Trigo bash-2

Relevant Wiki: Titu's Lemma .

You don't need to bash the problem at all.

By Titu's Lemma, $\dfrac{(\sec^2 \alpha )^2}{a} + \dfrac{(- \tan^2 \alpha )^2}{b} \geq \dfrac{ (\sec^2 \alpha - \tan^2 \alpha)^2}{a+b}$ By given condition, equality holds in above equation. Thus $\dfrac{(\sec^2 \alpha )}{a} = \dfrac{(- \tan^2 \alpha )}{b}$ Write $\sec^2 \alpha =1+ \tan^2 \alpha$ , rearrange to get $\tan^2 \alpha = \dfrac{-b}{a+b} \geq 0.$ .

Now $a$ and $b$ cannot have the same sign (that is, they can't be both positive or both negative) since that would make $\tan^2 \alpha$ negative.

If $b >0$ , then $(a+b)<0 \implies a<0 \implies -a >b \implies |a|>|b|.$
Now if $b<0$ then $a+b>0 \implies a>-b>0 \implies |a|>|b|.$

Thus the correct answer is $\boxed{ |a|>|b|}.$

$\begin{aligned} \frac {\sec^4 \alpha}a + \frac {\tan^4 \alpha}b & = \frac 1{a+b} \\ \frac {(\tan^2 \alpha+1)^2}a + \frac {\tan^4 \alpha}b & = \frac 1{a+b} \\ \frac {\tan^4 \alpha + 2 \tan^2 \alpha +1}a + \frac {\tan^4 \alpha}b & = \frac 1{a+b} \\ (a+b)\tan^4 \alpha + 2b \tan^2 \alpha +b & = \frac {ab}{a+b} \\ (a+b)^2\tan^4 \alpha + 2b(a+b)\tan^2 \alpha +b^2 & = 0 \\ \left(\tan^2 \alpha + \frac b{a+b}\right) & = 0 \\ \implies \tan^2 \alpha & = - \frac b{a+b} \end{aligned}$

For $\alpha$ to be real, $-\dfrac b{a+b} > 0$ . Implying that $|a|$ can be smaller than, equal to or bigger than $|b|$ . Therefore, nothing can be said about the absolute values of $a$ and $b$ .