right)+\sin ^{2}\left(\frac{\pi}{2}+\frac{3 \pi}{8}\right)$

$=\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\cos^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}$

=1+1

=2

Question 16

सिद्ध करें कि

Prove that :

(i) $\sin ^{2} 6^{\circ}+\sin ^{2} 12^{\circ}+\sin ^{2} 18^{\circ}+\ldots+\sin ^{2} 84^{\circ}+\sin ^{2} 90^{\circ}=8$

Sol :

L.H.S

=$\sin ^{2} 6^{\circ}+\sin ^{2} 12^{\circ}+\sin ^{2} 18^{\circ}+\ldots+\sin ^{2} 84^{\circ}+\sin ^{2} 90^{\circ}$

$=\left(\sin ^{2} 6+\sin ^{2} 84\right)+\left(\sin ^{2} 12+\sin ^{2} 78\right)+\left(\sin ^{2} 18+\sin ^{2} 72\right)+\left(\sin ^{2} 24+\sin ^{2} 66\right)+\left(\sin ^{2} 30+\sin ^{2} 60\right)+\left(\sin ^{2} 36+\sin ^{2} 54\right)+\left(\sin ^{2} 42+\sin ^{2} 48\right)+\sin^{2}90$

$=\left(\sin ^{2} 6+\cos ^{2} 6\right)+\left(\sin ^{2} 12+\cos ^{2} 12\right)+\left(\sin ^{2} 18+\cos ^{2} 18\right)+\left(\sin ^{2} 24+\cos^{2} 24\right)+\left(\sin ^{2} 30+\cos ^{2} 30\right)+\left(\sin ^{2} 36+\cos ^{2} 36\right)+\left(\sin ^{2} 42+\cos ^{2} 42\right)+\sin^{2}90$

=1+1+1+1+1+1+1+1

=8

(ii) $\tan 9^{\circ} \tan 27^{\circ} \cdot \tan 45^{\circ} \cdot \tan 63^{\circ} \cdot \tan 81^{\circ}=1$

Sol :

L.H.S

=tan9°tan27°tan45°tan63°tan81°

=(tan9°tan81°)(tan27°tan63°)tan45°

=(tan9°cot9°)(tan27°cot27°)tan45°

=1×1×1

=1

(iii) $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$

Sol :

Sol :

L.H.S

$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\left(\cot\left(\frac{3 \pi}{2}-x\right)+\cos (2 \pi+x)\right]\right.$

$=\sin x \cdot \cos x[\tan x+\cot x]$

$=\sin x \cos x\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]$

$=\sin x \cos x \left[\frac{\sin ^{2} x+\cos ^{2} x}{\cos x \sin x}\right]$

=1

(iv) $\frac{\cos (\pi+\theta) \cdot \cos (-\theta)}{\sin (\pi-\theta) \cdot \cos \left(\frac{\pi}{2}+\theta\right)}=\cot ^{2} \theta$

Sol :

[$\cos (-\theta)=\cos \theta$]

L.H.S

=$\frac{\cos (\pi+\theta) \cos (-\theta)}{\sin (\pi-\theta) \cos \left(\frac{\pi}{2}+\theta\right)}$

$=\frac{(-\cos \theta) \times \cos \theta}{\sin \theta \times(-\sin \theta)}$

$=\frac{\cos ^{2} \theta}{\operatorname{sin}^{2} \theta}$

$=\cot^{2} \theta$

Question 17

निम्नलिखित का मान चिह्न बतलाएँ

Find the sign of the following :

(i) tan2500°

Sol :

$=\tan (27 \times 90+70)$

=-cot70 (negative)

(ii) cos(-2000°)

Sol :

=cos2000

$=\cos \left(22 \times 90+20\right)$

=-cos20(negative)

(iii) cosec(-3020°)

Sol :

Question 18

यदि n कोई पूर्णाक हो तो सिद्ध करें कि

[If n∈Z prove that]

(i) $\sin \{2 n+1) \pi+\theta\}=(-1)^{2 n+1} \cdot \sin \theta$

Sol :

L.H.S

=sin{(2n+1)𝜋+θ}

[$\sin (2 n \pi+\theta)=\sin \theta$]

=sin{2n𝜋+(𝜋+θ)}

=sin(𝜋+θ)

=sinθ

R.H.S

=$(-1)^{2 n+1} \sin \theta$

$=(-1)^{2 n} \times(-1)^{1} \sin \theta$

$=1 \times(-1) \times \sin \theta$

=-sinθ

(ii) $\left.\tan (n \pi+\theta)=\cot (2 n+1) \frac{\pi}{2}-\theta\right\}$

Sol :

L.H.S

[$\tan (n \pi+\theta)=\tan \theta$]

=$\tan (n \pi+\theta)$

=tanθ

R.H.S
[$\cot(n \pi+\theta)=\cot \theta$]
=$\cot\left\{(2 n+1) \frac{\pi}{2}-\theta\right\}$

$=\cot A\left[n \pi+\left(\frac{\pi}{2}-\theta\right)\right]$

$=\cos \left(\frac{\pi}{2}-\theta\right)$

=tanθ