Question

$\sin ^{-1}\left(\frac{x^2}{4}+\frac{y^2}{9}\right)+\cos ^{-1}\left(\frac{x}{2 \sqrt{2}}+\frac{y}{3 \sqrt{2}}-2\right)$ equals to :

Answer 1

$-1 \leq \frac{x^2}{4}+\frac{y^2}{9} \leq 1$ represents interior and the boundary of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Also $\quad-1 \leq \frac{x}{2 \sqrt{2}}+\frac{y}{3 \sqrt{2}}-2 \leq 1$

i.e. $\frac{x}{2 \sqrt{2}}+\frac{y}{3 \sqrt{2}} \geq 1$ and $\frac{x}{2 \sqrt{2}}+\frac{y}{3 \sqrt{2}} \leq 3$

$\frac{x}{2 \sqrt{2}}+\frac{y}{3 \sqrt{2}} \geq 1$ represents the portion of $x y$ plane

SOLUTION : which contains only one point viz: $\left(\sqrt{2}, \frac{3}{\sqrt{2}}\right)$ of $\frac{x^2}{4}+\frac{y^2}{9} \leq 1$

$\therefore \quad \sin ^{-1}\left(\frac{x^2}{4}+\frac{y^2}{9}\right)+\cos ^{-1}\left(\frac{x}{2 \sqrt{2}}+\frac{y}{3 \sqrt{2}}-2\right)=\sin ^{-1}\left(\frac{1}{2}+\frac{1}{2}\right)+\cos ^{-1}\left(\frac{1}{2}+\right. \\$

$= \sin ^{-1} 1+\cos ^{-1}(-1)=\frac{\pi}{2}+\pi=\frac{3 \pi}{2} \\ $