Question

The derivative of $\ln (\sec \theta +\tan \theta )$ with respect to $\sec \theta$ at $\theta =\frac{\pi }{4}$ is

Answer 1

Let $y=\ln (\sec \theta +\tan \theta ),z=\sec \theta$
Now,
$\frac{dy}{dz}=\frac{\frac{dy}{d\theta }}{\frac{dz}{d\theta }}\Rightarrow \frac{dy}{dz}=\frac{\frac{\sec \theta \tan \theta +{\sec }^2\theta }{\sec \theta +\tan \theta }}{\sec \theta \tan \theta }\Rightarrow \frac{dy}{dz}=\frac{\sec \theta }{\sec \theta \tan \theta }=\frac{1}{\tan \theta }\therefore {\left(\frac{dy}{dz}\right)}_{\theta =\pi /4}=1$