Question

if $\sec \theta +\tan \theta =x,$, then what is the value of $\sec \theta$ ?

Answer 1

Consider the given equation.

$\sec \theta +\tan \theta =x$ ……. (1)

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

$\frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }=x$

We know that

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

Therefore,

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

$\sec \theta -\tan \theta =\frac{1}{x}$ …….. (2)

From equation (1) and (2), we get

$2\sec \theta =x+\frac{1}{x}$

$2\sec \theta =\frac{1+x^2}{x}$

$\sec \theta =\frac{1+x^2}{2x}$

Hence, the value of $\sec \theta$ is $\frac{1+x^2}{2x}$.