Question

If $cos\left(si{n}^{-1}\frac{2}{5}+co{s}^{-1}x\right)=0$, then $x$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $0$
• D. $1$

Answer 1

Answer: (B)

$cos\left(si{n}^{-1}\frac{2}{5}+co{s}^{-1}x\right)=0$
$\Rightarrow si{n}^{-1}\frac{2}{5}+co{s}^{-1}x=co{s}^{-1}0=\frac{\pi }{2}$
$\Rightarrow si{n}^{-1}\frac{2}{5}+\frac{\pi }{2}-si{n}^{-1}x=\frac{\pi }{2}$
$\Rightarrow si{n}^{-1}x=si{n}^{-1}\left(\frac{2}{5}\right)$
$\Rightarrow x=\frac{2}{5}$