Question

Derivative of $log\left(sec\theta +tan\theta \right)$ with respect to $sec\theta$ at $\theta =\pi /4$ is

• A. $0$
• B. $1$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

Let $u=log(sec\theta +tan\theta )$ and $v=sec\theta$
On differentiating both sides w.r.t. $\theta ,$ we get
$\frac{du}{d\theta }=\frac{1}{(\sec \theta +\tan \theta )}\left(\sec \theta \tan \theta +{\sec }^2\theta \right)$
and $\frac{dv}{d\theta }=\sec \theta \tan \theta$
$\therefore \frac{du}{dv}=\frac{\frac{du}{d\theta }}{\frac{dv}{d\theta }}$
$=\frac{\sec \theta (\tan \theta +\sec \theta )}{(\sec \theta +\tan \theta )\times \sec \theta \tan \theta }=\cot \theta$
$\Rightarrow \frac{du}{dv\left(\theta =\frac{\pi }{4}\right)}=\cot \frac{\pi }{4}=1$