# 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!

Add 1 to both sides, and massage the result.

$$\cos 2A /1 = \sin^2 B/ \cos^2 B = \tan^2 B \tag 1$$

Apply Componendo Dividendo Rule

$$\dfrac{1- \cos 2A}{1+ \cos 2A}= \dfrac{\cos^2 B- \sin^2 B}{1}$$

$$\tan^2 A = \cos 2 B$$

$$\cos 2 B = \tan^2 A. \tag 2$$

$$\tan^2B=\cos^2 A- \sin^2 A$$

$$\iff\dfrac{\sin^2B}{\cos^2B}=\dfrac{\cos^2 A- \sin^2 A}{\cos^2 A+\sin^2 A}$$

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 $$\tan^2A=\frac{\cos^2B-\sin^2B}{2\cos^2B}\,2\cos^2B$$

The equation can be rewritten

$$(\cos^2(A)-(1-\cos^2(A)))\cos^2(B)=\sin^2(B)$$

or after simplification

$$2\cos^2(A)\cos^2(B)=1,$$

which is perfectly symmetrical.