Q66 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

$\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\tan ^{-1} \frac{16}{63}=$

  • Correct answer
    A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{6}$

Solution

$\cos ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{4}{3}\right)$ $\begin{aligned} & \sin ^{-1}\left(\frac{5}{13}\right)=\tan ^{-1}\left(\frac{5}{12}\right) \\ & \Rightarrow \cos ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\tan ^{-1}\left(\frac{16}{63}\right) \\ & =\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{16}{63}\right) \\ & =\tan ^{-1}\left[\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}\right]+\tan ^{-1}\left(\frac{16}{63}\right) \\ & =\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{16}{63}=\cot ^{-1}\left(\frac{16}{63}\right)+\tan ^{-1}\left(\frac{16}{63}\right)=\frac{\pi}{2} .\end{aligned}$

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

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