Question

If $\sec \theta =x+\frac{1}{4x}$, prove that $\sec \theta +\tan \theta =2x$ or $\frac{1}{2x}$

Answer 1

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

$\Rightarrow {\sec }^2\theta ={(x+\frac{1}{4x})}^2$

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

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

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

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

$\Rightarrow \sec \theta +\tan \theta =x+\frac{1}{4x}+x-\frac{1}{4x}=2x$

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

$=\frac{2}{4x}$

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

Hence, the answer is proved.