Question

Find the value of x :
$2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{6}+{\tan }^{-1}\frac{1}{x}=\frac{\pi }{4}$

• A. $\frac{109}{25}$
• B. $\frac{19}{25}$
• C. $-\frac{109}{25}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{109}{25}$

$2{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{6}\right)+{\tan }^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{4}$

${\tan }^{-1}\left(\frac{1}{6}\right)+{\tan }^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{4}-2{\tan }^{-1}\left(\frac{1}{5}\right)$ - $\left(1\right)$

Now, let ${\tan }^{-1}\left(\frac{1}{6}\right)=\alpha +{\tan }^{-1}\left(\frac{1}{x}\right)=\beta$ and ${\tan }^{-1}\left(\frac{1}{5}\right)=\theta$

On taking $\tan$ on both the sides in equation $\left(1\right)$

we, get,

$\tan (\alpha +\beta )=\tan (\frac{\pi }{4}-2\theta )$ - $\left(2\right)$

We know $\begin{array}{c}\tan ({\theta }_1+{\theta }_2)=\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}\\ \tan ({\theta }_1-{\theta }_2)=\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}\end{array}\}formulae$

Using above formulae in equation $\left(2\right)$, we get

$\frac{\tan \left(\alpha \right)+\tan \beta }{1-\tan \alpha \tan \beta }=\frac{\tan \left(\frac{\pi }{4}\right)-\tan 2\theta }{1+\tan \left(\frac{\pi }{4}\right)\tan 2\theta }$

$\frac{\tan \left({\tan }^{-1}\frac{1}{6}\right)+\tan \left({\tan }^{-1}\left(\frac{1}{x}\right)\right)}{1-\tan \left({\tan }^{-1}\frac{1}{6}\right)\tan \left({\tan }^{-1}\left(\frac{1}{8}\right)\right)}=(\because \tan \left({\tan }^{-1}\right)=x)$

$\frac{\frac{1}{6}+\frac{1}{x}}{1-\frac{1}{6}-\frac{1}{x}}=\frac{1-\frac{2\tan \theta }{1-{\tan }^2\theta }}{1+\frac{2\tan \theta }{1+{\tan }^2\theta }}[\because \tan 2\theta =\frac{2\tan \theta }{1-\tan \theta }]$

$\frac{\frac{1}{6}+\frac{1}{x}}{1-\frac{1}{6x}}=\frac{1-{\tan }^2\left({\tan }^{-1}\left(\frac{1}{5}\right)\right)-2\tan \left({\tan }^{-1}\left(\frac{1}{5}\right)\right)}{1+{\tan }^2\left({\tan }^{-1}\left(\frac{1}{5}\right)\right)+2\tan \left({\tan }^{-1}\left(\frac{1}{5}\right)\right)}$

$\frac{x+6}{6x-1}=\frac{1-\frac{1}{25}-\frac{2}{5}}{1+\frac{1}{25}+\frac{2}{25}}$

Solving we get value of $x$ as

$x=\frac{109}{25}$