Question

$x=\frac{1}{2}$ and $(x+1)(y+1)=2$ , then the radian measure of ${\cot }^{-1}x+{\cot }^{-1}y$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is D $\frac{3\pi }{4}$

Given, $x=\frac{1}{2}$ and $(1+x)(1+y)=2$

$\therefore (\frac{1}{2}+1)(1+y)=2$

$\left(\frac{3}{2}\right)(1+y)=2$

$1+y=\frac{4}{3}$

$\therefore y=\frac{1}{3}$

Hence

${\cot }^{-1}\left(x\right)+{\cot }^{-1}\left(y\right)$

$=\pi -\left[{\tan }^{-1}\right(x)+{\tan }^{-1}(y\left)\right]$

$=\pi -\left[{\tan }^{-1}\right(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\left)\right]$

$=\pi -{\tan }^{-1}\left(\frac{5}{5}\right)$

$=\pi -\frac{\pi }{4}$

$=\frac{3\pi }{4}$