Question

If $ta{n}^{2}x=2ta{n}^{2}y+1$, show that $cos2x+si{n}^{2}y=0.$

Answer 1

We have,

${\tan }^{2}x=2{\tan }^{2}y+1$

On adding 1 both side and we get,

${\tan }^{2}x+1=2{\tan }^{2}y+1+1$

$\Rightarrow {\sec }^{2}x=2{\tan }^{2}y+2$

$\Rightarrow {\sec }^{2}x=2({\tan }^{2}y+1)$

$\Rightarrow {\sec }^{2}x=2{\sec }^{2}y$

$\Rightarrow {\cos }^{2}x=\frac{1}{2}{\cos }^{2}y$

$\Rightarrow 2{\cos }^{2}x={\cos }^{2}y$

Subtracting 1 from both side and we get,

$\Rightarrow 2{\cos }^{2}x-1={\cos }^{2}y-1$

$\Rightarrow \cos 2x=-(1-{\cos }^{2}y)$

$\Rightarrow \cos 2x=-{\sin }^{2}y$

So,

L.H.S.

$\cos 2x+{\sin }^{2}y$

$=(-{\sin }^{2}y)+{\sin }^{2}y$

$=0$

R.H.S.