Question

If $y = \tan^{ - 1}\left[ {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right]$. Then what is the value of $\dfrac{{dy}}{{dx}}$?
A. 0
B. $\dfrac{\pi }{4}$
C. 1
D. $\dfrac{1}{2}$

Answer 1

Hint: Simplify the given trigonometric equation by multiplying the numerator and denominator by $\cos x$. Convert the equation in terms of a trigonometric function tan. Then use the property of inverse trigonometric function. In the end, differentiate the equation with respect to $x$ and get the required answer.

Formula Used:
$\dfrac{{\sin A}}{{\cos A}} = \tan A$
$\tan\left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \tan B}}$

Complete step by step solution:
The given trigonometric equation is $y = \tan^{ - 1}\left[ {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right]$.
Let’s simplify the above equation.
Divide the numerator and denominator on the right-hand side by $\cos x$.
$y = \tan^{ - 1}\left[ {\dfrac{{\dfrac{{\sin x + \cos x}}{{\cos x}}}}{{\dfrac{{\cos x - \sin x}}{{\cos x}}}}} \right]$
$ \Rightarrow y = \tan^{ - 1}\left[ {\dfrac{{\tan x + 1}}{{1 - \tan x}}} \right]$ [Since $\dfrac{{sin A}}{{cos A}} = tan A$]
$ \Rightarrow y = \tan^{ - 1}\left[ {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right]$
$ \Rightarrow y = \tan^{ - 1}\left[ {\dfrac{{\tan\left( {\dfrac{\pi }{4}} \right) + \tan x}}{{1 - \tan\left( {\dfrac{\pi }{4}} \right)\tan x}}} \right]$ [Since $\tan\left( {\dfrac{\pi }{4}} \right) = 1$]
Now apply the trigonometric identity $\tan\left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \tan B}}$.
We get,
$y = \tan^{ - 1}\left[ {\tan\left( {\dfrac{\pi }{4} + x} \right)} \right]$
Apply the inverse trigonometry rule $\tan^{ - 1}\left( {\tan A} \right) = A$.
$y = \dfrac{\pi }{4} + x$

Differentiate the above equation with respect to $x$.
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} + x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \left( {0 + 1} \right)$ [Since the derivative of constant term is zero.]
$ \Rightarrow \dfrac{{dy}}{{dx}} = 1$

Option ‘C’ is correct

Note: Remember the formula of $\tan\left( {A + B} \right)$ is $\tan\left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \tan B}}$. The chain rule of differentiation can also be used to resolve the problem which makes it easier to examine the behavior of function step by step.