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If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, then what is $x$ equal to?
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Verified Answer
The correct answer is:
$\frac{1}{5}$
Let $\sin \left[\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x\right]=1$
$\Rightarrow \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\sin ^{-1} 1$
$\Rightarrow \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow x=\frac{1}{5}$
$\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\sin ^{-1} 1$
$\Rightarrow \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow x=\frac{1}{5}$
$\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$
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