Question

If $y=\frac{\sin \left(x+9\right)}{\cos x}$, then $\frac{dy}{dx}$ at $x=0$, is

Answer 1

$y=\frac{\sin (x+9)}{\cos x}$

$y=\frac{\sin x\cos 9+\cos x\sin 9}{\cos x}$

$y=\cos 9\tan x+\sin 9$

$\Rightarrow \frac{dy}{dx}=\cos 9{\sec }^{2}x$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{x=0}=\cos 9{\sec }^{2}0=\cos 9$