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

If $\tan \left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{x}{2}\right)$ then $\frac{3 \sin x+\sin ^3 x}{1+3 \sin ^2 x}=$
  1. 0
  2. 1
  3. $\sin 2 y$
  4. $\sin y$

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} $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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