Question

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

(a) $\frac{1}{5}$ (b) $\frac{2}{5}$ (c) 0 (d) 1

Answer 1

(b) We have, $cos(si{n}^{-1}\frac{2}{5}+co{s}^{-1}x)=0,$

$\Rightarrow si{n}^{-1}\frac{2}{5}+co{s}^{-1}x=co{s}^{-1}0\Rightarrow si{n}^{-1}\frac{2}{5}+co{s}^{-1}x=co{s}^{-1}cos\frac{\pi }{2}\Rightarrow si{n}^{-1}\frac{2}{5}+co{s}^{-1}x=\frac{\pi }{2}\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}\frac{2}{5}\Rightarrow co{s}^{-1}x=co{s}^{-1}\frac{2}{5}[\because co{s}^{-1}x+si{n}^{-1}x=\frac{\pi }{2}]\therefore x=\frac{2}{5}$