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
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