Exercise 8.2
TYPE-I : दिए कोणों के त्रिकोणमितीय अनुपातों से सम्बद्ध व्यंजको के मान
निकालने पर आधारित प्रश्न:
Question 1
निम्नलिखित के मान ज्ञात कीजिए:
(i) sin 30°+cos 60°
Sol :
$=\frac{1}{2}+\frac{1}{2}$
$=\frac{1+1}{2}=\frac{2}{2}$
=1
(ii) sin2 45°+cos2 45°
Sol :
$=\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=\frac{1}{2}+\frac{1}{2}$
$=\frac{1+1}{2}$
$=\frac{2}{2}=1$
(iii) sin 30°+cos 60°-tan 45°
Sol :
$=\frac{1}{2}+\frac{1}{2}-1$
$=\frac{1+1-2}{2}=\frac{2-2}{2}$
$=\frac{0}{2}=0$
(iv) $\sqrt{1+\tan ^{2} 60^{\circ}}$
Sol :
$=\sqrt{1+(\sqrt{3})^{2}}=\sqrt{1+3}$
$=\sqrt{4}=2$
(v) tan 60° ×cos 30°
Sol :
$=\sqrt{3}=\frac{\sqrt{3}}{2}=\frac{3}{2}$
Question 2
यदि θ=45° , तब इनके मान ज्ञात करें।
(i) $\tan ^{2} \theta+\frac{1}{\sin ^{2} \theta}$
Sol :
$\begin{aligned} & =\tan ^{2} 45^{\circ}+\frac{1}{\sin ^{2} 45^{\circ}} \\=& 1^{2}+\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}} \\=& 1^{2}+\frac{1}{\frac{1}{2}} \end{aligned}$
=1+2=3
(ii) cos2 θ-sin2 θ
Sol :
$=\cos ^{2} 45^{\circ}-\sin ^{2} 45^{\circ}$
$=\left(\frac{1}{\sqrt{2}}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=\frac{1}{2}-\frac{1}{2}$
$=\frac{1-1}{2}$
$=\frac{0}{2}$=0
Question 3
निम्नलिखित के सांख्यिक मान परिकलित करें:
(i) $\sin 45^{\circ} \cdot \cos 45^{\circ}-\sin ^{2} 30^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}-\left(\frac{1}{2}\right)^{2}$
$=\frac{1}{2}-\frac{1}{4}$
$=\frac{2-1}{4}=\frac{1}{4}$
(ii) $\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}$
Sol :
$=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{\frac{\sqrt{3}+1}{2}}=\frac{2 \sqrt{3}}{\sqrt{3}+1}$
परिमेयकरण करने पर
$=\frac{2 \sqrt{3}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{(2 \times 3-2 \sqrt{3})}{(\sqrt{3})^{2}-(1)^{2}}=\frac{6-2 \sqrt{3}}{3-1}$
$=\frac{2(3-\sqrt{3})}{2}=3-\sqrt{3}$
(iii) $\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 30^{\circ}}$
Sol :
$=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}$
$=\frac{\sqrt{3}}{\frac{2 \sqrt{3}}{2}}$
$=\frac{2 \sqrt{3}}{2 \sqrt{3}}=1$
(iv) $\frac{4}{\sin ^{2} 60^{\circ}}+\frac{3}{\cos ^{2} 60^{\circ}}$
Sol :
$\frac{4}{\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{\left(\frac{1}{2}\right)^{2}}$
$=\frac{4}{\frac{3}{4}}+\frac{3}{\frac{1}{4}}$
$=\frac{4 \times 4}{3}+\frac{4 \times 3}{1}=\frac{16}{3}+\frac{12}{1}$
$=\frac{16+36}{3}=\frac{52}{3}$
(v) $\sin ^{2} 60^{\circ}-\cos ^{2} 60^{\circ}$
Sol :
$=\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}=\frac{3}{4}-\frac{1}{4}$
$=\frac{3-1}{4}=\frac{2}{4}=\frac{1}{2}$
(vi) $4 \sin ^{2} 60^{\circ}+3 \tan ^{2} 30^{\circ}-8 \sin 45^{\circ} \cdot \cos 45^{\circ}$
Sol :
$=4 \times\left(\frac{\sqrt{3}}{2}\right)^{2}+3 \times\left(\frac{1}{\sqrt{3}}\right)^{2}-8\left(\frac{1}{\sqrt{2}}\right) \times \frac{1}{\sqrt{2}}$
$=4 \times \frac{3}{4}+3 \times \frac{1}{3}-8 \times \frac{1}{2}$
=3+1-4=0
(vii) $2 \sin ^{2} 30^{\circ}-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}$
Sol :
$=2 \times\left(\frac{1}{2}\right)^{2}-3 \times\left(\frac{1}{\sqrt{2}}\right)^{2}+(\sqrt{3})^{2}$
$=2 \times \frac{1}{4}-3 \times \frac{1}{2}+3$
$=\frac{1}{2}-\frac{3}{2}+\frac{3}{1}$
$=\frac{1-3+6}{2}=\frac{7-3}{2}$
$=\frac{4}{2}=2$
(viii) $\sin 90^{\circ}+\cos 0^{\circ}+\sin 30^{\circ}+\cos 60^{\circ}$
Sol :
$=1+1+\frac{1}{2}+\frac{1}{2}$
$=\frac{2+2+1+1}{2}=\frac{6}{2}$
=3
(ix) $\sin 90^{\circ}+\cos 0^{\circ}+\tan 0^{\circ}+\tan 45^{\circ}$
Sol :
=1+1+0+1=3
(x) $\cos ^{2} 0^{\circ}+\tan ^{2} \frac{\pi}{4}+\sin ^{2} \frac{\pi}{4}$ जहाँ, $\pi=180^{\circ}$
Sol :
$=\cos ^{2} 0^{\circ}+\tan ^{2} \frac{180^{\circ}}{4}+\sin ^{2} \frac{180^{\circ}}{4}$
$=\cos ^{2} 0^{\circ}+\tan ^{2} 45^{\circ}+\sin ^{2} 45^{\circ}$
$=(1)^{2}+(1)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=1+1+\frac{1}{2}=\frac{2+2+1}{2}=\frac{5}{2}$
(xi) $\frac{\cos 60^{\circ}}{\sin ^{2} 45^{\circ}}-3 \cot 45^{\circ}+2 \sin 90^{\circ}$
Sol :
$\frac{\frac{1}{2}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}-3 \times 1+2 \times 1$
$\frac{\frac{1}{2}}{\frac{1}{2}}=-3+2$
=1-3+3=3+3=0
(xii) $\frac{4}{\tan ^{2} 60^{\circ}}+\frac{1}{\cos ^{2} 30^{\circ}}-\sin ^{2} 45^{\circ}$
Sol :
$\frac{4}{(\sqrt{3})^{2}}+\frac{2}{\left(\frac{\sqrt{3}}{2}\right)^{2}}-\frac{1}{2}$
$=\frac{8+8-3}{6}=\frac{16-3}{6}=\frac{13}{6}$
(xiii) $\cos 60^{\circ} \cdot \cos 30^{\circ}-\sin 60^{\circ} \cdot \sin 30^{\circ}$
Sol :
$=\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=\frac{\sqrt{3}-\sqrt{3}}{4}$
$=\frac{0}{4}=0$
(xiv) $\frac{4\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)}{\tan ^{2} 30^{\circ}+\cos ^{2} 90^{\circ}}$
Sol :
$=\frac{4 \times\left(\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}\right)}{\left(\frac{1}{\sqrt{3}}\right)^{2}+(0)^{2}}=\frac{4 \times\left(\frac{3}{4}-\frac{1}{2}\right)}{\frac{1}{3}}$
$=12\left(\frac{3-2}{4}\right)=12\left(\frac{1}{4}\right)$
=3
Question 4
(i) $\sin 30^{\circ} \cdot \cos 45^{\circ}+\cos 30^{\circ} \cdot \sin 45^{\circ}$
Sol :
$=\frac{1}{2} \times \frac{1}{\sqrt{2}}+\times \frac{\sqrt{3}}{2} \times \frac{1}{2}$
$=\frac{1}{2 \sqrt{2}}+\frac{\sqrt{3}}{2 \sqrt{2}}$
$=\frac{1+\sqrt{3}}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
(ii) $\operatorname{cosec}^{2} 30^{\circ} \cdot \tan ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}$
Sol :
$(2)^{2} \times(1)^{2}-(2)^{2}$
=4×1-4
=4-4=0
(iii) $2 \sin ^{2} 30^{\circ} \cdot \tan 60^{\circ}-3 \cos ^{2} 60^{\circ} \cdot \sec ^{2} 30^{\circ}$
Sol :
$=2 \times\left(\frac{1}{2}\right)^{2} \times \sqrt{3}-3 \times\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{\sqrt{3}}\right)^{2}$
$=2 \times \frac{1}{4} \times \sqrt{3}-3 \times \frac{1}{4} \times \frac{4}{3}$
$=\frac{\sqrt{3}}{2}-\frac{1}{1}$
$=\frac{\sqrt{3}-2}{2}$
(iv) $\tan 60^{\circ} \cdot \operatorname{cosec}^{2} 45^{\circ}+\sec ^{2} 60^{\circ} \cdot \tan 45^{\circ}$
Sol :
$\begin{aligned} & =\sqrt{3} \times(\sqrt{2})^{2}+(2)^{2} \times 1 \\ & =\sqrt{3} \times 2+4 \\=& 2 \sqrt{3}+4 \\=& 2(2+\sqrt{3}) \end{aligned}$
(v) $\tan 30^{\circ} \cdot \sec 45^{\circ}+\tan 60^{\circ} \cdot \sec 30^{\circ}$
Sol :
$=\frac{1}{\sqrt{3}} \times \sqrt{2}+\sqrt{3} \times \frac{2}{\sqrt{3}}$
$=\frac{\sqrt{2}}{\sqrt{3}}+\frac{2}{1}$
$=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}+2$
$=\frac{\sqrt{6}}{3}+2$
(vi) $\cos 30^{\circ} \cdot \cos 45^{\circ}-\sin 30^{\circ} \cdot \sin 45^{\circ}$
Sol :
$=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times \frac{1}{\sqrt{2}}$
$=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
परिमेयकरण करने पर
$\frac{\sqrt{3}-1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}(\sqrt{3}-1)}{4}$
(vii) $\frac{4}{3} \tan ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}-3 \cos ^{2} 60^{\circ}+\frac{3}{4} \tan ^{2} 60^{\circ}-2 \tan ^{2} 45^{\circ}$
Sol :
$=\frac{4}{3} \times\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-3 \times\left(\frac{1}{2}\right)^{2}+\frac{3}{4} \times(\sqrt{2})^{2}-2 \times(1)^{2}$
$=\frac{4}{3} \times \frac{1}{3}+\frac{3}{4}-\frac{3}{4}+\frac{3}{4} \times 3-2$
$=\frac{4}{9}+\frac{3}{4}-\frac{3}{4}+\frac{9}{4}-\frac{2}{1}$
$=\frac{4}{9}+\frac{9}{4}-\frac{2}{1}$
$=\frac{16+81-72}{36}=\frac{97-72}{36}=\frac{25}{36}$
(viii) $\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$
Sol :
$=\frac{(\sqrt{3})^{2}+4 \times\left(\frac{1}{\sqrt{2}}\right)^{2}+3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}+5 \times(0)^{2}}{2+2-(\sqrt{3})^{2}}$
$=\frac{3+4 \times \frac{1}{2}+3 \times \frac{4}{3}+0}{4-3}$
$=\frac{3+2+4}{1}$
=9
(ix) $\frac{5 \sin ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}-4 \tan ^{2} 30^{\circ}}{2 \sin 30^{\circ} \cdot \cos 30^{\circ}+\tan 45^{\circ}}$
Sol :
$=\frac{5 \times\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}-4 \times\left(\frac{1}{\sqrt{3}}\right)^{2}}{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}+1}$
$=\frac{5 \times \frac{1}{4}+\frac{1}{2}-4 \times \frac{1}{3}}{\frac{\sqrt{3}}{2}+\frac{1}{1}}$
$=\frac{\frac{5}{4}+\frac{1}{2}-\frac{4}{3}}{\frac{\sqrt{3}+2}{2}}$
$=\frac{\frac{15+6-16}{12}}{\frac{\sqrt{3}+2}{2^{2}}}$
$=\frac{\frac{21-16}{6}}{\sqrt{3}+2}$
$=\frac{\frac{5}{6}}{\sqrt{3}+2}$
Question 5
निम्नलिखित को सिद्ध कीजिए :
(i) $\frac{(1-\cos B)(1+\cos B)}{(1-\sin B)(1+\sin B)}=\frac{1}{3}$ जब $B=30^{\circ}$
Sol :
LHS
$=\frac{\left(1^{2}-\cos ^{2} B\right)}{\left(1^{2}-\sin ^{2} B\right)}=\frac{1-\cos ^{2} B}{1-\sin ^{2} B}$
$=\frac{\sin ^{2} B}{\cos ^{2} B}=\frac{\sin ^{2} 30^{\circ}}{\cos ^{2} 30^{\circ}}$
$\frac{\left(\frac{1}{2}\right)^{2}}{\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
LHS=RHS
(ii) $\frac{(1-\cos \alpha)(1+\cos \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}=3$ जब $\alpha=60^{\circ}$
Sol :
LHS
$=\frac{\left(1^{2}-\cos ^{2} \alpha\right)}{\left(1^{2}-\sin ^{2} \alpha\right)}=\frac{1-\cos ^{2} \alpha}{1-\sin ^{2} \alpha}$
$=\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}=\frac{\sin ^{2} 60^{\circ}}{\cos ^{2} 60^{\circ}}$
$=\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{\left(\frac{1}{2}\right)^{2}}=\frac{\frac{3}{4}}{\frac{4}{4}}=3$
LHS=RHS
(iii) $\cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B$ यदि $A=B=60^{\circ}$
Sol :
LHS
cos(60°-60°)=cos60°×cos60°+sin60°×sin60°
$\cos 0^{\circ}=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$
$1=\frac{1}{4}+\frac{3}{4}$
$1=\frac{1+3}{4}=\frac{4}{4}$
1=1 proved
(iv) $\left.4 \sin ^{4} 30^{\circ}+\cos ^{4} 60^{\circ}\right)-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)=2$
Sol :
LHS
$4\left[\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{4}\right]-3\left[\left(\frac{1}{\sqrt{2}}\right)^{2}+(1)^{2}\right]=2$
$4\left[\frac{1}{16}+\frac{1}{16}\right]-3\left[\frac{1}{2}+1\right]=2$
$4\left[\frac{1+1}{16}\right]-3\left[\frac{1-2}{2}\right]=2$
$4 \times \frac{2}{16}+\frac{3}{2}=2$
$\frac{2+6}{4}=2$
$\frac{8}{4}=2$
2=2 proved
(v) $\sin 90^{\circ}=2 \sin 45^{\circ} \cdot \cos 45^{\circ}$
Sol :
LHS
$1=2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$1=2 \times \frac{1}{\frac{2}{2}}$
1=1
Proved
(vi) $\cos 60^{\circ}=2 \cos ^{2} 30^{\circ}-1=1-2 \sin ^{2} 30^{\circ}$
Sol :
$\frac{1}{2}=2 \times\left(\frac{\sqrt{3}}{2}\right)^{2}-1=1-2\left(\frac{1}{2}\right)^{2}$
$\frac{1}{2}=2 \times \frac{3}{4}-1=1-2 \frac{1}{4}$
$\frac{1}{2}=\frac{3}{2}-1=1-\frac{1}{2}$
$\frac{1}{2}=\frac{3-2}{2}=\frac{2-1}{2}$
$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$
(vii) $\cos 90^{\circ}=1-2 \sin ^{2} 45^{\circ}=2 \cos ^{2} 45^{\circ}-1$
Sol :
$\begin{aligned} 0 &=1-2 \times\left(\frac{1}{\sqrt{2}}\right)^{2}=2\left(\frac{1}{\sqrt{2}}\right)^{2}-1 \\ 0 &=1-2 \times \frac{3}{2}=2 \times \frac{1}{2} \\ 0 &=\frac{3}{2}-1=1-\frac{1}{2} \\ 0 &=\frac{1-1}{1}=\frac{1-1}{1} \end{aligned}$
0=0=0
(viii) $\sin 30^{\circ} \cdot \cos 60^{\circ}+\cos 30^{\circ} \cdot \sin 60^{\circ}=\sin 90^{\circ}$
Sol :
LHS
$\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=1$
$\frac{1}{4} \times \frac{3}{4}=1$
$\frac{1+3}{4}=1$
$\frac{4}{4}=1$
1=1 proved
(ix) $\cos 60^{\circ} \cdot \cos 30^{\circ}-\sin 60^{\circ} \cdot \sin 30^{\circ}=\cos 90^{\circ}$
Sol :
$\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}=0$
0=0
$\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$
0=0 proved
(x) $ \cos 60^{\circ}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
Sol :
LHS
$\frac{1}{2}=\frac{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$
$\frac{1}{2}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2}=\frac{\frac{3-1}{\frac{3}{3+1}}}{\frac{3}{3}}$
$\frac{1}{2}=\frac{3-1}{3+1}=\frac{1}{2}=\frac{2}{4}$
$\frac{1}{2}=\frac{1}{2}$=RHS
(xi) $\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \cdot \tan 30^{\circ}}=\tan 30^{\circ}$
Sol :
LHS
$\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}=\frac{1}{\sqrt{3}}$
$\frac{\frac{3-1}{\sqrt{3}}}{1+1}=\frac{1}{\sqrt{3}}$
$\frac{\frac{2}{\sqrt{3}}}{2}=\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
(xii) $\frac{1-\tan 30^{\circ}}{1+\tan 30^{\circ}}=\frac{1-\sin 60^{\circ}}{\cos 60^{\circ}}$
Sol :
$\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}$
$\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}=\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}$
$\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$
परिमेयकरण करने पर
$\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=2-\sqrt{3}$
$\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}=2-\sqrt{3}$
$\frac{(\sqrt{3})^{2}-(1)^{2}-2 \times \sqrt{3} \times 1}{3-1}=2-\sqrt{3}$
$\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
$\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}$
$2 \times \frac{(2-\sqrt{3})}{{2}}=2-\sqrt{3}$
$2-\sqrt{3}=2-\sqrt{3}=$ proved
(xiii) $\frac{\sin 60^{\circ}+\cos 30^{\circ}}{\sin 30^{\circ}+\cos 60^{\circ}+1}=\cos 30^{\circ}$
Sol :
LHS
$\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{\frac{1}{2}+\frac{1}{2}+1}=\frac{\sqrt{3}}{2}$
$\frac{\frac{\sqrt{3}+\sqrt{3}}{2}}{\frac{1+1+2}{2}}=\frac{\sqrt{3}}{2}$
$\frac{\frac{2 \sqrt{3}}{2}}{\frac{4}{2}}=\frac{\sqrt{3}}{2}$
$\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
$2 \times \frac{(\sqrt{3})}{4}=\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}=$ R.H.S
(xiv) $\sin 60^{\circ}=2 \sin 30^{\circ} \cdot \cos 30^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$
Sol :
$\frac{\sqrt{3}}{2}=2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$
$=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$
Type II: किसी सर्वसमिका को सत्यापित करने पर आधारित प्रश्न:
Question 6
यदि $\mathrm{A}=60^{\circ}$ और $\mathrm{B}=30^{\circ}$, तब निम्नलिखित को सत्यापित करें :
(i) $\cos (A+B)=\cos A \cos B-\sin A \sin B$
Sol :
LHS
cos(60°+30°)=cos60°×cos30°-sin60°×sin30°
$\cos 90^{\circ}=\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}$
$0=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
0=0 RHS
(ii) $\sin (A-B)=\sin A \cos B-\cos A \sin B$
Sol :
LHS
sin(60°-30°)=sin60°×cos30°-cos60°×sin30°
$\sin 30^{\circ}=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$
$\frac{1}{2}=\frac{3}{4}-\frac{1}{4}$
$\frac{1}{2}=\frac{3-1}{4}$
$\frac{1}{2}=\frac{2}{4}$
$\frac{1}{2}=\frac{1}{2}=$ proved
(iii) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
Sol :
LHS
$\tan \left(60^{\circ}-30^{\circ}\right)=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \times \tan 30^{\circ}}$
$\tan 30^{\circ}=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}$
$\frac{1}{\sqrt{3}}=\frac{\frac{3-1}{\sqrt{3}}}{1+1}$
$\frac{1}{\sqrt{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{1}{2}}$
$\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ proved
Question 7
यदि $A=30^{\circ}$, तब निम्नलिखित को सत्यापित करें :
(i) sin2A=2sinAcosA
Sol :
LHS
$\sin 2\left(30^{\circ}\right)=2 \sin 30^{\circ} \times \cos 30^{\circ}$
$\sin 60^{\circ}=2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$ proved
(ii) $\cos 2 A=1-2 \sin ^{2} A=2 \cos ^{2} A-1$
Sol :
$\cos 2\left(30^{\circ}\right)=1-2 \sin ^{2} 30^{\circ}=2 \cos ^{2} 30^{\circ}-1$
$\cos 60^{\circ}=1-2 \times\left(\frac{1}{2}\right)^{2}=2 \times\left(\frac{\sqrt{3}}{2}\right)^{2}-1$
$\frac{1}{2}=1-2 \times \frac{1}{4}=2 \times \frac{3}{4}-1$
$\frac{1}{2}=1-\frac{1}{2}=\frac{3}{2}-1$
$\frac{1}{2}=\frac{2-1}{2}=\frac{3-2}{2}$
$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$
Question 8
यदि $\theta=30^{\circ}$, तब निम्न को सत्यापित करें :
(i) $\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta$
Sol:
$\sin 3\left(30^{\circ}\right)=3 \sin \left(30^{\circ}\right)-4 \sin ^{3}\left(30^{\circ}\right)$
$\sin 90^{\circ}=3 \sin 30^{\circ}-4 \sin ^{3} 30^{\circ}$
$1=3 \times \frac{1}{2}-4 \times\left(\frac{1}{2}\right)^{3}$
$1=\frac{3}{2}-4 \times \frac{1}{8}$
$1=\frac{3}{2}-\frac{1}{2}$
$1=\frac{3-1}{2}$
$1=\frac{2}{2}$
1=1 proved
(ii) $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$
Sol:
$\cos 3\left(30^{\circ}\right)=4 \cos ^{3} 30^{\circ}-3 \cos 30^{\circ}$
$\cos 90^{\circ}=4 \cos ^{3} 30^{\circ}-3 \cos 30^{\circ}$
$0=4 \times\left(\frac{\sqrt{3}}{2}\right)^{3}-3 \times \frac{\sqrt{3}}{2}$
$0=4 \times \frac{3 \sqrt{3}}{8}-\frac{3 \sqrt{3}}{2}$
$0=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}$
0=0 proved
TYPE III : समीकरण के हल पर आधारित प्रश्न :
Question 9
यदि sin (A+B)=1 और $\cos (A-B)=\frac{\sqrt{3}}{2}$, तब Aऔर B के मान ज्ञात करें ।
Sol :
sin (A+B)=1
sin (A+B)=sin 90°
A+B=90°...(i)
$\cos (A-B)=\frac{\sqrt{3}}{2}$
cos(A-B)=cos 30°
A-B=30°....(ii)
समीकरण (i) तथा (ii) से
$\begin{aligned}&A+B=90^{\circ} \\&A-B=30^{\circ} \\&\hline 2 A=120^{\circ}\end{aligned}$
$A=\frac{120^{\circ}}{2}$
$A=60^{\circ}$
A का मान समीकरण (i) में रखने पर
$\begin{aligned}&A+B=90^{\circ} \\&60^{\circ}+B=90^{\circ} \\&B=90^{\circ}-60^{\circ} \\&B=30^{\circ} \\&A=60^{\circ}, B=30^{\circ}\end{aligned}$
Question 10
यदि $\sin (A+B)=1$ और $\cos (A-B)=1$, तब $A$ और $B$ के मान निकालें ।
Sol :
sin(A+B)=1
sin(A+B)=sin 90°
A+B=90°......(i)
cos(A-B)=1
cos(A-B)=cos 0°
A-B=0°.......(ii)
समीकरण (i) तथा (ii) से ,
$\begin{aligned}&A+B=90^{\circ} \\&A-B=0^{\circ} \\ \hline&2 A=90^{\circ}\end{aligned}$
$A=\frac{90^{\circ}}{2}$
$A=45^{\circ}$
A का मान समीकरण (i) में रखने पर
$\begin{aligned}&A+B=90^{\circ} \\&45^{\circ}+B=90^{\circ}\\&B=90^{\circ}-45^{\circ}\\&B=45^{\circ} \\&A=45^{\circ}, B=45^{\circ}\end{aligned}$
Question 11
यदि $\sin (A-B)=\cos (A+B)=\frac{1}{2}$, तब A और B ज्ञात कीजिए ।
Sol :
$\begin{aligned}&\sin (A+B)=\frac{1}{2} \\&\sin (A+B)=\sin 30^{\circ}....(i) \\&A+B=30^{\circ} \\&\cos (A-B)=\frac{1}{2} \\&\cos (A-B)=\cos 60^{\circ}...(ii) \\&A-B=60^{\circ}\end{aligned}$
समीकरण (i) तथा (ii) से
$\begin{aligned}&A+B=30^{\circ} \\&\frac{A-B=60^{\circ}}{2 A=90^{\circ}}\end{aligned}$
$A=\frac{90^{\circ}}{2}$
$A=45^{\circ}$
A का मान समीकरण (i) में रखने पर
$A+B=60^{\circ}$
$45^{\circ}+B=90^{\circ}$
$B=60^{\circ}-45^{\circ}$
$B=15^{\circ}$
$A=45^{\circ}, B=15^{\circ}$
Question 12
यदि $(A-B)=\frac{1}{2}, \cos (A+B)=\frac{1}{2} ;$ 0°<A+B<90°; A>B तब A और B ज्ञात करें।
Sol :
$\sin (A+B)=\frac{1}{2}$
$\sin (A+B)=\sin 30^{\circ}$
$A+B=30^{\circ}$....(i)
$\cos (A-B)=\frac{1}{2}$
$\cos (A-B)=\cos 60^{\circ}$
$A-B=60^{\circ}$....(ii)
समीकरण (i) तथा (ii)सें
$\begin{aligned}&A+B=30^{\circ} \\&\frac{A-B=60^{\circ}}{2 A=90^{\circ}}\end{aligned}$
$A=\frac{90^{\circ}}{2}$
$A=45^{\circ}$
A का मान समीकरण (i) में रखने पर
$A+B=60^{\circ}$
$45^{\circ}+B=90^{\circ}$
$B=60^{\circ}-45^{\circ}$
$B=15^{\circ}$
$A=45^{\circ}, B=15^{\circ}$
Question 13
एक उदाहरण द्वारा दिखाइए कि $\cos \mathrm{A}-\cos \mathrm{B} \neq \cos (\mathrm{A}-\mathrm{B})$
(ii) $\cos \mathrm{C}+\cos \mathrm{D} \neq \cos (\mathrm{C}+\mathrm{D})$
(iii) $\sin \mathrm{A}+\sin \mathrm{B} \neq \sin (\mathrm{A}+\mathrm{B})$
(iv) $\sin A-\sin B \neq \sin (A-B)$
Sol :
(i)
cos A-cos B≠cos(A-B)
L.H.S
A=60° , B=30°
cos 60°-cos30°≠cos(60°-30°)
$\frac{1}{2}-\frac{\sqrt{3}}{2} \neq \cos 30^{\circ}$
$\frac{1}{2}-\frac{\sqrt{3}}{2} \neq \frac{\sqrt{3}}{2}$
$\frac{1-\sqrt{3}}{2} \neq \frac{\sqrt{3}}{2} $ proved
(ii)
LHS
$C=60^{\circ}, D=30^{\circ}$
$\cos 60^{\circ}+\cos 30^{\circ} \neq \cos \left(60^{\circ}+30^{\circ}\right)$
$\cos 60^{\circ}+\cos 30^{\circ} \neq \cos \left(90^{\circ}\right)$
$\frac{1}{2}+\frac{\sqrt{3}}{2} \neq 0$
$\frac{1+\sqrt{3}}{2} \neq 0$ proved
(iii)
$A=60^{\circ}, B=30^{\circ}$
$\sin 60^{\circ}+\sin 30^{\circ} \neq \sin \left(60^{\circ}+30^{\circ}\right)$
$\sin 60^{\circ}+\sin 30^{\circ} \neq \sin \left(90^{\circ}\right)$
$\frac{\sqrt{3}}{2}+\frac{1}{2} \neq 1$
$\frac{\sqrt{3}+1}{2} \neq 0$ proved
(iv)
LHS
$A=60^{\circ}, B=30^{\circ}$
$\sin 60^{\circ}-\sin 30^{\circ} \neq \sin \left(60^{\circ}-30^{\circ}\right)$
$\sin 60^{\circ}-\sin 30^{\circ} \neq \sin \left(30^{\circ}\right)$
$\frac{\sqrt{3}}{2}-\frac{1}{2} \neq \frac{1}{2}$
$\frac{\sqrt{3}-1}{2} \neq \frac{1}{2} \quad$ proved
TYPE IV: किसी समकोण त्रिभुज में एक भुजा और एक न्यूनकोण ज्ञात रहने पर दूसरी भुजा ज्ञात करने पर आधारित प्रश्न :
Question 14
एक समकोण $\triangle \mathrm{ABC}$ में कर्ण $\mathrm{AC}=10 \mathrm{~cm}$ और $\angle \mathrm{A}=60^{\circ}$, तब अन्य भुजाओं को ज्ञात कीजिए।
Sol :
sin 60°=लम्ब/कर्ण
$\sin 60^{\circ}=\frac{B C}{A C}$
$\frac{\sqrt{3}}{2}=\frac{B C}{10}$
$\mathrm{BC}=5 \sqrt{5} \mathrm{~cm}$
cos 60°=आधार /कर्ण
$\cos 60^{\circ}=\frac{A B}{A C}$
$\frac{1}{2}=\frac{A B}{10}$
$\mathrm{BC}=5 \mathrm{~cm}$
Question 15
एक आयत $\mathrm{ABCD}$ में $\mathrm{BD}: \mathrm{BC}=2: \sqrt{3}$, तब $\angle \mathrm{BDC}$ का मान डिग्री में ज्ञात करें।
Sol :
cosec θ=कर्ण /लम्ब
$\operatorname{cosec} \theta=\frac{B \mathrm{D}}{\mathrm{B} C}$
$\operatorname{cosec} \theta=\frac{2}{\sqrt{3}}$
$\operatorname{cosec} \theta=\operatorname{cosec} 60^{\circ}$
$\theta=60^{\circ}$
कोण $\mathrm{BDC}=60^{\circ}$
Mst h
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