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

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

Answer 1

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

$se{c}^{-1}t={\cos }^{-1}\frac{1}{t}$ and ${\sin }^{-1}z+{\cos }^{-1}z=\frac{\pi }{2}$
$\because {\sin }^{-1}\sqrt{x^2+2x+1}+{\cos }^{-1}\frac{1}{\sqrt{x^2+2x+1}}=\frac{\pi }{2}$
Above implies that $\sqrt{x^2+2x+1}=\frac{1}{\sqrt{x^2+2x+1}}$
or $x^2+2x+1=1$ or $x(x+2)=0$
$\therefore x=0,-2\therefore x=-2asx\ne 0$
$2se{c}^{-1}\frac{x}{2}+{\sin }^{-1}\frac{x}{2}$
$=2se{c}^{-1}(-1)+{\sin }^{-1}(-1)$
$=2{\cos }^{-1}(-1)+{\sin }^{-1}(-1)$
$={\cos }^{-1}(-1)+{\cos }^{-1}(-1)+{\sin }^{-1}(-1)$
$=\pi +\frac{\pi }{2}=\frac{3\pi }{2}$