Question

Equation of the image of the line $x+y={\sin }^{-1}\left(a^6+1\right)+{\cos }^{-1}\left(a^4+1\right)-{\tan }^{-1}\left(a^2+1\right),a\in R$ about $x$ axis is given by

• A. $x-y=0$
• B. $x-y=\frac{\pi }{2}$
• C. $x-y=\pi$
• D. $x-y=\frac{\pi }{4}$

Answer 1

Answer: (D)

$\therefore {\sin }^{-1}\text{ is defined for }[-1,1]$
$\therefore a=0$
$\therefore x+y={\sin }^{-1}1+{\cos }^{-1}1-{\tan }^{-1}1=\frac{\pi }{4}$
Clearly image about $x$ axis will be $x-y=\frac{\pi }{4}$