Question

If $\sec \theta =x+\frac{1}{4x}$
$P.T.\sec \theta +\tan \theta =2x$ or $\frac{1}{2x}$

Answer 1

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

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

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

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

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

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

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

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

When $\tan \theta =-(x-\frac{1}{4x}),\sec \theta +\tan \theta =x+\frac{1}{4x}-(x-\frac{1}{4x})=\frac{1}{2x}$

Therefore,

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