Question

If tan $y=\frac{sin\,x+cos\,x}{cos\,x-sin\,x}$ , then $\frac{dy}{dx}=$

• A. $1$
• B. $cosx$
• C. $sinx$
• D. $-1$

Answer 1

Answer: (A)

$1$

Answer 2

We have, $tan\,y = \frac{sin\,x + cos\,x}{cos\,x - sin\,x}$
$= \frac {1 + tan\,x}{1-tan\,x} = tan(\frac{\pi}{4} + x)$
$\Rightarrow y = tan^{-1} (tan(\frac{\pi}{4} + x))$
$= \frac {\pi}{4} + x$
Differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 1$