Q60 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics
If $\tan \left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{x}{2}\right)$ then $\frac{3 \sin x+\sin ^3 x}{1+3 \sin ^2 x}=$
- A0
- B1
- C$\sin 2 y$
- Correct answerD$\sin y$
Solution
$$
\begin{aligned}
& \text { Given, } \tan \left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{x}{2}\right) \\
& \Rightarrow \quad \frac{1+\tan \frac{y}{2}}{1-\tan \frac{y}{2}}=\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)^3 \\
& \Rightarrow \quad \frac{\cos \frac{y}{2}+\sin \frac{y}{2}}{\cos \frac{y}{2}-\sin \frac{y}{2}}=\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)^3
\end{aligned}
$$
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
$$
\begin{array}{ll}
& \frac{2}{2 \sin y}=\frac{(1+\sin x)^3+(1-\sin x)^3}{(1+\sin x)^3-(1-\sin x)^3} \\
\Rightarrow \quad & \sin y=\frac{3 \sin x+\sin ^3 x}{1+3 \sin ^2 x} \\
\text { So, } & \frac{3 \sin x+\sin ^3 x}{1+3 \sin ^2 x}=\sin y
\end{array}
$$
Hence, the correct answer is option: 'D'.