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  4. If (9 π/8)-(9/4) sin -1 (1/3)=(9/4) sin -1 x, then the value of x is

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Q. If $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} x$, then the value of $x$ is

Inverse Trigonometric Functions

Solution:

$LHS =\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)=\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{3}\right)\right)$
$=\frac{9}{4}\left(\cos ^{-1}\left(\frac{1}{3}\right)\right) \left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$
$=\frac{9}{4} \sin ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^2}\left(\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^2}\right)$
$=\frac{9}{4} \sin ^{-1} \sqrt{1-\frac{1}{9}}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$
$\therefore \frac{9}{4} \sin ^{-1} x=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$
$\Rightarrow x=\frac{2 \sqrt{2}}{3}$