$sin^{-1}\sqrt{x^2+2x+1} +sec^{-1} \sqrt{x^2+2x+1} =\frac{\pi}{2}, x ≠ 0,$ then the value of $ 2 sec^{-1}\frac{x}{2} + sin^{-1}\frac{x}{2}$ is equal to
Answer & explanation
Correct answer: option 3
We have,
$sin^{-1}\sqrt{x^2+2x+1} +sec^{-1} \sqrt{x^2+2x+1} =\frac{\pi}{2}$
$sin^{-1}\sqrt{x^2+2x+1} +cos^{-1}\frac{1}{\sqrt{x^2+2x+1}}=\frac{\pi}{2}$
$⇒ \sqrt{x^2+2x+1}=\frac{1}{\sqrt{x^2+2x+1}}$
$⇒ x^2+2x+1 =1 ⇒ x = 0, - 2$.
For x = 0, we find that $sec^{-1}\frac{x}{2}$ is not defined.
For x = -2, we have
$2sec^{-1}\frac{x}{2}+sin^{-1}\frac{x}{2}=2sec^{-1}(-1)+sin^{-1} (-1)$
$= 2 × π -\frac{\pi}{2}=\frac{3\pi}{2}$