Exercise 3.6 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 3.6
Question 1.
Express each of the following as a sum or difference
(i) $\sin 35^{\circ} \cos 28^{\circ}$
(ii) $\sin 4 \mathrm{x} \cos 2 \mathrm{x}$
(iii) $2 \sin 10 \theta \cos 2 \theta$
(iv) $\cos 5 \theta \cos 2 \theta$
(v) $\sin 5 \theta \sin 4 \theta$.
Solution:
(i)
$
\begin{aligned}
\sin 35^{\circ} \cos 28^{\circ} & =\frac{1}{2}\left[2 \sin 35^{\circ} \cos 28^{\circ}\right] \\
& =\frac{1}{2}\left[\sin \left(35^{\circ}+28^{\circ}\right)+\sin \left(35^{\circ}-28^{\circ}\right)\right] \\
& =\frac{1}{2}\left[\sin 63^{\circ}+\sin 7^{\circ}\right]
\end{aligned}
$
(ii)
$
\begin{aligned}
\sin 4 x \cos 2 x & =\frac{1}{2}[2 \sin 4 x \cos 2 x] \\
& =\frac{1}{2}[\sin (4 x+2 x)+\sin (4 x-2 x)] \\
& =\frac{1}{2}(\sin 6 x+\sin 2 x)
\end{aligned}
$
(iii)
$
\begin{aligned}
2 \sin 10 \theta \cos 2 \theta & =\frac{1}{2}[2 \sin 10 \theta \cos 2 \theta] \\
& =\frac{1}{2}[\sin (10 \theta+2 \theta)+\sin (10 \theta-2 \theta] \\
& =\frac{1}{2}[\sin 12 \theta+\sin 8 \theta]
\end{aligned}
$

(iv)
$
\begin{aligned}
\cos 5 \theta \cos 2 \theta & =\frac{1}{2}[2 \cos 5 \theta \cos 2 \theta] \\
& =\frac{1}{2}[\cos (5 \theta+2 \theta)+\cos (5 \theta-2 \theta)] \\
& =\frac{1}{2}[\cos 7 \theta+\cos 3 \theta)
\end{aligned}
$
(v)
$
\begin{aligned}
\sin 5 \theta \sin 4 \theta & =\frac{1}{2}[2 \sin 5 \theta \sin 4 \theta] \\
& =\frac{1}{2}[\cos (5 \theta-4 \theta)-\cos (5 \theta+4 \theta)] \\
& =\frac{1}{2}[\cos \theta-\cos 9 \theta)
\end{aligned}
$

Question 2.
Express each of the following as a product
(i) $\sin 75^{\circ}-\sin 35^{\circ}$
(ii) $\cos 65^{\circ}+\cos 15^{\circ}$
(iii) $\sin 50^{\circ}+\sin 40^{\circ}$
(iv) $\cos 35^{\circ}-\cos 75^{\circ}$.
Solution:
(i)
$
\begin{aligned}
\sin 75^{\circ}-\sin 35^{\circ} & =2 \cos \left(\frac{75^{\circ}+35^{\circ}}{2}\right) \sin \left(\frac{75^{\circ}-35^{\circ}}{2}\right) \\
& =2 \cos 55^{\circ} \sin 20^{\circ}
\end{aligned}
$
(ii)
$
\begin{aligned}
\cos 65^{\circ}+\cos 15^{\circ} & =2 \cos \left(\frac{65^{\circ}+15^{\circ}}{2}\right) \cos \left(\frac{65^{\circ}-15^{\circ}}{2}\right) \\
& =2 \cos 40^{\circ} \cos 25^{\circ}
\end{aligned}
$
(iii)
$
\begin{aligned}
\sin 50^{\circ}+\sin 40^{\circ} & =2 \sin \left(\frac{50^{\circ}+40^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}-40^{\circ}}{2}\right) \\
& =2 \sin 45^{\circ} \cos 5^{\circ}
\end{aligned}
$
(iv)
$
\begin{aligned}
\cos 35^{\circ}-\cos 75^{\circ} & =2 \sin \left(\frac{35^{\circ}+75^{\circ}}{2}\right) \sin \left(\frac{75^{\circ}-35^{\circ}}{2}\right) \\
& =2 \sin 55^{\circ} \sin 20^{\circ}
\end{aligned}
$

Question 3.
Show that $\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ}=\frac{1}{8}$.
Solution:
$
\begin{aligned}
\mathrm{LHS} & =\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ} \\
& =\frac{1}{2}\left[2 \sin 12^{\circ} \sin 48^{\circ}\right] \sin 54^{\circ} \\
& \left.=\frac{1}{2}\left[\cos \left(48^{\circ}-12^{\circ}\right)-\cos \left(48^{\circ}+12^{\circ}\right)\right] \sin \left(90^{\circ}-36^{\circ}\right)\right]
\end{aligned}
$

$\begin{aligned}
& =\frac{1}{2}\left[\cos 36^{\circ}-\cos 60^{\circ}\right] \sin 54^{\circ} \\
& =\frac{1}{2}\left[\cos 36^{\circ}-\frac{1}{2}\right] \cos 36^{\circ} \\
& =\frac{1}{2}\left(\cos 36^{\circ}\right)^2-\frac{1}{4}\left(\cos 36^{\circ}\right) \\
& =\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right)^2-\frac{1}{4}\left(\frac{\sqrt{5}+1}{4}\right) \\
& =\frac{1}{32}(5+1+2 \sqrt{5})-\frac{1}{16}(\sqrt{5}+1)
\end{aligned}$

$\begin{aligned}
& \text { Since } \\
& * 2 \sin x \sin y=\cos (x-y)-\cos (x+y) \\
& * \sin 54^{\circ}=\cos 36^{\circ} \\
& * \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}
\end{aligned}$

$
\begin{aligned}
& =\frac{1}{32}(6+2 \sqrt{5})-\frac{1}{16}(\sqrt{5}+1)=\frac{1}{16}[(3+\sqrt{5})-(\sqrt{5}+1)] \\
& =\frac{1}{16}[3+\sqrt{5}-\sqrt{5}-1]=\frac{1}{16}[2]=\frac{1}{8}=\text { RHS }
\end{aligned}
$
Question 4.
Show that $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{3 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{6 \pi}{15} \cos \frac{7 \pi}{15}=\frac{1}{128}$
Solution:

$
\begin{aligned}
& \left(\pi / 15=12^{\circ}\right) \\
& \text { LHS }=\cos 12^{\circ} \cos 24^{\circ} \cos 36^{\circ} \cos 48^{\circ} \cos 60^{\circ} \cos 72^{\circ} \cos 84^{\circ} \\
& \text { consider (we know that) } \\
& \cos \mathrm{A} \cos \left(60^{\circ}+\mathrm{A}\right) \cos \left(60^{\circ}-\mathrm{A}\right) \\
& =\cos \mathrm{A}\left[\cos ^2 60^{\circ}-\sin ^2 \mathrm{~A}\right] \\
& =\cos A\left[\frac{1}{4}-\left(1-\cos ^2 A\right)\right] \\
& \cos \mathrm{A} \cos \left(60^{\circ}+\mathrm{A}\right) \cos \left(60^{\circ}-\mathrm{A}\right)=\frac{1}{4} \cos 3 \mathrm{~A} \\
& =\cos \mathrm{A}\left[\cos ^2 \mathrm{~A}-\frac{3}{4}\right] \\
& =\frac{4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{A}}{4} \\
& \cos 12^{\circ} \cos 72^{\circ} \cos 48^{\circ}=\frac{1}{4} \cos 3\left(12^{\circ}\right)=\frac{1}{4} \cos 36^{\circ}=\frac{1}{4}\left[\frac{\sqrt{5}+1}{4}\right] \\
& \cos 24^{\circ} \cos 84^{\circ} \cos 36^{\circ}=\frac{1}{4} \cos 3\left(24^{\circ}\right)=\frac{1}{4} \cos 72^{\circ}=\frac{1}{4} \cos \left(90^{\circ}-18^{\circ}\right) \\
& =\frac{1}{4} \sin 18^{\circ}=\frac{1}{4}\left[\frac{\sqrt{5}-1}{4}\right] \\
&
\end{aligned}
$
$
\begin{aligned}
(1) \Rightarrow \quad \text { LHS } & =\frac{1}{4}\left[\frac{\sqrt{5}+1}{4}\right] \cdot \frac{1}{4}\left[\frac{\sqrt{5}-1}{4}\right] \cdot \frac{1}{2}=\frac{1}{4}\left(\frac{\sqrt{5}+1}{4} \cdot \frac{\sqrt{5}-1}{4}\right) \cdot \frac{1}{2} \\
& =\frac{5-1}{128 \times 4}=\frac{1}{128}
\end{aligned}
$
Question 5.
Show that $\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}=\tan 2 x$
Solution:

$\begin{aligned}
\sin 8 x \cos x & =\frac{1}{2}[2 \sin 8 x \cos x] \\
& =\frac{1}{2}[\sin (8 x+x)+(\sin (8 x-x)] \\
& =\frac{1}{2}[\sin 9 x+\sin 7 x] \\
\sin 6 x \cos 3 x & =\frac{1}{2}[2 \sin 6 x \cos 3 x] \\
& =\frac{1}{2}[\sin (6 x+3 x)+\sin (6 x-3 x] \\
& =\frac{1}{2}[\sin 9 x+\sin 3 x] \\
\cos 2 x \cos x & =\frac{1}{2}[2 \cos 2 x \cos x] \\
& =\frac{1}{2}[\cos (2 x+x)+\cos (2 x-x] \\
& =\frac{1}{2}[\cos 3 x+\cos x]
\end{aligned}$

$\begin{aligned}
\sin 3 x \sin 4 x & =\frac{1}{2}[2 \sin 3 x \sin 4 x] \\
& =\frac{1}{2}[\cos (4 x-3 x)-\cos (4 x+3 x)] \\
& =\frac{1}{2}[\cos x-\cos 7 x] \\
\text { LHS } & =\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}
\end{aligned}$

$\begin{aligned}
& =\frac{\frac{1}{2}(\sin 9 x+\sin 7 x)-\frac{1}{2}(\sin 9 x+\sin 3 x)}{\frac{1}{2}(\cos 3 x+\cos x)-\frac{1}{2}(\cos x-\cos 7 x)} \\
& =\frac{\frac{1}{2}[(\sin 9 x+\sin 7 x)-(\sin 9 x+\sin 3 x)]}{\frac{1}{2}[(\cos 3 x+\cos x)-(\cos x-\cos 7 x)]} \\
& =\frac{\sin 7 x-\sin 3 x}{\cos 3 x+\cos 7 x} \\
& =\frac{2 \cos \frac{7 x+3 x}{2} \sin \frac{7 x-3 x}{2}}{2 \cos \frac{7 x+3 x}{2} \cos \frac{7 x-3 x}{2}}=\frac{\sin 2 x}{\cos 2 x}=\tan 2 x=\text { RHS } \\
&
\end{aligned}$

Question 6.
Show that $\frac{(\cos \theta-\cos 3 \theta)(\sin 8 \theta+\sin 2 \theta)}{(\sin 5 \theta-\sin \theta)(\cos 4 \theta-\cos 6 \theta)}=1$
Solution:
$
\begin{aligned}
\cos \theta-\cos 3 \theta & =2 \sin \frac{\theta+3 \theta}{2} \sin \frac{3 \theta-\theta}{2} \\
& =2 \sin 2 \theta \sin \theta \\
\sin 8 \theta+\sin 2 \theta & =2 \sin \frac{8 \theta+2 \theta}{2} \cos \frac{8 \theta-2 \theta}{2} \\
& =2 \sin 5 \theta \cos 3 \theta \\
\sin 5 \theta-\sin \theta & =2 \cos \frac{(5 \theta+\theta)}{2} \sin \frac{(5 \theta-\theta)}{2} \\
& =2 \cos 3 \theta \sin 2 \theta \\
\cos 4 \theta-\cos 6 \theta & =2 \sin \frac{4 \theta+6 \theta}{2} \sin \frac{6 \theta-4 \theta}{2} \\
& =2 \sin 5 \theta \sin \theta \\
\therefore \text { LHS } & =\frac{2 \sin 2 \theta \sin \theta \cdot 2 \sin 5 \theta \cos 3 \theta}{2 \cos 3 \theta \sin 2 \theta \cdot 2 \sin 5 \theta \sin \theta}=1=\text { RHS }
\end{aligned}
$

Question 7.
Prove that $\sin \mathrm{x}+\sin 2 \mathrm{x}+\sin 3 \mathrm{x}=\sin 2 \mathrm{x}(1+2 \cos \mathrm{x})$.
Solution:
$
\begin{aligned}
\sin x+\sin 3 x & =2 \sin \frac{x+3 x}{2} \cos \frac{3 x-x}{2} \\
& =2 \sin 2 x \cos x \\
\text { LHS }=\sin x+\sin 3 x+\sin 2 x & =2 \sin 2 x \cos x+\sin 2 x \\
& =\sin 2 x(2 \cos x+1)=\text { RHS }
\end{aligned}
$
Question 8.
Prove that $\frac{\sin 4 x+\sin 2 x}{\cos 4 x+\cos 2 x}=\tan 3 x$
Solution:
$
\begin{aligned}
\sin 4 x+\sin 2 x & =2 \sin \frac{4 x+2 x}{2} \cos \frac{4 x-2 x}{2} \\
& =2 \sin 3 x \cos x \\
\cos 4 x+\cos 2 x & =2 \cos \frac{4 x+2 x}{2} \cos \frac{4 x-2 x}{2}=2 \cos 3 x \cos x \\
\therefore \text { LHS }=\frac{\sin 4 x+\sin 2 x}{\cos 4 x+\cos 2 x} & =\frac{2 \sin 3 x \cos x}{2 \cos 3 x \cos x}=\tan 3 x=\text { RHS }
\end{aligned}
$

Question 9.
Prove that $1+\cos 2 \mathrm{x}+\cos 4 \mathrm{x}+\cos 6 x=4 \cos \mathrm{x} \cos 2 \mathrm{x} \cos 3 \mathrm{x}$.
Solution:
$
\begin{aligned}
& 1+\cos 6 x=2 \cos ^2 3 x \\
& \cos 2 x+\cos 4 x=2 \cos \frac{2 x+4 x}{2} \cos \frac{4 x-2 x}{2} \\
& =2 \cos 3 x \cos x \\
& \text { LHS }=1+\cos 2 x+\cos 4 x+\cos 6 x \\
& =(1+\cos 6 x)+(\cos 2 \mathrm{x}+\cos 4 \mathrm{x}) \\
& =2 \cos ^2 3 x+2 \cos 3 x \cos x \\
& =2 \cos 3 \mathrm{x}(\cos 3 \mathrm{x}+\cos \mathrm{x}) \\
& =2 \cos 3 x\left[2 \cos \frac{3 x+x}{2} \cos \frac{3 x-x}{2}\right] \\
& =2 \cos 3 x(2 \cos 2 x \cos x) \\
& =4 \cos x \cos 2 x \cos 3 x=\mathrm{RHS} \\
&
\end{aligned}
$

Question 10 .
Prove that $\sin \frac{\theta}{2} \sin \frac{7 \theta}{2}+\sin \frac{3 \theta}{2} \sin \frac{11 \theta}{2}=\sin 2 \theta \sin 5 \theta$.
Solution:
$
\begin{aligned}
& \sin \frac{\theta}{2} \sin \frac{7 \theta}{2}=\frac{1}{2}\left[2 \sin \frac{\theta}{2} \sin \frac{7 \theta}{2}\right] \\
& =\frac{1}{2}\left[\cos \frac{7 \theta-\theta}{2}-\cos \frac{7 \theta+\theta}{2}\right] \\
& =\frac{1}{2}(\cos 3 \theta-\cos 4 \theta) \\
& \sin \frac{3 \theta}{2} \sin \frac{11 \theta}{2}=\frac{1}{2}\left[2 \sin \frac{3 \theta}{2} \sin \frac{11 \theta}{2}\right] \\
& =\frac{1}{2}\left[\cos \frac{11 \theta-3 \theta}{2}-\cos \frac{11 \theta+3 \theta}{2}\right] \\
& =\frac{1}{2}[\cos 4 \theta-\cos 7 \theta] \\
& \text { LHS }=\sin \frac{\theta}{2} \sin \frac{7 \theta}{2}+\sin \frac{3 \theta}{2} \sin \frac{11 \theta}{2} \\
& =\frac{1}{2}[\cos 3 \theta-\cos 4 \theta)+\frac{1}{2}[\cos 4 \theta-\cos 7 \theta] \\
& =\frac{1}{2}[\cos 3 \theta-\cos 4 \theta+\cos 4 \theta-\cos 7 \theta] \\
& =\frac{1}{2}[\cos 3 \theta-\cos 7 \theta] \\
& =\frac{1}{2}\left[2 \sin \frac{7 \theta+3 \theta}{2} \sin \frac{7 \theta-3 \theta}{2}\right] \\
& =\sin 5 \theta \sin 2 \theta=\text { RHS } \\
&
\end{aligned}
$

Question 11.
Prove that $\cos \left(30^{\circ}-A\right) \cos \left(30^{\circ}+A\right)+\cos \left(45^{\circ}-A\right) \cos \left(45^{\circ}+A\right)=\cos 2 A+\frac{1}{4}$.
Solution:
$
\begin{aligned}
& \cos \left(30^{\circ}-\mathrm{A}\right) \cos \left(30^{\circ}+\mathrm{A}\right)=\left(\cos 30^{\circ} \cos \mathrm{A}+\sin 30^{\circ} \sin \mathrm{A}\right)\left(\cos 30^{\circ} \cos \mathrm{A}-\sin 30^{\circ} \sin \mathrm{A}\right) \\
& =\cos ^2 30^{\circ} \cos ^2 \mathrm{~A}-\sin ^2 30^{\circ} \sin ^2 \mathrm{~A} \\
& =\left(\frac{\sqrt{3}}{2}\right)^2 \cos ^2 \mathrm{~A}-\left(\frac{1}{2}\right)^2 \sin ^2 \mathrm{~A} \\
& =\frac{3}{4} \cos ^2 \mathrm{~A}-\frac{1}{4} \sin ^2 \mathrm{~A} \\
& \cos \left(45^{\circ}-\mathrm{A}\right) \cos \left(45^{\circ}+\mathrm{A}\right)=\left(\cos 45^{\circ} \cos \mathrm{A}+\sin 45^{\circ} \sin \mathrm{A}\right)\left(\cos 45^{\circ} \cos \mathrm{A}-\sin 45^{\circ} \sin \mathrm{A}\right) \\
& =\cos ^2 45^{\circ} \cos ^2 \mathrm{~A}-\sin ^2 45^{\circ} \sin ^2 \mathrm{~A} \\
& =\left(\frac{1}{\sqrt{2}}\right)^2 \cos ^2 A-\left(\frac{1}{\sqrt{2}}\right)^2 \sin ^2 A \\
& =\frac{1}{2} \cos ^2 \mathrm{~A}-\frac{1}{2} \sin ^2 \mathrm{~A} \\
& \mathrm{LHS}=(1)+(2)=\frac{3}{4} \cos ^2 \mathrm{~A}-\frac{1}{4} \sin ^2 \mathrm{~A}+\frac{1}{2} \cos ^2 \mathrm{~A}-\frac{1}{2} \sin ^2 \mathrm{~A} \\
& =\frac{5}{4} \cos ^2 \mathrm{~A}-\frac{3}{4} \sin ^2 \mathrm{~A} \\
& =\cos ^2 \mathrm{~A}-\frac{3}{4} \sin ^2 \mathrm{~A}-\frac{1}{4} \sin ^2 \mathrm{~A}+\frac{1}{4} \sin ^2 \mathrm{~A}+\frac{1}{4} \cos ^2 \mathrm{~A} \\
& =\left(\cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~A}\right)+\frac{1}{4}\left(\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right) \\
& =\cos 2 \mathrm{~A}+\frac{1}{4}(1)=\cos 2 \mathrm{~A}+\frac{1}{4}=\text { RHS } \\
&
\end{aligned}
$

Question 12 .
Prove that $\frac{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}=\tan 4 x$
Solution:
$
\begin{aligned}
& \text { Nr: }(\sin x+\sin 7 x)+(\sin 3 x+\sin 5 x) \\
& =\left[2 \sin \frac{7 x+x}{2} \cos \frac{7 x-x}{2}\right]+\left[2 \sin \frac{5 x+3 x}{2} \cos \frac{5 x-3 x}{2}\right] \\
& =2 \sin 4 x \cos 3 x+2 \sin 4 x \cos x \\
& =2 \sin 4 x(\cos 3 x+\cos x) \\
& \text { Dr. }(\cos x+\cos 7 x)+(\cos 3 x+\cos 5 x) \\
& =\left[2 \cos \frac{7 x+x}{2} \cos \frac{7 x-x}{2}\right]+\left[2 \cos \frac{5 x+3 x}{2} \cos \frac{5 x-3 x}{2}\right] \\
& =2 \cos 4 x \cos 3 x+2 \cos 4 x \cos x \\
& =2 \cos 4 x(\cos 3 x+\cos x) \\
& \text { LHS }=\frac{(1)}{(2)}=\frac{2 \sin 4 x(\cos 3 x+\cos x)}{2 \cos 4 x(\cos 3 x+\cos x)} \\
& =\tan 4 x=\text { RHS } \\
&
\end{aligned}
$

Question 13.
Prove that $\frac{\sin (4 A-2 B)+\sin (4 B-2 A)}{\cos (4 A-2 B)+\cos (4 B-2 A)}=\tan (A+B)$
Solution:
$
\begin{aligned}
\mathrm{LHS} & =\frac{2 \sin \left(\frac{4 A-2 B+4 B-2 A}{2}\right) \cos \left(\frac{4 A-2 B-4 B+2 A}{2}\right)}{2 \cos \left(\frac{4 A-2 B+4 B-2 A}{2}\right) \cos \left(\frac{4 A-2 B-4 B+2 A}{2}\right)} \\
& =\frac{2 \sin (A+B) \cos (3 A-3 B)}{2 \cos (A+B) \cos (3 A-3 B)}=\tan (A+B)=\text { RHS }
\end{aligned}
$
Question 14.
Show that $\cot \left(A+15^{\circ}\right)-\tan \left(A-15^{\circ}\right)=\frac{4 \cos 2 A}{1+2 \sin 2 A}$

Solution:
$
\begin{aligned}
\text { LHS } & =\frac{\cos \left(\mathrm{A}+15^{\circ}\right)}{\sin \left(\mathrm{A}+15^{\circ}\right)}-\frac{\sin \left(\mathrm{A}-15^{\circ}\right)}{\cos \left(\mathrm{A}-15^{\circ}\right)} \\
& =\frac{\cos \left(\mathrm{A}+15^{\circ}\right) \cos \left(\mathrm{A}-15^{\circ}\right)-\sin \left(\mathrm{A}+15^{\circ}\right) \sin \left(\mathrm{A}-15^{\circ}\right)}{\sin \left(\mathrm{A}+15^{\circ}\right) \cos \left(\mathrm{A}-15^{\circ}\right)} \\
& =\frac{\cos \left(\mathrm{A}+15^{\circ}+\mathrm{A}-15^{\circ}\right)}{\frac{1}{2}\left[\sin \left(\mathrm{A}+15^{\circ}+\mathrm{A}-15^{\circ}\right)+\sin \left(\mathrm{A}+15^{\circ}-\mathrm{A}+15^{\circ}\right)\right]} \\
& =\frac{2 \cos 2 \mathrm{~A}}{\sin 2 \mathrm{~A}+\sin 30^{\circ}}=\frac{2 \cos 2 \mathrm{~A}}{\frac{1}{2}+\sin 2 \mathrm{~A}} \\
& =\frac{4 \cos 2 \mathrm{~A}}{1+2 \sin 2 \mathrm{~A}}=\text { RHS }
\end{aligned}
$