# If $2(\sin a+\cos a)\sin b =3-\cos b$, then find $3\tan^2a+4\tan^2b$

This problem is on a contest $$10$$ years ago (I cannot remember its name); it is a common team contest. While checking it, I found one problem with trigonometry that I was unable to solve.

Let $$a,b$$ be real numbers such that:$$2(\sin a+\cos a)\sin b=3-\cos b$$ Find$$3\tan^2a+4\tan^2b$$

I doubt there is a solution on the internet now, but they do have a number $$35$$ as the final answer.

My tiny little progress:

Use simple simplification, $$3\tan^2a+4\tan^2b={3\over \cos^2a}+{4\over \cos^2 b}-7$$

Intuitively, in order to get this, the given equation should be squared. But that is where I got stuck, because the term $$\cos b$$ seems hard to handle; it just seems different from other terms.

• $(sin a +cos a)^2=2sin a.cos a$ – Fareed Abi Farraj Mar 26 '19 at 19:21

Rewrite it like this $$\underbrace{2(\sin a+\cos a)\sin b+\cos b}_{E}=3$$
Since we have by Cauchy inequality $$\sin a+\cos a \leq \sqrt{(\sin ^2a+\cos ^2a)(1^2+1^2)}=\sqrt{2}$$ we have always $$E \leq \sqrt{8}\sin b+ \cos b$$
Now again by Cauchy inequality $$\sqrt{8}\sin b+ \cos b \leq 3$$
Since we have equality case we have $$\sin b :\cos b = \sqrt{8}:1$$ so $$\tan b = \sqrt{8}$$ . Similary we have $$\tan a =1$$. So the result is $$35$$.