Question 12 Marks
$\sqrt{ } 3 \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4$
AnswerL.H.S
$\begin{aligned}
\mathrm{S} . & =\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} \\
& =\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}} \\
& =2\left(\frac{\frac{\sqrt{3}}{2}}{\sin 20^{\circ}}-\frac{\frac{1}{2}}{\cos 20^{\circ}}\right)
\end{aligned}$
$\begin{aligned}
& =2\left(\frac{\sin 60^{\circ}}{\sin 20^{\circ}}-\frac{\cos 60^{\circ}}{\cos 20^{\circ}}\right) \\
& =2\left(\frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right) \\
& =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)}{\frac{1}{2}\left(2 \sin 20^{\circ} \cos 20^{\circ}\right)} \\
& =\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} \\
& =4\\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 22 Marks
$\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}=\cos 7^{\circ}$
Answer$\begin{aligned}
& \text { L.H.S. }=\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ} \\
& =\left(\sin 47^{\circ}-\sin 25^{\circ}\right)+\left(\sin 61^{\circ}-\sin 11^{\circ}\right) \\
& =2 \cos \left(\frac{47^{\circ}+25^{\circ}}{2}\right) \sin \left(\frac{47^{\circ}-25^{\circ}}{2}\right) \\
& \quad+2 \cos \left(\frac{61^{\circ}+11^{\circ}}{2}\right) \sin \left(\frac{61^{\circ}-11^{\circ}}{2}\right) \\
& =2 \cos 36^{\circ} \sin 11^{\circ}+2 \cos 36^{\circ} \sin 25^{\circ} \\
& =2 \cos 36^{\circ}\left(\sin 11^{\circ}+\sin 25^{\circ}\right) \\
& =2 \cos 36^{\circ}\left[2 \sin \left(\frac{25^{\circ}+11^{\circ}}{2}\right) \cos \left(\frac{25^{\circ}-11^{\circ}}{2}\right)\right] \\
& =2 \cos 36^{\circ}\left(2 \sin 18^{\circ} \cos 7^{\circ}\right) \\
& =4\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right) \cos 7^{\circ} \\
& =\frac{4(5-1)}{16} \cos 7^{\circ} \\
& =\cos 7^{\circ} \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 32 Marks
$\tan \frac{\pi}{8}=\sqrt{2}-1$
AnswerWe know that,
$\begin{aligned}
& \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\
& \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}
\end{aligned}$
Let $\tan \frac{\pi}{8}=\mathrm{t}$
$\begin{array}{ll}
\therefore & \frac{2 \mathrm{t}}{1-\mathrm{t}^2}=1 \\
\therefore & 2 \mathrm{t}=1-\mathrm{t}^2 \\
\therefore & \mathrm{t}^2+2 \mathrm{t}-1=0 \\
\therefore & \mathrm{t}=\frac{-2 \pm \sqrt{4+4}}{2} \\
& =\frac{-2 \pm 2 \sqrt{2}}{2} \\
& =-1 \pm \sqrt{2} \\
& \mathrm{t}=\tan \frac{\pi}{8}>0 \\
\therefore & \tan \frac{\pi}{8}=\sqrt{2}-1
\end{array}$
View full question & answer→Question 42 Marks
$\sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4}$
AnswerWe know that, $\sin ^2 \theta=1-\cos ^2 \theta$
$\begin{aligned}
& \sin ^2 36^{\circ}=1-\cos ^2 36^{\circ} \\
& =1-\left(\frac{\sqrt{5}+1}{4}\right)^2 \\
& =\frac{16-(5+1+2 \sqrt{5})}{16} \\
& =\frac{10-2 \sqrt{5}}{16} \\
& \therefore \sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4} \ldots \ldots\left[\because \sin 36^{\circ} \text { is positive }\right]
\end{aligned}$
View full question & answer→Question 52 Marks
$\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$
AnswerWe know that,
$\begin{aligned}
& \cos 2 \theta=1-2 \sin ^2 \theta \\
& \cos 36^{\circ}=\cos 2\left(18^{\circ}\right) \\
& =1-2 \sin ^2 18^{\circ} \\
& =1-2\left(\frac{\sqrt{5}-1}{4}\right)^2 \\
& =\frac{8-(5+1-2 \sqrt{5})}{8} \\
& =\frac{8-(6-2 \sqrt{5})}{8} \\
& =\frac{2+2 \sqrt{5}}{8} \\
& \therefore \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}
\end{aligned}$
View full question & answer→Question 62 Marks
$\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
Answer$\begin{aligned}
\text { L.H.S. } & =\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x} \\
& =\frac{2 \sin \left(\frac{9 x+5 x}{2}\right) \sin \left(\frac{5 x-9 x}{2}\right)}{2 \cos \left(\frac{17 x+3 x}{2}\right) \sin \left(\frac{17 x-3 x}{2}\right)} \\
& =\frac{2 \sin 7 x \sin (-2 x)}{2 \cos 10 x \sin 7 x} \\
& =\frac{-\sin 2 x}{\cos 10 x} \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 72 Marks
$\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
Answer$\begin{aligned}
\text { L.H.S. } & =\cot 4 x(\sin 5 x+\sin 3 x) \\
& =\cot 4 x\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right] \\
& =\frac{\cos 4 x}{\sin 4 x}(2 \sin 4 x \cos x) \\
& =2 \cos 4 x \cos x \\
\text { R.H.S. } & =\cot x(\sin 5 x-\sin 3 x) \\
& =\cot x\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right] \\
& =\frac{\cos x}{\sin x}(2 \cos 4 x \sin x) \\
& =2 \cos 4 x \cos x
\end{aligned}$
From (i) and (ii), we get L.H.S. $=$ R.H.S.
View full question & answer→Question 82 Marks
$\cos ^2 2 x-\cos ^2 6 x=\sin 4 x \sin 8 x$
Answer$\begin{aligned}
\text { L.H.S. } & =\cos ^2 2 x-\cos ^2 6 x \\
& =(\cos 2 x)^2-(\cos 6 x)^2 \\
& =(\cos 2 x+\cos 6 x)(\cos 2 x-\cos 6 x) \\
& =\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right] \\
& \cdot\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \left(\frac{6 x-2 x}{2}\right)\right] \\
& =[2 \cos 4 x \cos (-2 x)][2 \sin 4 x \sin 2 x] \\
& =(2 \cos 4 x \cos 2 x)(2 \sin 4 x \sin 2 x) \\
& =(2 \sin 2 x \cos 2 x)(2 \sin 4 x \cos 4 x) \\
& =\sin 4 x \sin 8 x \\
& =\text { R.H.S.}
\end{aligned}$
View full question & answer→Question 92 Marks
$\sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x$
Answer$\begin{aligned}
\text { L.H.S. } & =\sin ^2 6 x-\sin ^2 4 x \\
& =(\sin 6 x)^2-(\sin 4 x)^2 \\
& =(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x) \\
& =\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right] \\
& {\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \sin \left(\frac{6 x-4 x}{2}\right)\right] } \\
& =(2 \sin 5 x \cos x)(2 \cos 5 x \sin x) \\
& =(2 \sin x \cos x)(2 \sin 5 x \cos 5 x) \\
& =\sin 2 x \sin 10 x \\
& =\text { R.H.S.}
\end{aligned}$
View full question & answer→Question 102 Marks
$\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x$
AnswerL.H.S.
$\begin{aligned}
& =\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right) \\
& =2 \cos \left(\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}\right) \cos \left(\frac{\frac{\pi}{4}+x-\left(\frac{\pi}{4}-x\right)}{2}\right) \\
& =2 \cos \frac{\pi}{4} \cos x \\
& =2\left(\frac{1}{\sqrt{2}}\right) \cos x \\
& =\sqrt{2} \cos x \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 112 Marks
If $\sin \alpha \sin \beta-\cos \alpha \cos \beta+1=0$, then prove that $\cot \alpha \tan \beta=-1$.
Answer$\begin{aligned}
& \sin \alpha \sin \beta-\cos \alpha \cos \beta+1=0 \\
& \therefore \cos \alpha \cos \beta-\sin \alpha \sin \beta=1 \\
& \therefore \cos (\alpha+\beta)=1 \\
& \therefore \alpha+\beta=0 \ldots \ldots[\because \cos 0=1] \\
& \therefore \beta=-\alpha \\
& \text { L.H.S. }=\cot \alpha \tan \beta \\
& =\cot \alpha \tan (-\alpha) \\
& =-\cot \alpha \tan \alpha \\
& =-1 \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 122 Marks
If $\tan \alpha=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \beta=\frac{\sqrt{x}}{\sqrt{x^2+x+1}} \text { and } \tan \gamma=\sqrt{x^{-3}+x^{-2}+x^{-1}}$ then show that $\propto+\beta=\gamma$
View full question & answer→Question 132 Marks
Show that $(\cos \theta+i \sin \theta)^3$ $=\cos 3 \theta+i \sin 3 \theta$, where $i^2=-1$.
View full question & answer→Question 142 Marks
If $\tan A -\tan B =x$ and $\cot B -\cot A =y$ then show that $\cot ( A - B )=\frac{1}{x}+\frac{1}{y}$
View full question & answer→Question 152 Marks
Prove that $\frac{\tan 5 A+\tan 3 A}{\tan 5 A-\tan 3 A}=4 \cos ^2 A \cdot \cos 4 A$
View full question & answer→Question 162 Marks
Prove that following : $\sin 6 \theta+\sin 4 \theta-\sin 2 \theta=4 \cos \theta \sin 2 \theta \cos 3 \theta$
View full question & answer→Question 172 Marks
Prove that following : $\frac{\cos (7 x-5 y)+\cos (7 y-5 x)}{\sin (7 x-5 y)+\sin (7 y-5 x)}=\cot (x+y)$
View full question & answer→Question 182 Marks
Prove that $2 \operatorname{cosec} 2 x+\operatorname{cosec} x=\sec x \cdot \cot (x / 2)$
View full question & answer→Question 192 Marks
Prove the following : $\cos ^2\left(\frac{\pi}{10}\right)+\cos ^2\left(\frac{2 \pi}{5}\right)+\cos ^2\left(\frac{3 \pi}{5}\right)+\cos ^2\left(\frac{9 \pi}{10}\right)=2$
View full question & answer→Question 202 Marks
Prove the following : $\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}=2$
View full question & answer→Question 212 Marks
Show that : $\frac{\sin 2 \alpha+\sin 2 \beta}{\sin 2 \alpha-\sin 2 \beta}=\frac{\tan (\alpha+\beta)}{\tan (\alpha-\beta)}$
View full question & answer→Question 222 Marks
Show that $\frac{\sin 8 x+\sin 2 x}{\cos 2 x-\cos 8 x}=\cos 3 x$
View full question & answer→Question 232 Marks
Show that : $\frac{\cot \left(\frac{\pi}{2}+\theta\right) \sin (-\theta) \cot (\pi-\theta)}{\cos (2 \pi-\theta) \sin (\pi+\theta) \tan (2 \pi-\theta)}=-\operatorname{cosec} \theta$
View full question & answer→Question 242 Marks
Show that : $\frac{\operatorname{cosec}\left(90^{\circ}-\theta\right) \cdot \sin \left(180^{\circ}-\theta\right) \cot \left(360^{\circ}-\theta\right)}{\sec \left(180^{\circ}+\theta\right) \tan \left(90^{\circ}+\theta\right) \sin (-\theta)}=1$
View full question & answer→Question 252 Marks
Show that:
$\text { } \cos 24^{\circ}+\cos 55^{\circ}+\cos 125^{\circ}+\cos 204^{\circ}+\cos 300^{\circ}=\frac{1}{2}$
View full question & answer→Question 262 Marks
Prove the following : $\cos 40^{\circ}+\cos 50^{\circ}+\cos 70^{\circ}+\cos 80^{\circ} \ =\cos 20^{\circ}+\cos 10^{\circ}$
View full question & answer→Question 272 Marks
Prove the following : $\sin 40^{\circ}-\cos 70^{\circ}=\sqrt{3} \cos 80^{\circ}$
View full question & answer→Question 282 Marks
Prove that $1+\tan \theta \tan \frac{\theta}{2}=\sec \theta$
View full question & answer→Question 292 Marks
In $\triangle A B C$ prove that
$\cot A \cot B+\cot B \cot +\cot C \cot A=1$
AnswerIn $\triangle A B C, \mathrm{~A}+\mathrm{B}+\mathrm{C}=\pi$
$
\begin{aligned}
& \therefore \mathrm{A}+\mathrm{B}=\pi-\mathrm{C} \\
& \therefore \tan (A+B)=\tan (\pi-\mathrm{C}) \\
& \therefore \frac{\tan A+\tan B}{1-\tan A \tan B}=\tan (\pi-\mathrm{C}) \\
& \therefore \tan A+\tan B=-\tan C+\tan A \tan B \tan C \\
& \therefore \tan A+\tan B+\tan C=\tan A \tan B \tan C \\
& \therefore \frac{1}{\cot A}+\frac{1}{\cot B}+\frac{1}{\cot C}=\frac{1}{\cot A} \cdot \frac{1}{\cot B} \cdot \frac{1}{\cot C} \\
& \therefore \cot A \cot B+\cot B \cot C+\cot C \cot A=1
\end{aligned}
$
View full question & answer→Question 302 Marks
In $\triangle A B C$ prove that
$\sin 2 A+\sin 2 B-\sin 2 C=4 \cos A \cos B \sin C $
Answer$
\begin{aligned}
& \text { L.H.S. }=\sin 2 A+\sin 2 B-\sin 2 C \\
&= 2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)-\sin 2 \mathrm{C} \\
&= 2 \sin (A+B) \cos (A-B)-2 \sin \mathrm{C} \cos \mathrm{C} \\
&= 2 \sin (\pi-C) \cos (A-B)-2 \sin C \cos [\pi-(A+B)] \\
&= 2 \sin C \cos (A-B)+2 \sin C \cos (A+B) \\
&= 2 \sin \mathrm{C}[\cos (A-B)+\cos (A+B) \\
&= 2 \sin \mathrm{C} \cdot 2 \cos \left(\frac{A-B+A+B}{2}\right) \cos \left(\frac{A-B-A-B}{2}\right) \\
&= 4 \sin \mathrm{C} \cos A \cos B \\
&= 4 \cos A \cos B \sin C \\
&= \text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 312 Marks
Find the value of $\tan \frac{\pi}{8}$
View full question & answer→Question 322 Marks
In $\triangle A B C, A+B+C=\pi$ show that
$\tan 2 A+\tan 2 B+\tan 2 C=\tan 2 A \tan 2 B \tan 2 C$
Answer$\ln \triangle A B C$
$\begin{aligned}
& A+B+C=\pi \\
& \therefore 2 A+2 B+2 C=2 \pi \\
& \therefore 2 A+2 B=2 \pi-2 C \\
& \therefore \tan (2 A+2 B)=\tan (2 \pi-2 C) \\
& \therefore \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A-\tan 2 B}=-\tan 2 C \\
& \therefore \tan 2 A+\tan 2 B=-\tan 2 C \cdot(1-\tan 2 A \cdot \tan 2 B) \\
& \therefore \tan 2 A+\tan 2 B=-\tan 2 C+\tan 2 A \cdot \tan 2 B \cdot \tan 2 C \\
& \therefore \tan 2 A+\tan 2 B+\tan 2 C=\tan 2 A \cdot \tan 2 B \cdot \tan 2 C
\end{aligned}$
View full question & answer→Question 332 Marks
Express the following as a sum or difference of two trigonometric functions : $2 \cos 35^{\circ} \cos 75^{\circ}$
Answer$2 \cos 35^{\circ} \cos 75^{\circ}$
$\begin{aligned}
& =\cos \left(35^{\circ}+75^{\circ}\right)+\cos \left(35^{\circ}-75^{\circ}\right) \\
& =\cos 110^{\circ}+\cos (-40)^{\circ} \\
& =\cos 110^{\circ}+\cos 40^{\circ} \ldots[\because \cos (-\theta)=\cos \theta]
\end{aligned}$
View full question & answer→Question 342 Marks
Express the following as a sum or difference of two trigonometric functions : $2 \cos 4 \theta \cos 2 \theta$
Answer$\begin{aligned}
& \text2 \cos 4 \theta \cos 2 \theta=\cos (4 \theta+2 \theta)+\cos (4 \theta-2 \theta) \\
& =\cos 6 \theta+\cos 2 \theta
\end{aligned}$
View full question & answer→Question 352 Marks
Express the following as a sum or difference of two trigonometric functions : $2 \sin \frac{2 \pi}{3} \cos \frac{\pi}{2}$
Answer$\begin{aligned}
2 \sin \frac{2 \pi}{3} \cos \frac{\pi}{2} & =\sin \left(\frac{2 \pi}{3}+\frac{\pi}{2}\right)+\sin \left(\frac{2 \pi}{3}-\frac{\pi}{2}\right) \\
& =\sin \frac{7 \pi}{6}+\sin \frac{\pi}{6}
\end{aligned}
$
[Note: Answer given in the textbook is $\sin \frac{7 \pi}{12}+\sin \frac{\pi}{12}$ However, as per our calculation it is $\sin \frac{7 \pi}{6}+\sin \frac{\pi}{6}$
View full question & answer→Question 362 Marks
Express the following as a sum or difference of two trigonometric functions : $2 \sin 4 x \cos 2 x$
Answer$\begin{aligned}
& \text { i. } 2 \sin 4 x \cos 2 x=\sin (4 x+2 x)+\sin (4 x-2 x) \\
& =\sin 6 x+\sin 2 x
\end{aligned}$
View full question & answer→Question 372 Marks
Prove the following : $\sin 18^{\circ} \cos 39^{\circ}+\sin 6^{\circ} \cos 15^{\circ}=\sin 24^{\circ} \cos 33^{\circ}$
Answer$\begin{aligned}
& \text { L.H.S. }=\sin 18^{\circ} \cdot \cos 39^{\circ}+\sin 6^{\circ} \cdot \cos 15^{\circ} \\
& =\frac{1}{2}\left(2 \cos 39^{\circ} \sin 18^{\circ}+2 \cdot \cos 15^{\circ} \cdot \sin 6^{\circ}\right) \\
& =\frac{1}{2}\left[\sin \left(39^{\circ}+18^{\circ}\right)-\sin \left(39^{\circ}-18^{\circ}\right)+\sin \left(15^{\circ}+6^{\circ}\right)-\sin \left(15^{\circ}-6^{\circ}\right)\right] \\
& =\frac{1}{2}\left(\sin 57^{\circ}-\sin 21^{\circ}+\sin 21^{\circ}-\sin 9^{\circ}\right) \\
& =\frac{1}{2}\left(\sin 57^{\circ}-\sin 9^{\circ}\right) \\
& =\frac{1}{2} \times 2 \cdot \cos \left(\frac{57^{\circ}+9^{\circ}}{2}\right) \cdot \sin \left(\frac{57^{\circ}-9^{\circ}}{2}\right) \\
& =\cos 33^{\circ} \cdot \sin 24^{\circ} \\
& =\sin 24^{\circ} \cdot \cos 33^{\circ} \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 382 Marks
Prove the following : $\sin 6 x+\sin 4 x-\sin 2 x=4 \cos x \sin 2 x \cos 3 x$
Answer$\begin{aligned}
& \text { L.H.S. }=\sin 6 x+\sin 4 x-\sin 2 x \\
& =2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)-2 \sin x \cos x \\
& =2 \sin 5 x \cos x-2 \sin x \cos x \\
& =2 \cos x(\sin 5 x-\sin x) \\
& =2 \cos \left[2 \cos \left(\frac{5 x+x}{2}\right) \sin \left(\frac{5 x-x}{2}\right)\right] \\
& =2 \cos x(2 \cos 3 x \sin 2 x) \\
& =4 \cos x \sin 2 x \cos 3 x \\
& =\text { R.H.S. }
\end{aligned}$
[Note: The question has been modified.]
View full question & answer→Question 392 Marks
Prove the following : $\frac{\sin 2 x+\sin 2 y}{\sin 2 x-\sin 2 y}=\frac{\tan (x+y)}{\tan (x-y)}$
Answer$\begin{aligned}
\text { L.H.S. } & =\frac{\sin 2 x+\sin 2 y}{\sin 2 x-\sin 2 y} \\
& =\frac{2 \sin \left(\frac{2 x+2 y}{2}\right) \cos \left(\frac{2 x-2 y}{2}\right)}{2 \cos \left(\frac{2 x+2 y}{2}\right) \sin \left(\frac{2 x-2 y}{2}\right)} \\
& =\frac{\sin (x+y) \cos (x-y)}{\cos (x+y) \sin (x-y)} \\
& =\tan (x+y) \cdot \cot (x-y) \\
& =\tan (x+y) \cdot \frac{1}{\tan (x-y)} \\
& =\frac{\tan (x+y)}{\tan (x-y)} \\
& =\mathrm{R} . \mathrm{H} . \mathrm{S} \text {. }
\end{aligned}$
View full question & answer→Question 402 Marks
Prove the following : $\sin x \tan \frac{x}{2}+2 \cos x=\frac{2}{1+\tan ^2\left(\frac{x}{2}\right)}$
Answer$\begin{aligned}
& \text { L.H.S. }=\sin x \tan (x / 2)+2 \cos x \\
& =\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)+2 \cos x \\
& =\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(\frac{\sin \frac{2}{2}}{\cos \frac{x}{2}}\right)+2 \cos x \\
& =2 \sin ^2 x / 2+2 \cos x \\
& =1-\cos x+2 \cos x \\
& =1+\cos x \\
& =2 \cos ^2 x / 2 \\
& =\frac{2}{\sec ^2 \frac{x}{2}}=\frac{2}{1+\tan ^2 \frac{x}{2}}=\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 412 Marks
Prove the following : $2 \operatorname{cosec} 2 x+\operatorname{cosec} x=\sec \cot \frac{x}{2}$
Answer$\begin{aligned}
\text { L.H.S. } & =2 \operatorname{cosec} 2 x+\operatorname{cosec} x \\
& =\frac{2}{\sin 2 x}+\frac{1}{\sin x} \\
& =\frac{2}{2 \sin x \cos x}+\frac{1}{\sin x} \\
& =\frac{1}{\sin x \cos x}+\frac{1}{\sin x} \\
& =\frac{1+\cos x}{\sin x \cos x}
\end{aligned}$
$\begin{aligned}
& =\frac{2 \cos ^2 \frac{x}{2}}{\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right) \cos x} \\
& =\frac{1}{\cos x} \cdot \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} \\
& =\sec x \cot \left(\frac{x}{2}\right) \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 422 Marks
Prove the following : $\frac{1}{\tan 3 \mathrm{~A}-\tan A}-\frac{1}{\cot 3 A-\cot A}=\cot 2 \mathrm{~A}$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A} \\
& =\frac{1}{\tan 3 A-\tan A}-\frac{1}{\frac{1}{\tan 3 A}-\frac{1}{\tan A}} \\
& =\frac{1}{\tan 3 A-\tan A}-\frac{\tan 3 A \cdot \tan A}{\tan A-\tan 3 A} \\
& =\frac{1}{\tan 3 A-\tan A}+\frac{\tan 3 A \cdot \tan A}{\tan 3 A-\tan A} \\
& =\frac{1+\tan 3 A \cdot \tan A}{\tan 3 A-\tan A} \\
& =\frac{1}{\frac{\tan 3 A-\tan A}{1+\tan 3 A \cdot \tan A}} \\
& =\frac{1}{\tan (3 A-A)} \\
& =\frac{1}{\tan 2 \mathrm{~A}} \\
& =\cot 2 \mathrm{~A}=\text { R.H.S. } \\
\end{aligned}$
View full question & answer→Question 432 Marks
Prove the following : $\frac{\tan \left(\frac{\theta}{2}\right)+\cot \left(\frac{\theta}{2}\right)}{\cot \left(\frac{\theta}{2}\right)-\tan \left(\frac{\theta}{2}\right)}=\sec \theta$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{\tan \frac{\theta}{2}+\cot \frac{\theta}{2}}{\cot \frac{\theta}{2}-\tan \frac{\theta}{2}} \\
& =\frac{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}} \\
& =\frac{\sin ^2 \frac{\theta}{2}+\cos ^2 \frac{\theta}{2}}{\cos ^2 \frac{\theta}{2}-\sin ^2 \frac{\theta}{2}} \\
& =\frac{1}{\cos \theta} \\
& =\sec \theta=\text { R.H.S. } \\
\end{aligned}$
View full question & answer→Question 442 Marks
Prove the following : $\frac{\cos x}{1+\sin x}=\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1}$
Answer$\begin{aligned} & \text { L.H.S. }=\frac{\cos x}{1+\sin x} \\ & =\frac{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}{\cos ^2\left(\frac{x}{2}\right)+\sin ^2\left(\frac{x}{2}\right)+2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} \\ & =\frac{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2} \\ & =\frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)} \\ & =\frac{\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}-\frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}}{\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}+\frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}} \\ & =\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1} \\ & =\text { R.H.S. } \\ & \end{aligned}$
View full question & answer→Question 452 Marks
Prove the following : $\frac{\sin 3 x}{\cos x}+\frac{\cos 3 x}{\sin x}=2 \cot 2 x$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{\sin 3 x}{\cos x}+\frac{\cos 3 x}{\sin x} \\
& =\frac{\sin 3 x \cdot \sin x+\cos 3 x \cdot \cos x}{\sin x \cdot \cos x} \\
& =\frac{\cos (3 x-x)}{\sin x \cdot \cos x} \\
& =\frac{2 \cos 2 x}{2 \sin x \cdot \cos x} \\
& =\frac{2 \cos 2 x}{\sin 2 x} \\
& =2 \cot 2 x \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 462 Marks
Prove the following : 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θ
AnswerL.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ
= 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ
= 8sin 2θ cos 2θ cos 4θ cos 8θ
= 4(2sin 2θ cos 2θ) cos 4θ cos 8θ
= 4sin 4θ cos 4θ cos 8θ
= 2(2sin 4θ cos 4θ) cos 8θ
= 2sin 8θ cos 8θ
= sin 16θ
= R.H.S.
View full question & answer→Question 472 Marks
Prove the following : $\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 x}}}=2$ $\cos x$
Answer$\begin{aligned}
& \text { L.H.S. }=\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 x}}} \\
&=\sqrt{2+\sqrt{2+\sqrt{2[1+\cos 2(4 x)]}}} \\
&=\sqrt{2+\sqrt{2+\sqrt{2 \times 2 \cos ^2 4 x}}} \\
&=\sqrt{2+\sqrt{2+2 \cos 4 x}} \\
&=\sqrt{2+\sqrt{2[1+\cos 2(2 x)]}} \\
&=\sqrt{2+\sqrt{2 \times 2 \cos ^2 2 x}} \\
&=\sqrt{2+2 \cos ^{2 x}}=\sqrt{2(1+\cos 2 x)} \\
&=\sqrt{2 \times 2 \cos ^2 x} \\
&=2 \text { cos } x \\
&=\text { R.H.S. }
\end{aligned}$
[Note: The question has been modified.]
View full question & answer→Question 482 Marks
Prove the following : $\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2 \tan 2 \mathrm{x}$
AnswerL.H.S. $=\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}$
$=\frac{(\cos x+\sin x)^2-(\cos x-\sin x)^2}{(\cos x-\sin x)(\cos x+\sin x)}$
$=\frac{\left(\cos ^2 x+\sin ^2 x+2 \sin x \cos x\right)-\left(\cos ^2 x+\sin ^2 x-2 \sin x \cos x\right)}{\cos ^2 x-\sin ^2 x}$
$=\frac{1+2 \sin x \cos x-1+2 \sin x \cos x}{\cos ^2 x-\sin ^2 x}$
$=\frac{2(2 \sin x \cos x)}{\cos ^2 x-\sin ^2 x}$
$=\frac{2 \sin 2 x}{\cos 2 x}$
$=2 \tan 2 x$
$=\text { R.H.S. }$
View full question & answer→Question 492 Marks
Prove the following : $\tan x+\cot x=2 \operatorname{cosec} 2 x$
Answer$ \text { L.H.S. }=\tan x+\cot x$
$=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
$=\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}$
$=\frac{1}{\sin x \cos x}=\frac{2}{2 \sin x \cos x}$
$=\frac{2}{\sin 2 x}$
$=2 \operatorname{cosec} 2 x=\text { R.H.S. }$
View full question & answer→Question 502 Marks
Prove the following : $(\cos x-\cos y)^2+(\sin x-\sin y)^2=4 \sin ^2\left(\frac{x-y}{2}\right)$
Answer$\text { L.H.S. }=(\cos x-\cos y)^2+(\sin x-\sin y)^2$
$=\cos ^2 x+\cos ^2 y+2 \cos x \cdot \cos y+\sin ^2 x+\sin ^2 y+2 \sin x \cdot \sin y$
$=\left(\cos ^2 x+\sin ^2 x\right)+\left(\cos ^2 y+\sin ^2 y\right)-2(\cos x \cdot \cos y+\sin x \cdot \sin y)$
$=1+1-2 \cos (x-y)$
$=2-2 \cos (x-y)$
$=2[1-\cos (x-y)]$
$=2\left[2 \sin ^2\left[\left(\left(\frac{x-y}{2}\right)\right)\right] \ldots\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right]\right.$
$=4 \sin ^2\left(\frac{x-y}{2}\right)$
$=\text { R.H.S. }$
View full question & answer→