Question

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

• A. $\frac{1}{2}(p-\frac{1}{p})$
• B. $\frac{1}{2}(p+\frac{1}{p})$
• C. $(p-\frac{1}{p})$
• D. $(p+\frac{1}{p})$

Answer 1

The correct option is A $\frac{1}{2}(p-\frac{1}{p})$

$\sec \theta +\tan \theta =p$ ...(1)

Now $1+{\tan }^2\theta ={\sec }^2\theta \Rightarrow {\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow (\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$

$\Rightarrow \sec \theta -\tan \theta =\frac{1}{\sec \theta +\tan \theta }$

$\Rightarrow \sec \theta -\tan \theta =\frac{1}{p}$ ...(2)

Subtracting (2) from (1), we get
$\sec \theta +\tan \theta -\sec \theta +\tan \theta =p-\frac{1}{p}$

$\Rightarrow 2\tan \theta =(p-\frac{1}{p})\Rightarrow \tan \theta =\frac{1}{2}(p-\frac{1}{p})$