Question

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

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

Answer 1

Answer: (D)

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