Trigonometry identity proof sum I am normally good at solving trigonometry sums. However I stumbled upon this and could not figure out how to do it. I solved a few lines and then I got stuck. If anyone can teach me how to prove the equation, I'll be highly grateful. 

If $\cos^2 A- \sin^2 A = \tan^2 B$, prove that $\cos^2 B - \sin^2 B = \tan^2 A$ 

So here is what I did: 
$\cos^2 B = \dfrac{\sin^2 B}{\cos^2 A - \sin^2 A}$
So going by the second R.T.P, 
$\cos^2 B - \sin^2 B = \dfrac{\sin^2 B}{\cos^2 A - \sin^2 A} - \sin^2 B$
which unfortunately ends up nowhere close to the original proving required. 
Thanks for any possible help! 

From $\cos^2A-\sin^2A=\tan^2B$ you get
$$
\cos^2A=\frac{1+\tan^2B}{2}=\frac{1}{2\cos^2B}
$$
and
$$
\sin^2A=\frac{1-\tan^2B}{2}=\frac{\cos^2B-\sin^2B}{2\cos^2B}
$$
Thus
$$
\tan^2A=\frac{\cos^2B-\sin^2B}{2\cos^2B}\,2\cos^2B
$$
