Question

If $\tan\left(\frac{\pi}{4} + \frac{y }{2}\right) = \tan^{3 } \left(\frac{\pi}{4} + \frac{x}{2}\right) ,$ then $\frac{3\sin x + \sin^{3} x}{1+ 3 \sin^{2} x} = $

• A. 0
• B. 1
• C. $\sin \, 2y $
• D. $\sin \, y $

Answer 1

Answer: (D)

Given, $\tan \left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan ^{3}\left(\frac{\pi}{4}+\frac{x}{2}\right)$
$\Rightarrow \frac{1+\tan \frac{y}{2}}{1-\tan \frac{y}{2}}=\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)^{3}$
$\Rightarrow \frac{\cos \frac{y}{2}+\sin \frac{y}{2}}{\cos \frac{y}{2}-\sin \frac{y}{2}}=\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)^{3}$
On squaring both sides, we are getting
$\frac{1+\sin y}{1-\sin y}=\left(\frac{1+\sin x}{1-\sin x}\right)^{3}$
On applying componentdo and dividendo rule
$\frac{2}{2 \sin y}=\frac{(1+\sin x)^{3}+(1-\sin x)^{3}}{(1+\sin x)^{3}-(1-\sin x)^{3}}$
$\Rightarrow \sin y=\frac{3 \sin x+\sin ^{3} x}{1+3 \sin ^{2} x}$
So, $\frac{3 \sin x+\sin ^{3} x}{1+3 \sin ^{2} x}=\sin y$