Question

If $sec\theta =x+\frac{1}{4x}$, then $sec\theta +tan\theta =$

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

Answer 1

The correct option is B

$2x,\frac{1}{2x}$

$2x,\frac{1}{2x}$

We have,

$sec\theta =x+\frac{1}{4x}$

$\Rightarrow se{c}^2\theta =x^2+\frac{1}{16x^2}+\frac{1}{2}$

$\Rightarrow 1+ta{n}^2\theta =1+x^2+\frac{1}{16x^2}-\frac{1}{2}$

$\Rightarrow ta{n}^2\theta =x^2+\frac{1}{16x^2}-\frac{1}{2}$

$\Rightarrow ta{n}^2\theta ={(x-\frac{1}{4x})}^2$

$\therefore tan\theta =\pm .(x-\frac{1}{4x})$

$\Rightarrow sec\theta -tan\theta$

$=(x+\frac{1}{4x})-(x-\frac{1}{4x})$

or $(x+\frac{1}{4x})-[-(x-\frac{1}{4x})]$

$=2x,\frac{1}{2x}$