Question

If $\tan (\frac{\pi }{4}+\frac{y}{2})$ = ${\tan }^3(\frac{\pi }{4}+\frac{x}{2})$, prove that - $\sin y=\sin x.\frac{3+{\sin }^{2}x}{1+3{\sin }^{2}x}.$

Answer 1

We know that $tan(\frac{\pi }{4}+\alpha )=\frac{1+tan\alpha }{1-tan\alpha }=\frac{cos\alpha +sin\alpha }{cos\alpha -sin\alpha }$
$\therefore ta{n}^2(\frac{\pi }{4}+\alpha )=\frac{1+sin2\alpha }{1-sin2\alpha }\because 2sin\alpha cos\alpha =sin2\alpha$
Squaring the given relation, we have to prove that
$\frac{1+siny}{1-siny}=\frac{{(1+sinx)}^3}{{(1-sinx)}^3}$.Apply comp. and divi.
$\frac{2siny}{2}=\frac{2(3sinx+si{n}^{3}x)}{2(1+3si{n}^{2}x)}$
$\therefore siny=\frac{sinx(3+si{n}^{2}x)}{1+3si{n}^{2}x}$