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]
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)]
[(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
=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$
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
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
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
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}$
$=\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}$
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$
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$
(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{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(\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$
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{\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}}$
$=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 \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
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+$\frac{\sin^2 A+\cos ^{2} A}{\cos A \sin A}$-cosecAsecA
=2+secAcosecA-cosecAsecA
=2
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
$B=\sqrt{5^{2}-3^{2}}$
$=\sqrt{25-9}=\sqrt{16}$
=4
$=\sqrt{1+4}=\sqrt{5}$
$\sin \alpha=\frac{3}{5}$
$\tan \alpha=\frac{P}{B}=-\frac{3}{4}$
$\tan \beta=\frac{1}{2}$
$\sec \beta=\frac{H}{B}=-\frac{\sqrt{5}}{2}$
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
$\sin^{2} \theta=\left(x+\frac{P}{x}\right)^2$
$=x^{2}+\left(\frac{P}{x}\right)^{2}+2 \cdot x \cdot \frac{p}{x}$
$=\left(x-\frac{p}{x}\right)^{2}+2 \cdot x \cdot \frac{p}{x}+2 p$
$=\left(x-\frac{p}{2}\right)^{2}+4 p$
∴$\left(x-\frac{p}{x}\right)^{2} \geqslant 0$
∴$P \leq \frac{1}{4}$
Question 9
$=2+\sin ^{2} \theta \cdot \tan ^{2} \theta$
$\sin ^{2} \theta \tan ^{2} \theta$≥0
$\cos ^{2} \theta+\sec ^{2} \theta \geqslant 2$
Question 10
$\because \sin ^{2} \theta \leq 1$
$\frac{(x+y)^{2}}{4x y} \leq 1$
$(x+y)^{2} \leq 4 x y$
$(x+y)^{2}-4 x y \leq 0$
$x^{2}+y^{2}+2 x y-4 x y \leq 0$
$x^{2}+y^{2}-2 x y \leq 0$
$(x-y)^{2} \leq 0$
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
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
$=\sqrt{25-9}=\sqrt{16}$
=4
Diagram
$H=\sqrt{1^{2}+2^{2}}$
$\sin \alpha=\frac{3}{5}$
$\tan \alpha=\frac{P}{B}=-\frac{3}{4}$
$\tan \beta=\frac{1}{2}$
$\sec \beta=\frac{H}{B}=-\frac{\sqrt{5}}{2}$
L.H.S
$8 \tan \alpha-\sqrt{5} \sec \beta$
$=8\left(-\frac{3}{4}\right)-\sqrt{5} \times\left(-\frac{\sqrt{5}}{2}\right)$
$=-6+\frac{5}{2}$
$=\frac{-12+5}{2}=\frac{-7}{2}$
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
$=x^{2}+\left(\frac{P}{x}\right)^{2}+2 \cdot x \cdot \frac{p}{x}$
$=\left(x-\frac{p}{x}\right)^{2}+2 \cdot x \cdot \frac{p}{x}+2 p$
$=\left(x-\frac{p}{2}\right)^{2}+4 p$
∴$\left(x-\frac{p}{x}\right)^{2} \geqslant 0$
∴$P \leq \frac{1}{4}$
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
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$
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
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}$
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
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
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 :
$=\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\cos^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}$
=1+1
=2
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
=$\sin ^{2} 6^{\circ}+\sin ^{2} 12^{\circ}+\sin ^{2} 18^{\circ}+\ldots+\sin ^{2} 84^{\circ}+\sin ^{2} 90^{\circ}$
$=\left(\sin ^{2} 6+\cos ^{2} 6\right)+\left(\sin ^{2} 12+\cos ^{2} 12\right)+\left(\sin ^{2} 18+\cos ^{2} 18\right)+\left(\sin ^{2} 24+\cos^{2} 24\right)+\left(\sin ^{2} 30+\cos ^{2} 30\right)+\left(\sin ^{2} 36+\cos ^{2} 36\right)+\left(\sin ^{2} 42+\cos ^{2} 42\right)+\sin^{2}90$
=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
=$\frac{\cos (\pi+\theta) \cos (-\theta)}{\sin (\pi-\theta) \cos \left(\frac{\pi}{2}+\theta\right)}$
$=\frac{(-\cos \theta) \times \cos \theta}{\sin \theta \times(-\sin \theta)}$
$=\frac{\cos ^{2} \theta}{\operatorname{sin}^{2} \theta}$
$=\cot^{2} \theta$
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 :
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
[$\cot(n \pi+\theta)=\cot \theta$]
=$\cot\left\{(2 n+1) \frac{\pi}{2}-\theta\right\}$
$=\cot A\left[n \pi+\left(\frac{\pi}{2}-\theta\right)\right]$
$=\cos \left(\frac{\pi}{2}-\theta\right)$
=tanθ
Hii
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