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Question: \(4\tan^{- 1}\frac{1}{5} - \tan^{- 1}\frac{1}{239}\) is equal to...

4tan⁡−115−tan⁡−112394\tan^{- 1}\frac{1}{5} - \tan^{- 1}\frac{1}{239} is equal to

A

π\pi

B

π2\frac{\pi}{2}

C

π3\frac{\pi}{3}

D

π4\frac{\pi}{4}

Answer

π4\frac{\pi}{4}

Explanation

Solution

Since, 2tan⁡−1x=tan⁡−12x1−x22\tan^{- 1}x = \tan^{- 1}\frac{2x}{1 - x^{2}}

∴4tan⁡−115=2[2tan⁡−115]=2tan⁡−1251−125=2tan⁡−11024=tan⁡−120241−100576=tan⁡−1120119\therefore 4\tan^{- 1}\frac{1}{5} = 2\left\lbrack 2\tan^{- 1}\frac{1}{5} \right\rbrack = 2\tan^{- 1}\frac{\frac{2}{5}}{1 - \frac{1}{25}} = 2\tan^{- 1}\frac{10}{24} = \tan^{- 1}\frac{\frac{20}{24}}{1 - \frac{100}{576}} = \tan^{- 1}\frac{120}{119}

So,4tan⁡−115−tan⁡−11239=tan⁡−1120119−tan⁡−112394\tan^{- 1}\frac{1}{5} - \tan^{- 1}\frac{1}{239} = \tan^{- 1}\frac{120}{119} - \tan^{- 1}\frac{1}{239} =tan⁡−1120119−12391+120119.1239=tan⁡−1(120×239)−119(119×239)+120=tan⁡−11=π4= \tan^{- 1}\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119}.\frac{1}{239}} = \tan^{- 1}\frac{(120 \times 239) - 119}{(119 \times 239) + 120} = \tan^{- 1}1 = \frac{\pi}{4}