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If $\sin ^{-1} x+\cot ^{-1}(1 / 2)=\pi / 2$, then what is the value of $x$ ?
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Verified Answer
The correct answer is:
$1 / \sqrt{5}$
Let $\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$
As we know $\cot ^{-1} x=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)$
$\therefore \sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1}\left(\frac{1}{\sqrt{1+\frac{1}{4}}}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\frac{\pi}{2}$
Now, $\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}$
$\therefore \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\cos ^{-1} \sqrt{1-\frac{4}{5}}=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$
From equation (1), we have
$\sin ^{-1} x+\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{\pi}{2}$
since, $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$x=\frac{1}{\sqrt{5}}$
As we know $\cot ^{-1} x=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)$
$\therefore \sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1}\left(\frac{1}{\sqrt{1+\frac{1}{4}}}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\frac{\pi}{2}$
Now, $\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}$
$\therefore \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\cos ^{-1} \sqrt{1-\frac{4}{5}}=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$
From equation (1), we have
$\sin ^{-1} x+\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{\pi}{2}$
since, $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$x=\frac{1}{\sqrt{5}}$
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