Question

If $\sec \theta =x+\frac{1}{4x},x\in R,x\ne 0$, then the value of $\sec \theta +\tan \theta$ is

• A. 2x
• B. $\frac{1}{2x}$
• C. $2x$ or $\frac{1}{2x}$
• D. none of these

Answer 1

Answer: (C)

${\tan }^2\theta ={\sec }^2\theta -1$
$={\left(x+\frac{1}{4x}\right)}^2-1={\left(x-\frac{1}{4x}\right)}^2$
$\Rightarrow \tan \theta =\pm \left(x-\frac{1}{4x}\right)$
$\Rightarrow \sec \theta +\tan \theta =x+\frac{1}{4x}\pm \left(x-\frac{1}{4x}\right)$