<s>[INST] If $\frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Then prove that $ab\leq 0$ 
If $\displaystyle \frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$  Prove that $ab\leq 0$

$\bf{My\; Try::}$ I am Trying To solve it Using Inequality.
Using $\bf{Cauchy\; Schwartz\; Inequality::}$
$$\frac{(\sec^4 \theta)^2}{a}+\frac{(\tan^4 \theta)^2}{b}\geq \frac{(\sec^4 \theta+\tan^4\theta)^2}{a+b}$$
and equality hold when $$\frac{\sec^4 \theta}{a}=\frac{\tan^4 \theta}{b}$$
Now I did not understand How can I solve after that
Help me
Thanks
 [/INST] $$\frac{\sec^2 \theta}{a}+\frac{\tan^2 \theta}{b}=\frac{1}{a+b} = \frac{\sec^2 \theta-\tan^2 \theta}{a+b}$$
$$\Rightarrow \frac{\sec^2 \theta}{a(a+b)}\left[(a+b)\sec^2 \theta-a\right]+\frac{\tan^2 \theta}{(a+b)a}\left[(a+b)\tan^2 \theta+b\right]=0$$
$$\Rightarrow\tan^2 \theta+b\sec^2 \theta = 0\Rightarrow \sin^2 \theta=-\frac{b}{a}\geq 0$$
$$\Rightarrow -\frac{b}{a}\geq 0\Rightarrow -ab\geq 0\Rightarrow ab\leq 0$$
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