Practicing Success
$sin^{-1}\sqrt{x^2+2x+1} +sec^{-1} \sqrt{x^2+2x+1} =\frac{\pi}{2}, x ≠ 0,$ then the value of $ 2 sec^{-1}\frac{x}{2} + sin^{-1}\frac{x}{2}$ is equal to |
$-\frac{\pi}{2}$ only $\begin{Bmatrix}-\frac{3\pi}{2} ,\frac{\pi}{2}\end{Bmatrix}$ $\frac{3\pi}{2}$ only $-\frac{3\pi}{2}$ only |
$\frac{3\pi}{2}$ only |
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 $2sec^{-1}\frac{x}{2}+sin^{-1}\frac{x}{2}=2sec^{-1}(-1)+sin^{-1} (-1)$ $= 2 × π -\frac{\pi}{2}=\frac{3\pi}{2}$ |