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Mathematics Question on Inverse Trigonometric Functions

4 tan−115−tan−1170+tan−11994\,tan^{-1} \frac{1}{5} -tan^{-1} \frac{1}{70} + tan^{-1} \frac{1}{99} is equal to

A

π/6\pi/6

B

π/4\pi/4

C

π/3\pi/3

D

π/2\pi/2

Answer

π/4\pi/4

Explanation

Solution

We have, 4tan−115−tan−1170+tan−11994tan^{-1} \frac{1}{5} -tan^{-1} \frac{1}{70} + tan^{-1} \frac{1}{99} = 2\left\\{2\,tan^{-1} \frac{1}{5}\right\\} - \left\\{tan^{-1} \frac{1}{70} -tan^{-1} \frac{1}{99}\right\\} = 2\left\\{tan^{-1}\left(\frac{2\times1/5 }{1-\left(1/5\right)^{2}}\right)\right\\} -tan^{-1}\left\\{\frac{\frac{1}{70}-\frac{1}{99}}{1+\frac{1}{70}\times \frac{1}{99}}\right\\} = 2\,tan^{-1} \frac{5}{12}\left\\{tan^{-1}\left(\frac{29}{6931}\right)\right\\} = tan^{-1}\left\\{\frac{2\times 5/12 }{1-\left(5/12\right)^{2}}\right\\} - tan^{-1}\left(\frac{1}{239}\right) =tan−1120119−tan−11239= tan^{-1} \frac{120}{119} - tan^{-1} \frac{1}{239} = tan^{-1} \left\\{\frac{\frac{120}{119} -\frac{1}{239} }{1+\frac{120}{119}\times\frac{1}{239}}\right\\} =tan−11=π4= tan^{-1} 1 = \frac{\pi}{4}