Question

If $y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$, then $\frac{dy}{dx}$ is

• A. $0$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$1$

Explanation for the correct option.

Find the value of $\frac{dy}{dx}$.

In the equation $y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$ divide numerator and denominator by $\cos x$ and simplify using trigonometric properties.

$\begin{array}{rcl}y& =& {\tan }^{-1}\left(\frac{{\displaystyle \frac{\sin x+\cos x}{\cos x}}}{{\displaystyle \frac{\cos x-\sin x}{\cos x}}}\right)\\ & =& {\tan }^{-1}\left(\frac{\tan x+1}{1-\tan x}\right)\\ & =& {\tan }^{-1}\left(\frac{\tan x+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}}{1-\tan x\tan {\displaystyle \frac{\mathrm{\pi }}{4}}}\right)\left[\tan \frac{\mathrm{\pi }}{4}=1\right]\\ & =& {\tan }^{-1}\left(\tan \left(x+\frac{\mathrm{\pi }}{4}\right)\right)\left[\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\right]\\ & =& x+\frac{\mathrm{\pi }}{4}\end{array}$

Now differentiate both sides $y=x+\frac{\mathrm{\pi }}{4}$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(x+\frac{\mathrm{\pi }}{4}\right)\\ & =& 1+0\\ & =& 1\end{array}$

Hence, the correct option is C.