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Mathematics Question on Inverse Trigonometric Functions

tan⁡⁡(2tan⁡−1⁡15+sec⁡−1⁡52+2tan⁡−1⁡18)\tan⁡(2\tan^{−1}\frac{⁡1}{5}+\sec^{−1}⁡\frac{\sqrt5}{2}+2\tan^{−1}⁡\frac{1}{8})
is equal to :

A

1

B

2

C

14\frac{1}{4}

D

54\frac{5}{4}

Answer

2

Explanation

Solution

tan⁡⁡(2tan⁡−1⁡15+sec⁡−1⁡52+2tan⁡−1⁡18)\tan⁡(2\tan^{−1}⁡\frac{1}{5} + \sec^{−1}\frac{\sqrt⁡5}{2} + 2\tan^{−1}⁡\frac{1}{8})
=tan⁡(2tan⁡−1(15+18)(1−15⋅18)+sec⁡−152)=\tan\left(2\tan^{-1}\frac{\left(\frac{1}{5}+\frac{1}{8}\right)}{\left(1-\frac{1}{5}\cdot\frac{1}{8}\right)} + \sec^{-1}{\frac{\sqrt5}{2}}\right)
=tan⁡[2tan⁡−1(13)+tan⁡−1(12)]=\tan\left[2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right)\right]
=tan⁡⁡[tan⁡−1⁡231−19+tan⁡−1⁡12]=\tan⁡[\tan^{−1}⁡ \frac{\frac{2}{3}}{1−\frac{1}{9}} + \tan^{−1}⁡\frac{1}{2}]
=tan⁡⁡[tan⁡−1⁡34+tan⁡−1⁡12]=\tan⁡[\tan^{−1}⁡ \frac{3}{4} + \tan^{−1} ⁡\frac{1}{2}]
=tan⁡⁡[tan⁡−1⁡34+121−38]=tan⁡⁡[tan⁡−1⁡5458]=\tan⁡[\tan^{−1} \frac{⁡\frac{3}{4} + \frac{1}{2} }{1−\frac{3}{8}}] = \tan⁡[\tan^{−1} ⁡\frac{\frac{5}{4}}{\frac{5}{8}}]
=tan⁡⁡[tan⁡−1⁡2]=\tan⁡[\tan^{−1} ⁡2]
= 2
So, the correct is (B): 2