KC Sinha Mathematics Solution Class 10 Chapter 8 त्रिकोणमितीय अनुपात एवम सर्वसमिकाए ( Trigonometry Ratios and Identities ) Exercise 8.2

  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 $\mathrm{BDC}=60^{\circ}$