Question

If $\sin ({\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x)=1$, then find the value of $x.$

Answer 1

$\sin ({\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x)=1$

$\sin ({\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x)=\sin \frac{\pi }{2}$

$\therefore {\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x=\frac{\pi }{2}$

Let ${\sin }^{-1}\frac{1}{5}=\theta$, so $\sin \theta =\frac{1}{5}$ ...(1)

$\therefore \theta +{\cos }^{-1}x=\frac{\pi }{2}$

${\cos }^{-1}x=\frac{\pi }{2}-\theta$

$x=\cos (\frac{\pi }{2}-\theta )$

$x=\sin \theta =\frac{1}{5}$ ...from (1)