$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=$

• A

  $2 \pi$
• Correct answer
  B

  $\pi$
• C

  0
• D

  $-\pi$

Solution

We have, $\begin{aligned} & \sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right) \\ & \text {Let } \theta=\sin ^{-1}\left(\frac{12}{13}\right) \Rightarrow \sin \theta=\frac{12}{13} \end{aligned}$ $\begin{aligned} & \Rightarrow \tan \theta=\frac{12}{5} \Rightarrow \theta=\tan ^{-1}\left(\frac{12}{5}\right) \\ & \text {and Let } \phi=\cos ^{-1}\left(\frac{4}{5}\right) \\ & \qquad \cos \phi=\frac{4}{5} \end{aligned}$ $\Rightarrow \tan \phi=\frac{3}{4} \Rightarrow \phi=\tan ^{-1}\left(\frac{3}{4}\right)$ Put in Eq. (i) $\begin{aligned} & \tan ^{-1} \frac{12}{5}+\tan ^{-1} \frac{3}{4}+\tan ^{-1}\left(\frac{63}{16}\right) \\ & =\tan ^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}\right)+\tan ^{-1}\left(\frac{63}{16}\right) \\ & {\left[\because \tan ^{-1} A+\tan ^{-1} B=\pi+\tan ^{-1}\left(\frac{A+B}{1-A B}\right)\right.} \\ & \left.=\pi+\tan ^{-1}\left(\frac{-63}{16}\right)+\tan ^{-1} \frac{63}{16} \text { if } A B>1\right] \\ & =\pi-\tan ^{-1}\left(\frac{63}{16}\right)+\tan ^{-1}\left(\frac{63}{16}\right) \end{aligned}$ $=\pi \left[\because \tan ^{-1}(-\theta)=-\tan ^{-1} \theta\right]$

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