Question

If $sin\{{\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x\}=1$, then $x$ is equal to

• A. $1$
• B. $0$
• C. $\frac{4}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is D $\frac{1}{5}$

Given, $sin(si{n}^{-1}\left(\frac{1}{5}\right)+co{s}^{-1}(x\left)\right)=1$

$si{n}^{-1}\left(\frac{1}{5}\right)+co{s}^{-1}\left(x\right)=si{n}^{-1}\left(1\right)$

$si{n}^{-1}\left(\frac{1}{5}\right)+co{s}^{-1}\left(x\right)=\frac{\pi }{2}$

$co{s}^{-1}\left(x\right)=\frac{\pi }{2}-si{n}^{-1}\left(\frac{1}{5}\right)$

$co{s}^{-1}\left(x\right)=co{s}^{-1}\left(\frac{1}{5}\right)$

$\therefore x=\frac{1}{5}$