# If $\cos\bigg(\sin^{-1}\frac{2}{5}+\cos^{-1}x\bigg)=0$,then x is equal to

$(A)\;\frac{1}{15} \\ (B)\;\frac{2}{5} \\ (C)\;0 \\ (D)\;1$

Toolbox:
• $$cos \large\frac{\pi}{2}=0\:\:or\:cos^{-1}0=\large\frac{\pi}{2}$$
• $$sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}$$
Ans (B) $$\frac{2}{5}$$

Given eqn. can be written as:
$$\Rightarrow sin^{-1}\large\frac{2}{5}+cos^{-1}x=cos^{-1}0=\large\frac{\pi}{2}$$
From the above formula we can say x=$$\large\frac{2}{5}$$

edited Mar 16, 2013