# Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

$\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]=?$

A. $\frac{7}{17}$

B. $\frac{-7}{17}$

C. $\frac{7}{12}$

D. $\frac{-7}{12}$

Solution:

To Find: The value of $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)$

Consider, $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(\tan ^{-1}\left(\frac{2\left(\frac{1}{5}\right)}{1-\left(\frac{1}{5}\right)^{2}}\right)-\frac{\pi}{4}\right)$

$\left(\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)$

$=\tan \left(\tan ^{-1}\left(\frac{\frac{2}{5}}{1 \frac{1}{25}}\right)-\frac{\pi}{4}\right)$

$=\tan \left(\tan ^{-1}\left(\frac{5}{12}\right)-\frac{\pi}{4}\right)$

$=\tan \left(\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)\right)\left(\because \tan \left(\frac{\pi}{4}\right)=1\right)$

$=\tan \left(\tan ^{-1}\left(\frac{\frac{5}{12}-1}{1 \frac{5}{12}}\right)\right)$

$\left(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right.$

$=\tan \left(\tan ^{-1}\left(\frac{-7}{17}\right)\right)$

$\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\frac{-7}{17}$