Question

Solve: $\sec [{\tan }^{-1}5+{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{3}{4}]$

Answer 1

We have,

$\sec [{\tan }^{-1}5+{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{3}{4}]$

$=\sec [{\tan }^{-1}\left(\frac{5+\frac{1}{5}}{1-5\times \frac{1}{5}}\right)+{\tan }^{-1}\frac{3}{4}]$

$=\sec [{\tan }^{-1}\left(\frac{\frac{26}{5}}{0}\right)+{\tan }^{-1}\frac{3}{4}]$

$=\sec [{\tan }^{-1}\infty +{\tan }^{-1}\frac{3}{4}]$

$=\sec [\frac{\pi }{2}+{\tan }^{-1}\frac{3}{4}]$

$=-\csc \left({\tan }^{-1}\frac{3}{4}\right)$

Let ${\tan }^{-1}\frac{3}{4}=x$

$\tan x=\frac{3}{4}$

$\tan x=\frac{3}{4}=\frac{p}{b}$

By Pythagoras theorem,

$h^2=p^2+b^2$

$h^2=3^2+4^2$

$h^2=25$

$h=5$

Then,

$\csc x=\frac{h}{p}=\frac{5}{3}$

$x={\csc }^{-1}\frac{5}{3}$

Now,

$=-\csc \left({\csc }^{-1}\left(\frac{5}{3}\right)\right)$

$=-\frac{5}{3}$

Hence, this is the answer.