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Question

Mathematics Question on Inverse Trigonometric Functions

What is the value of tan⁡−1(mn)−tan⁡−1(m−nm+n)?\tan^{-1} \left(\frac{m}{n}\right) - \tan^{-1} \left(\frac{m-n}{m+n}\right) ?

A

π\pi

B

π2\frac{\pi}{2}

C

π4\frac{\pi}{4}

D

π3\frac{\pi}{3}

Answer

π4\frac{\pi}{4}

Explanation

Solution

The given expression is : tan⁡−1(mn)−tan⁡−1(m−nm+n)\tan^{-1} \left(\frac{m}{n}\right) - \tan^{-1} \left(\frac{m-n}{m+n}\right) =tan⁡−1(mn)−tan⁡−1(1−nm1+nm) = \tan^{-1} \left(\frac{m}{n}\right) - \tan^{-1} \left(\frac{1- \frac{n}{m}}{1 + \frac{n}{m}}\right) =tan⁡−1(mn)−tan⁡−1(1)+tan⁡−1(mn) = \tan^{-1} \left(\frac{m}{n}\right) - \tan^{-1} \left(1\right) + \tan^{-1}\left(\frac{m}{n}\right) =tan⁡−1(mn)+cot⁡−1(mn)−π4= \tan^{-1} \left(\frac{m}{n}\right) + \cot^{-1} \left(\frac{m}{n}\right) - \frac{\pi}{4} =π2−π4=π4= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}