Question

Differentiate the following w.r.t.x:
${\tan ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} \right)$

Answer 1

Make use of he standard formula which says $\left[ {\dfrac{{\tan A + \tan B}}{{1 - \tan A\,\,\tan B}} = \tan \left( {A + B} \right)} \right]$

Complete step by step solution:
$y = {\tan ^{ - 1}}\left( {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} \right)$
Taking $\cos x$common in the numerator and denominator, we will get
$y = {\tan ^{ - 1}}\left[ {\dfrac{{\cos x\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)}}{{\cos x\left( {1 - \dfrac{{\sin x}}{{\cos x}}} \right)}}} \right]\,\,\,$
As we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$y = {\tan ^{ - 1}}\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)$
As we know that $\tan \dfrac{\pi }{4} = 1$
$\tan y = \dfrac{{\left( {\tan \dfrac{\pi }{4} + \tan x} \right)}}{{1 - \tan \dfrac{\pi }{4} \times \tan x}}$ $\left[ {\therefore \dfrac{{\tan A + \tan B}}{{1 - \tan A\,\,\tan B}} = \tan \left( {A + B} \right)} \right]\,$
Then, by using the formula $\left[ {\dfrac{{\tan A + \tan B}}{{1 - \tan A\,\,\tan B}} = \tan \left( {A + B} \right)} \right]\,$
$\tan y = \tan \left( {\dfrac{\pi }{4} + x} \right)$
Equating angles, when the trigonometric are the same
$y = \dfrac{\pi }{4} + x$
Now, by differentiating on both sides of the equation with respect to x, we will have.
$\dfrac{{dy}}{{dx}} = 0 + 1\;\;\;\;\;\;\;\;\;\left[ {\therefore \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4}} \right) = 0} \right] \\\ = 1\;\;\;\;\;\;\;\;\;\;\;\;\; \\\$

Note: The inverse trigonometric functions have suitably restricted domains. So, when solving these problems check if the domain of the function is asked/given and proceed accordingly