Question

Prove that
$\frac{{\sec }^2\theta -{\tan }^2\theta +\tan \theta }{\sec \theta }=\sin \theta +\cos \theta$

Answer 1

Now,

$\frac{{\sec }^2\theta -{\tan }^2\theta +\tan \theta }{\sec \theta }$

$=\frac{1+\tan \theta }{\sec \theta }$ [ Since ${\sec }^2\theta -{\tan }^2\theta =1$]

$=\cos \theta +\sin \theta$.