Question

If ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-y^2}{1+y^2}\right)=\frac{\pi }{2}$, where $xy<1$, then

• A. $x-y-xy=1$
• B. $x-y+xy=1$
• C. $x+y-xy=1$
• D. $x+y+xy=1$
• E. $y-x-xy=1$

Answer 1

The correct option is D $x+y+xy=1$

Given, ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-y^2}{1+y^2}\right)=\frac{\pi }{2}$
$\Rightarrow 2{\tan }^{-1}x+2{\tan }^{-1}y=\frac{\pi }{2}$
$\Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{x+y}{1-xy}=\tan \frac{\pi }{4}=1$
$\Rightarrow x+y=1-xy$

$\Rightarrow x+y+xy=1$