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

Option: 2
$\pi$

Option: 3
$\frac{\pi}{\sqrt{2}}$

Option: 4
$\frac{3\pi}{2}$

Answers (1)

best_answer

Domains and Ranges of Inverse Trigonometric Functions -

For $\sin ^{-1}x$

Domain $\epsilon \left [ -1, 1 \right ]$

Range $\epsilon \left [ -\frac{\pi }{2}, \frac{\pi }{2} \right ]$

-

Domains and Ranges of Inverse Trigonometric Functions -

For $\cos ^{-1}x$

Domain $\epsilon \left [ -1, 1 \right ]$

Range $\epsilon \left [ 0, \pi \right ]$

-

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

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

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

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

which contains only one point viz :

$\left ( \sqrt{2},\frac{3}{\sqrt{2}} \right ) \; of\;\frac{x^{2}}{4}+\frac{y^{2}}{9}< 1$

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

$=sin^{-1}\left ( \frac{1}{2} + \frac{1}{2} \right )+cos^{-1}\left ( \frac{1}{2} + \frac{1}{2}-2 \right )$

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