Q117 / 170 · Manipal (MET) 2018 · Mathematics

$\sin ^{-1} \frac{1}{\sqrt{5}}+\cos ^{-1} \frac{3}{\sqrt{10}}$ is equal to

  • A
    $\pi / 6$
  • Correct answer
    B
    $\pi / 4$
  • C
    $\pi / 3$
  • D
    $2 \pi / 3$

Solution

$\sin ^{-1} \frac{1}{\sqrt{5}}+\cos ^{-1} \frac{3}{\sqrt{10}}$ Now, $\sin ^{-1} \frac{1}{\sqrt{5}}=\tan ^{-1}\left(\frac{1 / \sqrt{5}}{\sqrt{1-\frac{1}{5}}}\right)=\tan ^{-1} \frac{1}{2}$ and $\cos ^{-1} \frac{3}{\sqrt{10}}=\tan ^{-1}\left(\frac{\sqrt{1-\frac{9}{10}}}{\frac{3}{\sqrt{10}}}\right)=\tan ^{-1} \frac{1}{3}$ $\therefore \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan ^{-1}(1)=\frac{\pi}{4}$

Hence, the correct answer is option: 'B'.

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