What is the value of $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$ is equal to
What is the value of $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$ is equal to
- $\pi$
- $\frac{\pi}{2}$
- $\frac{\pi}{6}$
- $\frac{3\pi}{4}$
Solution
$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)$
$\begin{array}{r}=\sin ^{-1}\left(\frac{12}{13}\right)+\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right) \\ \left\{\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right\} \\ =\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{4}{5}\right)+\frac{\pi}{2}+\sin ^{-1}\left(\frac{63}{\sqrt{(63)^2+(16)^2}}\right) \\ \quad\left\{\because \sin ^{-1} x \pm \sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^2} \pm y \sqrt{\left.1-x^2\right)}\right\}\right. \\ =\sin ^{-1}\left\{\frac{12}{13} \cdot \sqrt{1-\left(\frac{4}{5}\right)^2}-\frac{4}{5} \cdot \sqrt{1-\left(\frac{12}{13}\right)^2}\right\} \\ +\frac{\pi}{2}+\sin ^{-1}\left(\frac{63}{65}\right)\end{array}$
$\begin{aligned} & =\sin ^{-1}\left\{\frac{12}{13} \cdot \frac{3}{5}-\frac{4}{5} \cdot \frac{5}{13}\right\}+\frac{\pi}{2}+\sin ^{-1}\left(\frac{63}{65}\right) \\ & =\sin ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{63}{65}\right)+\frac{\pi}{2} \\ & =\sin ^{-1}\left\{\frac{16}{65} \cdot \sqrt{1-\left(\frac{63}{65}\right)^2}+\frac{63}{65} \cdot \sqrt{1-\left(\frac{16}{65}\right)^2}\right\}+\frac{\pi}{2} \\ & =\sin ^{-1}(1)+\frac{\pi}{2} \Rightarrow \frac{\pi}{2}+\frac{\pi}{2}=\pi\end{aligned}$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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