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The value of $\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cot ^{-1}(3)$ is
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The correct answer is:
$\frac{\pi}{4}$
Consider $\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cot ^{-1} 3$
We have, $\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\cot ^{-1} 2$
$\therefore$ From equation (i), we have
$\cos ^{-1} 2+\cot ^{-1} 3=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$
$=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}\right)$
$=\tan ^{-1}\left(\frac{5 / 6}{\frac{6-1}{6}}\right)=\tan ^{-1} 1=\frac{\pi}{4}$
We have, $\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\cot ^{-1} 2$
$\therefore$ From equation (i), we have
$\cos ^{-1} 2+\cot ^{-1} 3=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$
$=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}\right)$
$=\tan ^{-1}\left(\frac{5 / 6}{\frac{6-1}{6}}\right)=\tan ^{-1} 1=\frac{\pi}{4}$
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