Question

$\frac{1-\sec \theta +\tan \theta }{\sec \theta +\tan \theta -1}$ is equal to

• A. $\sec \theta +\tan \theta$
• B. $\sin \theta +\cos \theta$
• C. $\sec \theta -\tan \theta$
• D. $\sin \theta -\cos \theta$

Answer 1

The correct option is C $\sec \theta -\tan \theta$

$\frac{1-\sec \theta +\tan \theta }{\sec \theta +\tan \theta -1}$
$=\frac{({\sec }^2\theta -{\tan }^2\theta )-(\sec \theta -\tan \theta )}{\sec \theta +\tan \theta -1}$
$=\frac{(\sec \theta -\tan \theta )(\sec \theta +\tan \theta -1)}{(\sec \theta +\tan \theta -1)}$
$=\sec \theta -\tan \theta$