Answer:
We are given that:
[tex]\cos 2x=\tan ^2y---------(1)[/tex]
Now we are asked to show that:
[tex]\cos 2y=\tan ^2x[/tex]
We know that:
[tex]\cos 2y=\dfrac{1-\tan ^2y}{1+\tan ^2y}[/tex]
Hence using equation (1) we get:
[tex]\cos 2y=\dfrac{1-\cos 2x}{1+\cos 2x}--------(2)[/tex]
Also we know that:
[tex]\cos 2x=1-2 \sin ^2x[/tex]
and [tex]\cos 2x=2 \cos ^2x-1[/tex]
so using above two formula in equation (2) we get:
[tex]\cos 2y=\dfrac{1-(1-2\sin ^2x)}{1+2\cos ^2x-1}\\\\\cos 2y=\dfrac{1-1+2\sin ^2x}{2\cos ^2x}\\\\\cos 2y=\dfrac{2\sin ^2x}{2\cos ^2x}\\\\\cos 2y=\tan ^2x[/tex]
( since we know that:
[tex]\tan x=\dfrac{\sin x}{\cos x}[/tex]
)
Hence we have proved that:
[tex]\cos 2y=\tan ^2x[/tex]