KC Sinha Mathematics Solution Class 10 Chapter 8 त्रिकोणमितीय अनुपात एवम सर्वसमिकाए ( Trigonometry Ratios and Identities ) Exercise 8.3 (Q11-Q20)

Exercise 8.3

Question 11

किसी $\triangle \mathrm{ABC}$ में सिद्ध कीजिए कि-

(a) $\sin \frac{\mathrm{B}+\mathrm{C}}{2}=\cos \frac{\mathrm{A}}{2}$

Sol :

त्रिभुज के तीनों कोणो का मान $180^{\circ}$ होता है

$A+B+C=180^{\circ}$

$B+C=180^{\circ}-A$

LHS

$\sin \frac{B+C}{2}=\cos \frac{A}{2}$

$\sin \left(\frac{180^{\circ}-\mathrm{A}}{2}\right)=\cos \frac{\mathrm{A}}{2}$

$\sin \left(\frac{180^{\circ}}{2}-\frac{A}{2}\right)=\cos \frac{A}{2}$

$\sin \left(90^{\circ}-\frac{A}{2}\right)=\cos \frac{A}{2}$

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

(b) $\tan \frac{\mathrm{B}+\mathrm{C}}{2}=\cot \frac{\mathrm{A}}{2}$

Sol :

त्रिभुज के तीनों कोणो का मान $180^{\circ}$ होता है

$\begin{aligned}&A+B+C=180^{\circ} \\&B+C=180^{\circ}-A\end{aligned}$

LHS

$\tan \frac{B+C}{2}=\cot \frac{A}{2}$

$\tan \left(\frac{180^{\circ}-A}{2}\right)=\cot \frac{A}{2}$

$\tan \left(\frac{180^{\circ}}{2}-\frac{A}{2}\right)=\cot \frac{A}{2}$

$\tan \left(90^{\circ}-\frac{A}{2}\right)=\cot \frac{A}{2}$

$\cot \frac{A}{2}=\cot \frac{A}{2}$

(c) $\cos \frac{\mathrm{A}+\mathrm{B}}{2}=\sin \frac{\mathrm{C}}{2}$

Sol :

त्रिभुज के तीनों कोणो का मान $180^{\circ}$ होता है

$\begin{aligned}&A+B+C=180^{\circ} \\&A+B=180^{\circ}-A\end{aligned}$

L.H.S

$\begin{aligned}&\cos \frac{A+B}{2}=\sin \frac{A}{2} \\&\cos \left(\frac{180^{\circ}-C}{2}\right)=\sin \frac{C}{2}\end{aligned}$

$\cos \frac{A+B}{2}=\sin \frac{A}{2}$

$\cos \left(\frac{180^{\circ}-C}{2}\right)=\sin \frac{C}{2}$

$\cos \left(\frac{180^{\circ}}{2}-\frac{C}{2}\right)=\sin \frac{C}{2}$

$\cos \left(90^{\circ}-\frac{\mathrm{C}}{2}\right)=\sin \frac{\mathrm{C}}{2}$

$\sin \frac{\mathrm{C}}{2}=\sin \frac{\mathrm{C}}{2}$

Question 12

(i) यदि $\sin 3 \mathrm{~A}=\cos \left(\mathrm{A}-26^{\circ}\right)$, जहाँ $3 \mathrm{~A}$ एक न्यूनकोण है तब $\mathrm{A}$ का मान ज्ञात कीजिए ।

Sol :

$\cos \left(90^{\circ}-3 A\right)=\cos \left(A-26^{\circ}\right)$

$90^{\circ}-3 A=A-26^{\circ}$

$90^{\circ}+26^{\circ}=A+3 A$

$116^{\circ}=4 A$

$A=\frac{116}{4}$

$A=29^{\circ}$

(ii) यदि $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta$. जहाँ $\left(2 \theta+54^{\circ}\right)$ एक न्यूनकोण है तब $\theta$ का मान ज्ञात कीजिए ।

Sol :

$\cos \left(2 \theta+54^{\circ}\right)=\cos \left(90^{\circ}-\theta\right)$

$2 \theta+54^{\circ}=90^{\circ}-\theta$

$2 \theta+\theta=90^{\circ}-54^{\circ}$

$3 \theta=36^{\circ}$

$\theta=\frac{36^{\circ}}{3}$

$\theta=12^{\circ}$

(iii) यदि $\tan 3 \theta=\cot \left(\theta+18^{\circ}\right)$, जहाँ $3 \theta$ और $\theta+18^{\circ}$ न्यूनकोण हैं, तो $\theta$ का मान ज्ञात कीजिए ।

Sol :

$\cot =\left(90^{\circ}-3 \theta\right)=\cot \left(\theta+18^{\circ}\right)$

$90^{\circ}-3 \theta=\theta+18^{\circ}$

$-3 \theta-\theta=18^{\circ}-90^{\circ}$

$-4 \theta=-72^{\circ}$

$\theta=\frac{72}{4}$

$\theta=18^{\circ}$

(iv) यदि $\sec 5 \theta=\operatorname{cosec}\left(\theta-36^{\circ}\right)$, जहाँ 5θ एक न्यूनकोण है, तो θ का मान निकालिए ।

Sol :

$\operatorname{cosec}\left(90^{\circ}-5 \theta\right)=\operatorname{cosec}\left(\theta-36^{\circ}\right)$

$90^{\circ}-5 \theta=\theta-36^{\circ}$

$90^{\circ}+36^{\circ}=\theta+5 \theta$

$126^{\circ}=6 \theta$

$6 \theta=126^{\circ}$

$\theta=\frac{126}{6}$

$\theta=21^{\circ}$

सिद्ध कीजिए कि :

Question 13

$\sin 70^{\circ} \cdot \sec 20^{\circ}=1$

Sol :

L.H.S

$\sin 70^{\circ} . \operatorname{Sec} 20^{\circ}$

$\sin 70^{\circ} \cdot \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right)$

$\sin 70^{\circ} \cdot \operatorname{cosec} 70^{\circ}=1$ R.H.S

Question 14

$\sin \left(90^{\circ}-\theta\right) \tan \theta=\sin \theta$

Sol :

L.H.S

$\sin \left(90^{\circ}-\theta\right) \cdot \tan \theta$

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

$=\sin \theta$ R.H.S

Question 15

$\tan 63^{\circ} \cdot \tan 27^{\circ}=1$

Sol :

L.H.S

$\tan 63^{\circ} . \tan 27$

$\tan 63^{\circ} \cdot \cot \left(90^{\circ}-27^{\circ}\right)$

$\tan 63^{\circ} \cdot \cot 63^{\circ}$

=1 R.H.S

Question 16

$\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1=-\sin ^{2} \theta$

Sol :

L.H.S

$

\begin{aligned}

&\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1 \\

&\frac{\cos \theta \cdot \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1 \\

&\cos ^{2}-1 \\

&-\sin ^{2} \theta \text { R.H.S }

\end{aligned}

$

Question 17

$\sin 55^{\circ} \cdot \cos 48^{\circ}=\cos 35^{\circ} \cdot \sin 42^{\circ}$

Sol :

L.H.S

$\sin 55^{\circ} \cdot \cos 48^{\circ}$

$\cos \left(90^{\circ}-55^{\circ}\right) \cdot \sin \left(90^{\circ}-48^{\circ}\right)$

$\cos 35^{\circ} . \operatorname{Sin} 42^{\circ} \quad$ R.H.S

Question 18

$\sin 25^{\circ} \cdot \sin 65^{\circ}=\cos 25^{\circ} \cdot \cos 65^{\circ}$

Sol :

L.H.S

$\sin 25^{\circ} . \operatorname{Sin} 65^{\circ}$

$\cos \left(90^{\circ}-25^{\circ}\right) \cdot \cos \left(90^{\circ}-65^{\circ}\right)$

$\cos 25^{\circ} \cdot \cos 65^{\circ} \quad$ R.H.S

Question 19

$\sin 54^{\circ}+\cos 67^{\circ}=\sin 23^{\circ}+\cos 36^{\circ}$

Sol :

L.H.S

$\sin 54^{\circ} \cdot \cos 67^{\circ}$

$\cos \left(90^{\circ}-54^{\circ}\right) \cdot \sin \left(90^{\circ}-67^{\circ}\right)$

$\sin 23^{\circ} \cdot \cos 36^{\circ} \quad$ R.H.S

Question 20

$\cos 27^{\circ}+\sin 51^{\circ}=\sin 63^{\circ}+\cos 39^{\circ}$

Sol :

L.H.S

$\cos 27^{\circ} \cdot \sin 51^{\circ}$

$\sin \left(90^{\circ}-27^{\circ}\right) \cdot \cos \left(90^{\circ}-51^{\circ}\right)$

$\sin 63^{\circ} \cdot \cos 39^{\circ} \quad$ R.H.S