Question

if sin^-1x+sin^-1y=pi/2 then find cos^-1x+cos^-1y

Answer 1

$$\begin{gathered} {\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(y\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \cos e{c}^{-1}\left(\frac{1}{x}\right)+\cos e{c}^{-1}\left(\frac{1}{y}\right)=\frac{\mathrm{\pi }}{2} \end{gathered}$$

$\Rightarrow \frac{\mathrm{\pi }}{2}-se{c}^{-1}\left(\frac{1}{x}\right)+\frac{\mathrm{\pi }}{2}-se{c}^{-1}\left(\frac{1}{y}\right)=\frac{\mathrm{\pi }}{2}$ [Since, $\cos e{c}^{-1}\left(x\right)+se{c}^{-1}\left(x\right)=\frac{\mathrm{\pi }}{2}$]

$$\begin{gathered} \Rightarrow \mathrm{\pi }-se{c}^{-1}\left(\frac{1}{x}\right)-se{c}^{-1}\left(\frac{1}{y}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{\mathrm{\pi }}{2}=se{c}^{-1}\left(\frac{1}{x}\right)+se{c}^{-1}\left(\frac{1}{y}\right) \\ \Rightarrow \frac{\mathrm{\pi }}{2}={\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right) \end{gathered}$$

Thus, ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)=\frac{\mathrm{\pi }}{2}$