Question

If $\sec \theta +tan\theta =x$, then $\sec \theta =........$

• A. $\frac{x^2+1}{x}$
• B. $\frac{x^2+1}{2x}$
• C. $\frac{x^2-1}{2x}$
• D. $\frac{x^2-1}{x}$

Answer 1

The correct option is B $\frac{x^2+1}{2x}$

We have,

$\sec \theta +\tan \theta =x$ ……. (1)

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

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

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)

On adding the equation (1) and (2), we get

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

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

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

Hence, this is the answer.