IIT-JEE 2009 · Paper 1, Mathematics question 32

Cauchy-Schwarz. $(\sin^2x+\cos^2x)^2\le(2+3)\left(\dfrac{\sin^4x}2+\dfrac{\cos^4x}3\right)$, i.e. $\dfrac{\sin^4x}2+\dfrac{\cos^4x}3\ge\dfrac15$, with equality exactly when $\dfrac{\sin^2x}2=\dfrac{\cos^2x}3$.

So $\sin^2x=\frac25$, $\cos^2x=\frac35$, and $\tan^2x=\frac23$.

Check (B). $\dfrac{(2/5)^4}8+\dfrac{(3/5)^4}{27}=\dfrac{2}{625}+\dfrac{3}{625}=\dfrac1{125}$.

(A), (B)