KC Sinha: Exercise 8.4 – Mathematics Solution Class 10 Chapter 8 त्रिकोणमितीय अनुपात एवम सर्वसमिकाए

Type-I : त्रिकोणमितीय एवं बीजीय सुत्रों के सीधे प्रयोग पर आधारित प्रश्न :

Question 1

रिक्त स्थानों की पूर्ति करें :

(i) $\sin ^{2} \theta+\cos ^{2} \theta=\ldots \ldots \ldots .$

Sol :

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

(ii) $1+\tan ^{2} \theta=$….

Sol :

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

(iii) $\sin \theta \cdot \cot \theta$ का व्युत्क्रम =……..

Sol :

$\sin \theta \times \frac{\cos \theta}{\sin \theta}=\sec \theta$

(iv) $1-\ldots \ldots . . .=\cos ^{2} \theta$

Sol :

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

(v) $\tan \mathrm{A}=\frac{\ldots \ldots \ldots}{\cos \mathrm{A}}$

Sol :

$\tan A=\frac{\sin A}{\cos A}$

(vi) ….$=\frac{\cos A}{\sin A}$

Sol :

$\cot A=\frac{\cos }{\sin A}$

(vii) cos θ व्युत्क्रम है…….का ।

Sol :

$\cos \theta=\sec \theta$

(viii) sin θ का व्युत्क्रम है…….का ।

Sol :

$\sin \theta=\operatorname{cosec} \theta$

(ix) sin θ का मान cos θ=q, तो p और q में क्या सम्बन्ध है?

Sol :

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

(x) cos θ का मान sin θ के पदों में …. है।

Sol :

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

Question 2

यदि sin θ=p और cos θ=q, तो p और q में क्या सम्बन्ध है।

Sol :

$\sin ^{2} \theta+\cos ^{2} \theta=p^{2}+q^{2}$

$1=p^{2}+q^{2}$

$p^{2}+q^{2}=1$

Question 3

यदि cos A=x, तो sin A को x के पदों में व्यक्त करें।

Sol :

$\sin A=\sqrt{1-\cos ^{2} A}=\sqrt{1-x^{2}}$

Question 4

यदि $x \cos \theta=1$ और $y \sin \theta=1$, तो $\tan \theta$ का मान ज्ञात करें ।

Sol :

$\mathrm{Y} \sin \theta=1, x \cos \theta=1$

भाग देने पर

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

$\frac{x}{y} \tan \theta=1$

$\tan \theta=\frac{x}{y}$

Question 5

यदि $\cos 40^{\circ}=p$ तब $\sin 40^{\circ}$ का मान p के पदों में लिखें ।

Sol :

$\sin 40^{\circ}=\sqrt{1-\cos ^{2} 40^{\circ}}$

$=\sqrt{1-p^{2}}$

Question 6

यदि $\sin 77^{\circ}=x$, तो $\cos 77^{\circ}$ का मान x के पदों में लिखें ।

Sol :

$\cos 77^{\circ}=\sqrt{1-\sin ^{2} 77^{\circ}}$

$=\sqrt{1-x^{2}}$

Question 7

यदि $\cos 55^{\circ}=x^{2}$, तब $\sin 55^{\circ}$ का मान x के पदों में लिखें

Sol :

$\sin 55^{\circ}=\sqrt{1-\cos ^{2} 55^{\circ}}$

$=\sqrt{1-\left(x^{2}\right)^{2}}$

$=\sqrt{1-x^{4}}$

Question 8

यदि $\sin 50^{\circ}=a$. तब $\cos 50^{\circ}$ का मान a के पदों में लिखें ।

Sol :

$\cos 50^{\circ}=\sqrt{1-\sin ^{2} 50^{\circ}}$

$=\sqrt{1-a^{2}}$

Question 9

यदि xcos A=1 और tan A=y, तब $x^{2}-y^{2}$ का मान क्या है ।

Sol :

$X=\frac{1}{\cos A}=x=\sec A$

$Y=\tan A$

$x^{2}-y^{2}=\sec ^{2} \mathrm{~A}-\tan ^{2} \mathrm{~A}$

=1

निम्नलिखित सर्वसमिकाओं को सिद्ध करें ।

Question 10

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

Sol :

LHS

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

$1^{2}-\sin ^{2} \theta$

$1-\sin ^{2} \theta$

$=\cos ^{2} \theta$

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

Sol :

LHS

$(1+\cos \theta) \times(1-\cos \theta)$

$1^{2}-\cos ^{2} \theta$

$1-\cos ^{2} \theta$

$=\sin ^{2} \theta$

Question 11

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

Sol :

LHS

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

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

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

Question 12

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

Sol :

LHS

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

$=\sec \theta-\tan \theta=$ R.H.S

Question 13

$\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\sin \theta \cos \theta=1$

Sol :

$\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\sin \theta \times \cos \theta$

$a^{3}+b^{3}=(\mathrm{a}+\mathrm{b})\left(a^{2}-a b+b^{2}\right)$

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

$\sin ^{2} \theta-\sin \theta \times \cos \theta+\cos ^{2} \theta+\sin \theta \times \cos \theta$

$\sin ^{2} \theta+\cos ^{2} \theta-\sin \theta \times \cos \theta+\sin \theta \times \cos \theta$

$1-\sin \theta \times \cos \theta+\sin \theta \times \cos \theta$

=1 RHS

Type II : LCM लेकर सर्वसमिकाओं को सिद्ध करने पर आधारित प्रश्न :

Question 14

निम्नलिखित सर्वसमिकाओं को सिद्ध करें :

(i) $\sin \theta \cdot \cot \theta=\cos \theta$

Sol :

LHS

$\sin \theta \times \cot \theta$

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

RHS

(ii) $\sin ^{2} \theta\left(1+\cot ^{2} \theta\right)=1$

Sol :

LHS

$\sin ^{2} \theta\left(1+\cot ^{2} \theta\right)$

$\sin ^{2} \theta \times \operatorname{cosec}^{2} \theta=(\sin \theta \times \operatorname{cosec} \theta)^{2}$

$(1)^{2}=1$=RHS

(iii) $\cos ^{2} \mathrm{~A}\left(\tan ^{2} \mathrm{~A}+1\right)=1$

Sol :

LHS

$\cos ^{2} A\left(\tan ^{2} A+1\right)$

$\cos ^{2} A \times \sec ^{2} A$

$(\cos A \times \sec A)^{2}$

$(1)^{2}=1$ RHS

(iv) $\tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta$

Sol :

LHS

$\tan ^{4} \theta+\tan ^{2} \theta$

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

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

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

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

$\sec ^{4} \theta-\sec ^{4} \theta$

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

Sol :

LHS

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

$\frac{\sec ^{2} \theta \times \sin ^{2} \theta}{\tan \theta}$

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

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

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

Sol :

LHS

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

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

RHS

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

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

LHS=RHS

(vii) $\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}=3-\tan ^{2} \theta$

Sol :

LHS

$\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}$

$\frac{3}{\cos ^{2} \theta}-\frac{4 \sin ^{2} \theta}{\cos ^{2} \theta}$

$3 \sec ^{2} \theta-4 \tan ^{2} \theta$

$3 \tan ^{2} \theta-4 \tan ^{2} \theta$

$3-\tan ^{2} \theta $ R.H.S

(viii) $\left(1+\tan ^{2} \theta\right) \cos \theta \cdot \sin \theta=\tan \theta$

Sol :

LHS

$\left(1+\tan ^{2} \theta\right) \cos \theta \cdot \sin \theta$

$\sec ^{2} \theta \times \cos \theta \times \sin \theta$

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

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

R.H.S

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

Sol :

LHS

$\sin ^{2} \theta-\cos ^{2} \phi$

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

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

$-\cos ^{2} \theta+\sin ^{2} \phi$

$\sin ^{2} \phi-\cos ^{2} \theta $ RHS

(x) $\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta$

Sol :

LHS

$\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}$

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

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

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

RHS

Question 15

निम्नलिखित सर्वसमिकाओं को सिद्ध करें :

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

Sol :

LHS

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

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

$=\sin ^{2} \theta \times \operatorname{cosec}^{2} \theta$

=1

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

Sol :

LHS

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

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

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

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

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

RHS

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

Sol :

LHS

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

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

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

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

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

Sol :

LHS

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

$\sin ^{2} \theta+\cos ^{2} \theta-2 \times \sin \theta \times \cos \theta$

$\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \times \cos \theta$

$1-2 \sin \theta \times \cos \theta$

R.H.S 

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

Sol :

LHS

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

$\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \times \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \times \cos \theta$

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

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

$2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$

2×1=2 RHS

(vi) $(a \sin \theta+b \cos \theta)^{2}+(a \cos \theta-b \sin \theta)^{2}=a^{2}+b^{2}$

Sol :

LHS

$(a \sin \theta+b \cos \theta)^{2}+(a \cos \theta-b \sin \theta)^{2}$

$(a \sin \theta)^{2}+(b \cos \theta)^{2}+2 \times a \sin \theta \times b \cos \theta+(a \cos \theta)^{2}+(b \sin \theta)^{2}-2 a \cos \theta \times b \sin \theta$

$a^{2} \times \sin ^{2} \theta+b^{2} \times \cos ^{2} \theta+2 a \sin \theta \times b \cos \theta+a^{2} \times \cos ^{2} \theta+b^{2} \times \sin ^{2} \theta-2 a \cos \theta \times b \sin \theta$

$a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta$

$\left(a^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta\right)+\left(b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\right)$

$a^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+b^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$a^{2} \times 1+b^{2} \times 1$

$a^{2}+b^{2}$

RHS

(vii) $\cos ^{4} A+\sin ^{4} A+2 \sin ^{2} A \cdot \cos ^{2} A=1$

Sol :

LHS

$\cos ^{4} A+\sin ^{4} A+2 \sin ^{2} A \times \cos ^{2} A$

$\left(\cos ^{2} A\right)^{2}-\left(\sin ^{2} A\right)^{2}+2 \sin ^{2} A \times \cos ^{2} A$

$\left(\cos ^{2} A+\sin ^{2} A\right) 2 \sin ^{2} A \times \cos ^{2} A$

$=(1)^{2}$=1 RHS

(viii) $\sin ^{4} A-\cos ^{4} A=2 \sin ^{2} A-1=1-2 \cos ^{2} A=\sin ^{2} A-\cos ^{2} A$

Sol :

LHS

$\sin ^{4} A-\cos ^{4} A$

$\left(\sin ^{2} A\right)^{2}-\left(\cos ^{2} A\right)^{2}$

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

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

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

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

$=2 \sin ^{2} A-1$

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

$=2-2 \cos ^{2} A-1=1-2 \cos ^{2} A$

=RHS

(ix) $\cos ^{4} \theta-\sin ^{4} \theta=\cos ^{2} \theta-\sin ^{2} \theta=2 \cos ^{2} \theta-1$

Sol :

LHS

$\cos ^{4} \theta-\sin ^{4} \theta=\cos ^{2} \theta-\sin ^{2} \theta=2 \cos ^{2} \theta-1$

$\cos ^{4} \theta-\sin ^{4} \theta=\left(\cos ^{2} \theta\right)^{2}-\left(\sin ^{2} \theta\right)^{2}$

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

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

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

$\cos ^{2} \theta-1+\cos ^{2} \theta$

$2 \cos ^{2} \theta-1$

RHS

(x) $2 \cos ^{2} \theta-\cos ^{4} \theta+\sin ^{4} \theta=1$

Sol :

LHS

$2 \cos ^{2} \theta-\cos ^{4} \theta+\sin ^{4} \theta$

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

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

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

$2 \cos ^{2} \theta-\cos ^{2} \theta+1-\cos ^{2} \theta$

$2 \cos ^{2} \theta-2 \cos ^{2} \theta+1$

0+1=1

(xi) $1-2 \cos ^{2} \theta+\cos ^{4} \theta=\sin ^{4} \theta$

Sol :

LHS

$1-2 \cos ^{2} \theta+\cos ^{4} \theta$

$1^{2}-2 \times 1 \times \cos ^{2} \theta+\left(\cos ^{2}\right)^{2}$

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

$\left(\sin ^{2} \theta\right)^{4}=\sin ^{4} \theta$

RHS

(xii) $1-2 \sin ^{2} \theta+\sin ^{4} \theta=\cos ^{4} \theta$

Sol :

LHS

$1-2 \sin ^{2} \theta+\sin ^{4} \theta$

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

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

$\left(\cos ^{2} \theta\right)^{2}=\cos ^{4} \theta$

RHS

Question 16

निम्नलिखित सर्वसमिकाओं को सिद्ध करें :

(i) $\sec ^{2} \theta+\operatorname{cosec}^{2} \theta=\sec ^{2} \theta \cdot \operatorname{cosec}^{2} \theta$

Sol :

LHS

$\sec ^{2} \theta+\operatorname{cosec}^{2} \theta$

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

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

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

$=\frac{1}{\frac{1}{\sec ^{2} \theta} \times \frac{1}{\operatorname{cosec}^{2} \theta}}$

$\sec ^{2} \theta \times \operatorname{cosec}^{2} \theta$

RHS

(ii) $\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta=\operatorname{cosec} \theta$

Sol :

LHS

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

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

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

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

cosec θ

RHS

(iii) $\cot \theta+\tan \theta=\operatorname{cosec} \theta \cdot \sec \theta$

Sol :

LHS

$\cot \theta+\tan \theta$

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

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

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

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

$\operatorname{cosec} \theta \times \sec \theta$

RHS

Question 17

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

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

RHS

Question 18

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

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

Question 19

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

Sol :

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

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

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

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

$\frac{1+\cos \theta}{1-\cos \theta}$ R.H.S.

Question 20

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

Sol :

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

परिमेयकरण करने पर

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

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

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

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

Question 21

$\left(\sin ^{8} \theta-\cos ^{8} \theta\right)=\left(\sin ^{2} \theta-\cos ^{2} \theta\right)\left(1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta\right)$

Sol :

LHS

$\left(\sin ^{8} \theta-\cos ^{8} \theta\right)$

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

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

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

$\left(\sin ^{2} \theta-\cos ^{2} \theta\right) \times 1\left[(1)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]$

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

RHS

Question 22

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=0$

Sol :

LHS

$=2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1$

$=\left[\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}\right]-3\left[\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}\right]+1$

$=2\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta\right)^{2}-\sin ^{2} \theta \times \cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2}\right]-3\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]+1$

$=2 \times 1\left[\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}-\sin ^{2} \theta \times \cos ^{2} \theta\right]-3 \times\left[(1)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]+1$

$=2\left[\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]-3\left[(1)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]+1$

$=2\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta-\sin ^{2} \theta \times \cos ^{2} \theta\right]-3\left[1-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]+1$

$=2\left[(1)^{2}-2 \sin ^{2} \theta \times \cos ^{2} \theta-\sin ^{2} \theta \times \cos ^{2} \theta\right]-3\left[1-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]+1$

$=2\left[1-3 \sin ^{2} \theta \times \cos ^{2} \theta\right]-3\left[1-2 \sin ^{2} \theta \times \cos ^{2} \theta\right]+1$

$=2-6 \sin ^{2} \theta \times \cos ^{2} \theta-3+6 \sin ^{2} \theta \times \cos ^{2} \theta+1$

=2-3+1=3-3=0

Type-III : त्रिकोणमितीय व्यंजकों के वर्गमूल से सम्बद्ध सर्वसमिकाओं को सिद्ध करने पर आधारित प्रश्न :

निम्नलिखित सर्वसमिकाओं को सिद्ध करें :

Question 23

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

Sol :

LHS

$\frac{\cos \mathrm{A}}{1-\tan \mathrm{A}}+\frac{\sin \mathrm{A}}{1-\cot \mathrm{A}}$

$\frac{\cos \mathrm{A}}{1-\frac{\sin \mathrm{A}}{\cos \mathrm{A}}}+\frac{\sin \mathrm{A}}{1-\frac{\cos \mathrm{A}}{\sin \mathrm{A}}}$

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

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

$\frac{\cos ^{2} \mathrm{~A}}{\cos \mathrm{A}-\sin \mathrm{A}}-\frac{\sin ^{2} \mathrm{~A}}{\sin \mathrm{A}-\cos \mathrm{A}}$

$\frac{\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}}{\cos \mathrm{A}-\sin \mathrm{A}}$

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

cos A+sin A 

RHS

Question 24

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

$\frac{2}{\sin \theta}$ 

RHS

Question 25

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

Sol :

LHS

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

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

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

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

Proved

Question 26

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

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

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

$\frac{2}{\cos \theta}=2 \sec \theta$

RHS

Question 27

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

Sol :

LHS

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

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

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

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

$\frac{2}{\cos \theta}$ proved

Question 28

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

Sol :

LHS

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

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

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

$\frac{2}{\sin ^{2} \theta}$ proved

Question 29

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

Sol :

LHS

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

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

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

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

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

$\frac{2 \tan \theta}{1-\frac{1}{\cos \theta}}$

$\frac{2 \tan \theta}{\cos \theta}$ = proved

Question 30

$\cot ^{2} \theta-\cos ^{2} \theta=\cot ^{2} \theta \cdot \cos ^{2} \theta$

Sol :

LHS

$\cot ^{2} \theta-\cos ^{2} \theta$

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

$\cos ^{2} \theta\left[\frac{1}{\sin ^{2} \theta}-1\right]$

$\cos ^{2} \theta\left[\frac{1-\sin ^{2} \theta}{\sin ^{2} \theta}\right]$

$\cos ^{2} \theta\left[\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\right]$

$\cos ^{2} \theta \times \cot ^{2} \theta$ proved

Question 31

$\tan ^{2} \phi-\sin ^{2} \phi-\tan ^{2} \phi \cdot \sin ^{2} \phi=0$

Sol :

LHS

$\tan ^{2} \phi-\sin ^{2} \phi-\tan ^{2} \phi \times \sin ^{2} \phi$

$\frac{\sin ^{2} \phi}{\cos ^{2} \phi}-\sin ^{2} \phi-\frac{\sin ^{2} \phi}{\cos ^{2} \phi} \times \sin ^{2} \phi$

$\sin ^{2} \phi\left[\frac{1}{\cos ^{2} \phi}-1\right]-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

$\sin ^{2} \phi\left[\frac{1-\cos ^{2} \phi}{\cos ^{2} \phi}\right]-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

$\sin ^{2} \phi\left[\frac{\sin ^{2} \phi}{\cos ^{2} \phi}\right]-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

$\frac{\sin ^{4} \phi}{\cos ^{2} \phi}-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

=0 proved

Question 32

$\tan ^{2} \phi+\cot ^{2} \phi+2=\sec ^{2} \phi \cdot \operatorname{cosec}^{2} \phi$

LHS

$\tan ^{2} \phi+\cot ^{2} \phi+2$

$\sin ^{2} \phi-1+\operatorname{cosec}^{2} \phi-1+2$

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

$\frac{1}{\cos ^{2} \phi}+\frac{1}{\sin ^{2} \phi}-\not 2+\not 2$

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

$\sin ^{2} \phi \times \operatorname{cosec}^{2} \phi$ proved

Question 33

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

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

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

$\frac{(\operatorname{cosec} \theta+\cot \theta)-(\operatorname{cosec} \theta-\cot \theta) \times(\operatorname{cosec} \theta+\cot \theta)}{1-\operatorname{cosec} \theta+\cot \theta}$

$\frac{(\operatorname{cosec} \theta+\cot \theta) \times[1-(\operatorname{cosec} \theta-\cot \theta)]}{1-\operatorname{cosec} \theta+\cot \theta}$

$\frac{(\operatorname{cosec} \theta+\cot \theta) \times(1-\operatorname{cosec} \theta+\cot \theta)}{1-\operatorname{cosec} \theta+\cot \theta}$

$=\operatorname{cosec} \theta+\cot \theta$

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

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

Question 34

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

LHS

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

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

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

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

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

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

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

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

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

$1+\tan \theta+\cot \theta$ proved

Question 35

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

LHS

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

परिमेयकरण करने पर

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

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

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

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

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

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

निम्नलिखित सर्वसमिकाओं को सिद्ध करें :

Question 36

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

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

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

Question 37

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

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

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

Question 38

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

Sol :

LHS

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

परिमेयकरण करने पर

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

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

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

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

$\sec \theta-\tan \theta$=RHS

Question 39

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

Sol :

LHS

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

अंश का परिमेयकरण करने पर

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

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

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

$\frac{\cos \theta}{1+\sin \theta}=\text { R.H.S. }$

Question 40

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

Sol :

LHS

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

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

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

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

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

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

$2 \times \frac{1}{\cos \theta}=2 \sec \theta$=proved

Type IV : शंर्त वाले सर्वसमिकाओं पर आधारित प्रश्न :

Question 41

यदि $\sec \theta+\tan \theta=m$ और $\sec \theta-\tan \theta=n$, तो सिद्ध करें कि $\sqrt{m n}=1$

Sol :

LHS

$\sqrt{m n}=\sqrt{\sec \theta+\tan \theta(\sec \theta-\tan \theta)}$

$\sqrt{\sec ^{2} \theta-\sec \theta \times \tan \theta+\sec \theta \times \tan \theta-\tan ^{2} \theta}$

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

=1 proved

Question 42

यदि $\cos \theta+\sin \theta=1$, तब सिद्ध करें कि $\cos \theta-\sin \theta=\pm 1$

Sol :

माना $-\cos \theta-\sin \theta=\mathrm{x}$

दिया है – $\cos \theta+\sin \theta=1$

दोनों को वर्ग करने पर

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

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

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

$1+1=x^{2}+1$

$2=x^{2}+1$

$x^{2}+1=2$

$x^{2}=2-1$

$x=\sqrt{1}$

$x=\pm 1$

$\cos \theta+\sin \theta=x$

$\cos \theta+\sin \theta=\pm 1$

Question 43

यदि $\sin \theta+\sin ^{2} \theta=1$. तब सिद्ध करें कि $\cos ^{2} \theta+\cos ^{4} \theta=1$

Sol :

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

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

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

LHS

$\cos ^{2} \theta+\cos ^{4} \theta$

$\cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2}$

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

=1 proved

Question 44

यदि $\tan \theta+\sec \theta=x$. सिद्ध करें कि $\sin \theta=\frac{x^{2}-1}{x^{2}+1}$

Sol :

$\sec \theta+\tan \theta=\mathrm{x}$….(i)

हम जानते है कि $\sec ^{2} \theta-\tan ^{2} \theta=1$

$(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=1$

$(\sec \theta-\tan \theta) \cdot x=1$

$\sec \theta-\tan \theta=\frac{1}{x}$…(ii)

सामीकरण (i) तथा (ii) से

$\sec \theta+\tan \theta=x$

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

$2 \sec \theta=x+\frac{1}{x}$

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

$\sec \theta=\frac{x^{2}+1}{2 x}=\frac{\mathrm{AC}}{\mathrm{BC}}$

$\mathrm{AB}=\sqrt{(\mathrm{AC})^{2}-(\mathrm{BC})^{2}}$

$\mathrm{AB}=\sqrt{\left(x^{2}+1\right)^{2}-(2 \mathrm{x})^{2}}$

$\mathrm{AB}=\sqrt{\left(\left(x^{2}\right)^{2}+(1)^{2}+2 \cdot x^{2} \cdot 1\right)-4 x^{2}}$

$A B=\sqrt{\left(x^{2}\right)^{2}+1+2 x^{2}-4 x^{2}}$

$A B=\sqrt{\left(x^{2}\right)^{2}-2 x^{2}+1}$

$A B=\sqrt{\left(x^{2}-1\right)^{2}}$

$A B=x^{2}-1$

$\sin \theta=\frac{A B}{A C}=\frac{x^{2}-1}{x^{2}+1}$

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

Question 45

यदि $\sin \theta+\cos \theta=p$ और $\sec \theta+\operatorname{cosec} \theta=q$, तब सिद्ध करें कि $q\left(p^{2}-1\right)=2 p$

Sol :

LHS

$\mathrm{q}\left(\mathrm{p}^{2}-1\right)$

$(\sec \theta+\operatorname{cosec} \theta)\left[(\sin \theta+\cos \theta)^{2}-1\right]$

$\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\left[\sin ^{2} \theta+\cos \theta^{2}+2 \sin \theta \cdot \cos \theta-1\right]$

$\frac{\sin \theta+\cos \theta}{\cos \theta \cdot \sin \theta}[1+2 \sin \theta \cdot \cos \theta- 1]$

$\frac{\sin \theta+\cos \theta}{\cos \theta \cdot \sin \theta} \times 2 \sin \theta \cdot \cos \theta$

=2p proved

Question 46

यदि $x \cos \theta=a$ और $y=a \tan \theta$, तब सिद्ध करें कि $x^{2}-y^{2}=a^{2}$

Sol :

$x \cos \theta=a$

$x=\frac{a}{\cos \theta}$

$x=a \sec \theta, \quad y=a \tan \theta$

दोनों को वर्ग करके घटाने पर 

$x^{2}-y^{2}=(a \sec \theta)^{2}-(a \tan \theta)^{2}$

$x^{2}-y^{2}=a^{2} \sec ^{2} \theta-a^{2} \tan ^{2} \theta$

$x^{2}-y^{2}=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$

$x^{2}-y^{2}=a^{2}(1)^{2}$

$x^{2}-y^{2}=a^{2}$

Question 47 (to be fixed)

यदि $x=r \cos \alpha$. $\sin \beta, y=r \cos \alpha \cdot \cos \beta, z=r \sin \alpha$, तब सिद्ध करें कि $x^{2}+y^{2}+z^{2}=r$ ?

Sol :

LHS

$x^{2}+y^{2}+z^{2}=(r \cos \alpha \cdot \sin \beta)^{2}+(r \cos \alpha \cdot \cos \beta)^{2}+(r \sin \alpha)^{2}$

$x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha \cdot \sin ^{2} \beta+r^{2} \cos ^{2} \alpha \cdot \cos ^{2} \beta+r \sin ^{2} \alpha$

$x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+r^{2} \sin ^{2} \alpha$

$x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha(1)+r^{2} \cdot \sin ^{2} \alpha$

$x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha+r^{2} \cdot \sin ^{2} \alpha$

$x^{2}+y^{2}+z^{2}=r^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)$

$x^{2}+y^{2}+z^{2}=r^{2}(1)^{2}$

$x^{2}+y^{2}+z^{2}=r^{2}=$ R.H.S.

Question 48

यदि $\sec \theta-\tan \theta=x$, तब सिद्ध करें कि

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

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

Sol :

$\sec \theta-\tan \theta=\mathrm{x}$…(i)

हम जानते है कि $\left(\sec ^{2} \theta-\tan ^{2} \theta\right)=1$

$(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$

$(\sec \theta+\tan \theta) \cdot \mathrm{x}=1$

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

सामीकरण (i) तथा (ii) से

$\sec \theta-\tan \theta=x$

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

$2 \sec \theta=x+\frac{1}{x}$

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

$\sec \theta=\frac{x^{2}+1}{2 x}=\frac{\mathrm{AC}}{\mathrm{BC}}$

$A B=\sqrt{(A C)^{2}-(B C)^{2}}$

$A B=\sqrt{\left(x^{2}+1\right)^{2}-(2 x)^{2}}$

$A B=\sqrt{\left(x^{2}\right)^{2}+(1)^{2}+2 x^{2} \cdot 1-4 x^{2}}$

$A B=\sqrt{\left(x^{2}\right)^{2}+1+2 x^{2}-4 x^{2}}$

$A B=\sqrt{\left(x^{2}\right)^{2}+1-2 x^{2}}$

$A B=\sqrt{\left(x^{2}-1\right)^{2}}$

$A B=x^{2}-1$

$\cos \theta=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{2 x}{x^{2}+1}$

$\sin \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{x^{2}-1}{x^{2}+1}$

Question 49

यदि $a \cos \theta+b \sin \theta=c$, तब सिद्ध करें कि $a \sin \theta-b \cos \theta=\pm \sqrt{a^{2}+b^{2}-c^{2}}$

Sol :

माना $-\mathrm{a} \sin \theta-\mathrm{b} \cos \theta=\mathrm{x}$…(i)

दिया है – $\mathrm{a} \cos \theta+\mathrm{b} \sin \theta=\mathrm{c}$…(ii)

दोनों को वर्ग करके जोड़ने पर

$(a \sin \theta-b \cos \theta)^{2}+(a \cos \theta+b \sin \theta)^{2}=x^{2}+c^{2}$

$a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a \sin \theta \cdot b \cos \theta+a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+$ $2 a \cos \theta \cdot b \sin \theta=x^{2}+c^{2}$

$a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta=x^{2}+c^{2}$

$\left(a^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta\right)+\left(b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\right)=x^{2}+c^{2}$

$a^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+b^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=x^{2}+c^{2}$

$a^{2}=1+b^{2} \times 1=x^{2}+c^{2}$

$a^{2}+b^{2}=x^{2}+c^{2}$

$x^{2}+c^{2}=a^{2}+b^{2}$

$x^{2}=a^{2}+b^{2}-c^{2}$

$x=\pm \sqrt{a^{2}+b^{2}-c^{2}}$

$x=\pm \sqrt{a^{2}+b^{2}-c^{2}}$

Question 50

यदि $1+\sin ^{2} \theta=3 \sin \theta \cdot \cos \theta$, तब सिद्ध करें कि $\tan \theta=1$ या, $\frac{1}{2}$, जहाँ $\theta<90^{\circ}$

Sol :

$1+\sin ^{2} \theta-3 \sin \theta \cos \theta$

दोनों तरफ $\cos ^{2} \theta$ से भाग देने पर

$\frac{1}{\cos ^{2} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{3 \cdot \sin \theta \cos \theta}{\cos ^{2} \theta}$

$\sec ^{2} \theta+\tan ^{2} \theta=3 \tan \theta$

$1+\tan ^{2} \theta+\tan ^{2} \theta-3 \tan \theta=0$

$1+2 \tan ^{2} \theta-3 \tan \theta=0$

$2 \tan ^{2} \theta-3 \tan \theta+1=0$

$2 \tan ^{2} \theta-2 \tan \theta-\tan \theta+1=0$

$2 \tan \theta(\tan \theta-1)-1(\tan \theta-1)=0$

$\begin{array}{r|rl}(\tan \theta-1) & 2 \tan \theta & -1 \\\tan \theta=1 & 2 \tan \theta & =1 \\&\tan \theta &=\frac{1}{2}\end{array}$

$\tan \theta=1$ या $\frac{1}{2}$

Question 51

यदि $a \cos \theta-b \sin \theta=x$ और $a \sin \theta+b \cos \theta=y$, तो सिद्ध करें कि $a^{2}+b^{2}=x^{2}+y^{2}$

Sol :

$a \cos \theta-b \sin \theta=x$….(i)

$a \sin \theta+b \cos \theta=y$…..(ii)

दोनों समीकरण को वर्ग करके जोड़ने पर

$(a \cos \theta-b \sin \theta)^{2}+(a \sin \theta+b \cos \theta)=x^{2}+y^{2}$

$a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta-2 a \cos \theta \cdot b \sin \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+2 a \sin \theta \cdot b \cos \theta=x^{2}+y^{2}$

$a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta=x^{2}+y^{2}$

$\left(a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+\left(b^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta\right)=x^{2}+y^{2}$

$a^{2} \times 1+b^{2} \times 1=x^{2}+y^{2}$

$a^{2}+b^{2}=x^{2}+y^{2}$

Question 52

यदि $x=a \sec \theta+b \tan \theta$ और $y=a \tan \theta+b \sec \theta$, तो सिद्ध करें कि $x^{2}-y^{2}=a^{2}-b^{2}$

Sol :

$x=a \sec \theta+b \tan \theta$…(i)

$y=a \tan \theta+b \sec \theta$…(ii)

समीकरण (i) तथा (ii) को वर्ग करके घटाने पर

$x^{2}-y^{2}=(a \sec \theta-b \tan \theta)^{2}-(a \tan \theta+b \sec \theta)^{2}$

$x^{2}-y^{2}=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a \sec \theta \cdot b \tan \theta-\left(a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta+\right. 2 \operatorname{atan} \theta \cdot b \sec \theta)$

$x^{2}-y^{2}=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a \sec \theta \cdot b \tan \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta-2 a \tan \theta \cdot b \sec \theta$

$x^{2}-y^{2}=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta$

$x^{2}-y^{2}=a^{2} \sec ^{2} \theta-a^{2} \tan ^{2} \theta+b^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta$

$x^{2}-y^{2}=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)-b^{2}\left(\tan ^{2} \theta-\sec ^{2} \theta\right)$

$x^{2}-y^{2}=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)-b^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$

$x^{2}-y^{2}=a^{2}-b^{2}$

Question 53

यदि $\left(a^{2}-b^{2}\right) \sin \theta+2 a b \cos \theta=a^{2}+b^{2}$, तो सिद्ध करें कि $\tan \theta=\frac{a^{2}-b^{2}}{2 a b}$

Sol :

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

दोनों तरफ $\cos \theta$ से भाग देने पर

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

$a^{2}-b^{2} \tan \theta+2 a b=a^{2} \sec \theta+b^{2} \sec \theta$

$a^{2}-b^{2} \tan \theta+2 \mathrm{ab}=a^{2}+b^{2}(\sec \theta)$

$a^{2}-b^{2} \tan \theta+2 \mathrm{ab}=\left(a^{2}+b^{2}\right) \sqrt{1+\tan ^{2} \theta}$

दोनों तरफ वर्ग करने पर

$\left[\left(a^{2}-b^{2}\right) \tan \theta+2 a b\right]^{2}=\left[\left(a^{2}-b^{2}\right) \sqrt{1+\tan ^{2}} \theta\right]^{2}$

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

$\left(a^{2}-b^{2}\right) \times \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right) \times \tan \theta+4 a^{2} b^{2}=\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}+b^{2}\right)^{2} \times \tan ^{2} \theta$

$\left(a^{2}-b^{2}\right)^{2} \times \tan ^{2} \theta-\left(a^{2}-b^{2}\right)^{2} \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right) \tan \theta+4 a^{2} b^{2}-\left(a^{2}+b^{2}\right)^{2}=0$

$\left[a^{4}-2 a^{2} \cdot b^{2}+b^{4}-a^{4}-2 a^{2} \cdot b^{2}-b^{4}\right] \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right)\tan \theta+4 a^{2} \cdot b^{2}-a^{4}-b^{4} 2 a^{2} b^{2}=0$

$-4 a^{2} \cdot b^{2} \times \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right) \tan \theta-a^{2}-b^{4}+2 a^{2} \cdot b^{2}=0$

$4 a^{2} \cdot b^{2} \times \tan ^{2} \theta-4 a b\left(a^{2}-b^{2}\right) \tan \theta+a^{4}+b^{4}-2 a^{2} \cdot b^{2}=0$

$4 a^{2} \cdot b^{2} \times \tan ^{2} \theta-4 a b\left(a^{2}-b^{2}\right) \tan\theta+\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}-2 a^{2} \cdot b^{2}=0$

$(2 a b \tan \theta)^{2}-2 \cdot 2 a b \tan \theta\left(a^{2}-b^{2}\right)+\left(a^{2}-b^{2}\right)^{2}=0$

$\left[2 a b \tan \theta-\left(a^{2}-b^{2}\right)\right]^{2}=0$

$2 a b \tan \theta-\left(a^{2}-b^{2}\right)=0$

$2 a b \tan \theta=a^{2}-b^{2}$

$\tan \theta=\frac{a^{2}-b^{2}}{2 a b}$