Question

If $\sec \theta +\tan \theta =p$, then find the value of $\sec \theta -\tan \theta$.

• A. $\frac{1}{p}$
• B. $\frac{1}{p^2}$
• C. $p^2$
• D. None of these

Answer 1

The correct option is A $\frac{1}{p}$

We have,

$\sec \theta +\tan \theta =p$

$\frac{(\sec \theta +\tan \theta )}{\sec \theta -\tan \theta }$$(\sec \theta -\tan \theta )=p$

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

$\frac{1}{\sec \theta -\tan \theta }=p$

$\sec \theta -\tan \theta =\frac{1}{p}$

Hence, this is the answer.