If $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$,then prove that $\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}$
$\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$
$\frac{1+\tan\frac{y}{2}}{1-\tan\frac{y}{2}}=(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})^3$
Applying componendo and dividendo rule on both sides,
$\tan\frac{y}{2}=\frac{\tan\frac{x}{2}(3+\tan^2\frac{x}{2})}{(1+3\tan^2\frac{x}{2})}$
I am stuck here.
Best Answer
Now using Weierstrass substitution, $$\cos\left(\dfrac\pi2+A\right)=\dfrac{1-\tan^2\left(\dfrac\pi4+\dfrac A2\right)}{1+\tan^2\left(\dfrac\pi4+\dfrac A2\right)}$$
As $\sin A=-\cos\left(\dfrac\pi2+A\right)$ applying Componendo and Dividendo
$$\tan^2\left(\dfrac\pi4+\dfrac A2\right)=\dfrac{1+\sin A}{1-\sin A}$$
Replace the values of $\tan^2\left(\dfrac\pi4+\dfrac y2\right),\tan^2\left(\dfrac\pi4+\dfrac x2\right)$ in $$\tan^2\left(\dfrac\pi4+\dfrac y2\right)=\tan^6\left(\dfrac\pi4+\dfrac x2\right)=\left\{\tan^2\left(\dfrac\pi4+\dfrac x2\right)\right\}^3$$ and apply Componendo and Dividendo