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, $\frac{3\sin x+{\sin }^{3}x}{1+3{\sin }^{2}x}=\sin y$