Question

If $x>0,y>0$ and $x>y$, then ${\tan }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(\frac{x+y}{x-y}\right)$ is equal to

• A. $-\frac{\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. none of these

Answer 1

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

We can write ${\tan }^{-1}\left(\frac{x-y}{x+y}\right)$ $={\tan }^{-1}\left(\frac{1-\frac{x}{y}}{1+\frac{x}{y}}\right)$
$={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(\frac{x}{y}\right)$
Now $x>y$
Hence
$\pi +{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(\frac{x}{y}\right)$
$=\frac{3\pi }{4}-{\tan }^{-1}\left(\frac{x}{y}\right)$.
Adding ${\tan }^{-1}\left(\frac{x}{y}\right)$ in the above equation, we get $\frac{3\pi }{4}$