Question

The value of ${\tan }^{-1}\left(\frac{m}{n}\right)-{\tan }^{-1}\left(\frac{m-n}{m+n}\right)$ is

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

Answer 1

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

We have,

${\tan }^{-1}\left(\frac{m}{n}\right)-{\tan }^{-1}\left(\frac{m-n}{m+n}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{m}{n}-\frac{m-n}{m+n}}{1+\frac{m}{n}\times \frac{m-n}{m+n}}\right)[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)]$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{m(m+n)-n(m-n)}{n(m+n)}}{\frac{n(m+n)+m(m-n)}{n(m+n)}}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{m^2+mn-mn+n^2}{mn+n^2+m^2-mn}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{m^2+n^2}{n^2+m^2}\right)$

$\Rightarrow {\tan }^{-1}\left(1\right)$

$\Rightarrow {\tan }^{-1}\left(\tan \frac{\pi }{4}\right)$

$\Rightarrow \frac{\pi }{4}$

Hence, this is the answer.