Question

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

• A. 1
• B. 2
• C. \(\frac{1}{4}\)
• D. \(\frac{5}{4}\)

Answer 1

The Correct Option is B

\(\tan⁡(2\tan^{−1}⁡\frac{1}{5} + \sec^{−1}\frac{\sqrt⁡5}{2} + 2\tan^{−1}⁡\frac{1}{8})\)
\(=\tan\left(2\tan^{-1}\frac{\left(\frac{1}{5}+\frac{1}{8}\right)}{\left(1-\frac{1}{5}\cdot\frac{1}{8}\right)} + \sec^{-1}{\frac{\sqrt5}{2}}\right)\)
\(=\tan\left[2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right)\right]\)
\(=\tan⁡[\tan^{−1}⁡ \frac{\frac{2}{3}}{1−\frac{1}{9}} + \tan^{−1}⁡\frac{1}{2}]\)
\(=\tan⁡[\tan^{−1}⁡ \frac{3}{4} + \tan^{−1} ⁡\frac{1}{2}]\)
\(=\tan⁡[\tan^{−1} \frac{⁡\frac{3}{4} + \frac{1}{2} }{1−\frac{3}{8}}] = \tan⁡[\tan^{−1} ⁡\frac{\frac{5}{4}}{\frac{5}{8}}]\)
\(=\tan⁡[\tan^{−1} ⁡2]\)
= 2
So, the correct is (B): 2