Question

If $si{n}^{-1}\sqrt{x^2+2x+1}+se{c}^{-1}\sqrt{x^2+2x+1}=\frac{\pi }{2},x\ne 0$ then the value of $2se{c}^{-1}\left(\frac{x}{2}\right)+si{n}^{-1}\left(\frac{x}{2}\right)$ is equal to

• A. $-\frac{3\pi }{2}$
• B. $\frac{3\pi }{2}$
• C. $-\frac{\pi }{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{3\pi }{2}$

Given,$si{n}^{-1}\sqrt{x^2+2x+1}+se{c}^{-1}\sqrt{x^2+2x+1}=\frac{\pi }{2}$

$\Rightarrow si{n}^{-1}\sqrt{(x+1{)}^2}+se{c}^{-1}\sqrt{(x+1{)}^2}=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}(x+1)+{\sec }^{-1}(x+1)=\frac{\pi }{2}$

$\Rightarrow {\sec }^{-1}(x+1)=\frac{\pi }{2}-{\sin }^{-1}(x+1)$

$\Rightarrow {\sec }^{-1}(x+1)={\cos }^{-1}(x+1)$

$\Rightarrow {\cos }^{-1}\left(\frac{1}{x+1}\right)={\cos }^{-1}(x+1)$

$\Rightarrow \frac{1}{x+1}=x+1$

$\Rightarrow (x+1{)}^2=1$

$\Rightarrow x+1=\pm 1$

$x=0$ and $x=-2$

However as ${\cos }^{-1}(x+1)$ to be defined it should lie between $[-1,1]$

$\Rightarrow -1\le (x+1)\le 1$

$\Rightarrow -2\le x\le 0$

$2{\sec }^{-1}(-1)+{\sin }^{-1}(-1)$ ...since $x\ne 0$

$=2\pi -\frac{\pi }{2}$

$=\frac{3\pi }{2}$