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Question: \[4\tan^{- 1}\frac{1}{5} - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99} =\]...

4tan⁡−115−tan⁡−1170+tan⁡−1199=4\tan^{- 1}\frac{1}{5} - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99} =

A

π2\frac{\pi}{2}

B

π3\frac{\pi}{3}

C

π4\frac{\pi}{4}

D

None of these

Answer

π4\frac{\pi}{4}

Explanation

Solution

4tan⁡−115−tan⁡−1170+tan⁡−11994\tan^{- 1}\frac{1}{5} - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99}

= 2tan⁡−1[251−125]−tan⁡−1170+tan⁡−11992\tan^{- 1}\left\lbrack \frac{\frac{2}{5}}{1 - \frac{1}{25}} \right\rbrack - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99}

= 2tan⁡−1(512)−tan⁡−1170+tan⁡−11992\tan^{- 1}\left( \frac{5}{12} \right) - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99}

= tan⁡−1[561−25144]−tan⁡−1170+tan⁡−1199\tan^{- 1}\left\lbrack \frac{\frac{5}{6}}{1 - \frac{25}{144}} \right\rbrack - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99}

= tan⁡−1(120119)−tan⁡−1170+tan⁡−1199\tan^{- 1}\left( \frac{120}{119} \right) - \tan^{- 1}\frac{1}{70} + \tan^{- 1}\frac{1}{99}

= tan⁡−1120119+tan⁡−1[199−1701+199.170]\tan^{- 1}\frac{120}{119} + \tan^{- 1}\left\lbrack \frac{\frac{1}{99} - \frac{1}{70}}{1 + \frac{1}{99}.\frac{1}{70}} \right\rbrack= tan⁡−1120119+tan⁡−1(−296931)\tan^{- 1}\frac{120}{119} + \tan^{- 1}\left( - \frac{29}{6931} \right)

= tan⁡−1120119−tan⁡−1296931=tan⁡−1120119−tan⁡−11239\tan^{- 1}\frac{120}{119} - \tan^{- 1}\frac{29}{6931} = \tan^{- 1}\frac{120}{119} - \tan^{- 1}\frac{1}{239}

= tan⁡−1[120119−12391+120119×1239]=tan⁡−1(1)=π4\tan^{- 1}\left\lbrack \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}} \right\rbrack = \tan^{- 1}(1) = \frac{\pi}{4}