Question

If $co{s}^{-1}x+co{s}^{-1}y=\frac{\pi }{2}$ then prove that $co{s}^{-1}x=si{n}^{-1}y$

Answer 1

Given $co{s}^{-1}x+co{s}^{-1}y=\frac{\pi }{2}$

To prove: $co{s}^{-1}x=si{n}^{-1}y$

Consider $co{s}^{-1}x+co{s}^{-1}y=\frac{\pi }{2}$

$\Rightarrow co{s}^{-1}y=\frac{\pi }{2}-co{s}^{-1}x$

$\Rightarrow y=cos(\frac{\pi }{2}-co{s}^{-1}x)$

$\Rightarrow y=sin\left(co{s}^{-1}x\right)$ (Since $cos(\frac{\pi }{2}-\theta )=sin\theta$)

$\Rightarrow si{n}^{-1}y=co{s}^{-1}x$

Hence proved.