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Question

Mathematics Question on Inverse Trigonometric Functions

Solve tan−1(xy)−tan−1(x−yx+y)tan^{-1} (\frac xy)-tan^{-1}(\frac {x-y}{x+y}) is equal to

A

π2\frac {\pi}{2}

B

π3\frac {\pi}{3}

C

π4\frac {\pi}{4}

D

−3π2-\frac {3\pi}{2}

Answer

π4\frac {\pi}{4}

Explanation

Solution

Given :
tan⁡−1(xy)−tan⁡−1x−yx+y\tan^{-1}(\frac{x}{y})-\tan^{-1}\frac{x-y}{x+y}
=tan⁡−1[xy−x−yx+y1+(xy)(x−yx+y)]=\tan^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+(\frac{x}{y})(\frac{x-y}{x+y})}\right]

=tan⁡−1[x(x+y)−y(x−y)y(x+y)y(x+y)+x(x−y)y(x+y)]=\tan^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]
=tan⁡−1(x2+xy−xy+y2xy+y2+x2−xy)=\tan^{-1}\left(\frac{x^2+xy-xy+y^2}{xy+y^2+x^2-xy}\frac{}{}\right)
=tan⁡−1(x2+y2x2+y2)=\tan^{-1}\left(\frac{x^2+y^2}{x^2+y^2}\right)
=tan⁡−11=π4=\tan^{-1}1=\frac{\pi}{4}
So, the correct option is (C) : π4\frac{\pi}{4}.