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.

  • $\begingroup$ $(sin a +cos a)^2=2sin a.cos a$ $\endgroup$ – 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 $35$.

  • $\begingroup$ Ah,so I was just blind simplifying the equation, thank you. $\endgroup$ – StAKmod Mar 26 '19 at 19:27

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