Question

If point $P(x,y)$ completely lies in the first quadrant, then the range of ${\cos }^{-1}x+{\cot }^{-1}(-\frac{1}{y})+{\tan }^{-1}y$ is

• A. $[\pi ,\frac{3\pi }{2}]$
• B. $(\pi ,\frac{3\pi }{2})$
• C. $[0,\frac{\pi }{2}]$
• D. $(0,\frac{\pi }{2})$

Answer 1

The correct option is B $(\pi ,\frac{3\pi }{2})$

Given : $P(x,y)$ lies in first quadrant
$\Rightarrow x>0,y>0$
for ${\cos }^{-1}x$ to be defined $x\in (0,1)\Rightarrow {\cos }^{-1}x\in (0,\frac{\pi }{2})$
Now,
${\cos }^{-1}x+{\cot }^{-1}(-\frac{1}{y})+{\tan }^{-1}y$
$={\cos }^{-1}x+\pi +{\tan }^{-1}(-y)+{\tan }^{-1}y$
$(\because {\cot }^{-1}\left(\frac{1}{-y}\right)=\pi +{\tan }^{-1}(-y);y<0)$
$={\cos }^{-1}x+\pi -{\tan }^{-1}y+{\tan }^{-1}y$
$=\pi +{\cos }^{-1}x\in (\pi ,\frac{3\pi }{2})$