Question

Simplify ${\tan ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} \right)$

Answer 1

Hint:
It is given in the question that we have to simplify ${\tan ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} \right)$
Firstly, we will multiply whole equation with cos x then
After that applying property $\left( {\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right) = \tan \left( {A + B} \right)$ we will get answer.

Complete step by step solution:
It is given in the question that we have to simplify ${\tan ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} \right)$
 $\because {\tan ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} \right)$
Now, divide the whole equation by cos x,

 $\therefore {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\cos x}}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{\cos x}}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}}} \right)$
 $\therefore {\tan ^{ - 1}}\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)$
Since we can write 1 as tan $\dfrac{\pi }{4}$
 $\therefore {\tan ^{ - 1}}\left( {\dfrac{{\tan \dfrac{\pi }{4} + \tan x}}{{1 - \tan \dfrac{\pi }{4}\tan x}}} \right)$
Now, applying property of $\left( {\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right) = \tan \left( {A + B} \right)$
 $\therefore {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{\pi }{4} + x} \right)} \right]$

$\therefore \dfrac{\pi }{4} + x$

Note:
Some properties of $\tan \theta $ :
1) ${\tan ^2}\theta = {\sec ^2}\theta - 1$
2) $\tan \left( { - \theta } \right) = - \tan \theta $
3) $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
4) $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
5) $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$