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Mathematics Question on Inverse Trigonometric Functions

If sin−1(x5)+cos ec−1(54)=π2sin^{-1}\left(\frac{x}{5}\right)+cos\,ec^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2} then a value of xx is

A

11

B

33

C

44

D

55

Answer

33

Explanation

Solution

sin−1x5+sin−145=π2sin^{-1}\frac{x}{5}+sin^{-1}\frac{4}{5}=\frac{\pi}{2} ⇒sin−1x5=cos−145⇒sin−1x5=sin−135\Rightarrow sin^{-1} \frac{x}{5}=cos^{-1} \frac{4}{5} \Rightarrow sin^{-1} \frac{x}{5}=sin^{-1} \frac{3}{5} ∴x=3.\therefore x=3.