Question

If ${\sin }^{-1}\left(\frac{x}{5}\right)+co{\sec }^{-1}\left(\frac{5}{4}\right)=\frac{\pi }{2}$ then $a$ value of $x$ is:

• A. 1
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

let $cose{c}^{-1}\left(\frac{5}{4}\right)=y=>cosecy=\frac{5}{4}=>siny=\frac{4}{5}=>y=si{n}^{-1}\left(\frac{4}{5}\right)$

$si{n}^{-1}\left(\frac{x}{5}\right)+si{n}^{-1}\left(\frac{4}{5}\right)=\frac{\pi }{2}$

$si{n}^{-1}(\frac{x}{5}\sqrt{1-\frac{16}{25}}+\frac{4}{5}\sqrt{1-\frac{x^2}{25}})=\frac{\pi }{2}$

$si{n}^{-1}(\frac{3x}{25}+\frac{4\sqrt{25-x^2}}{25})=\frac{\pi }{2}$

apply sin on both sides

$3x+4\sqrt{25-x^2}=25$

by solving this we get $x^2-6x+9=0$

so $x=3$