Question

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

Answer 1

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

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

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

$=\sqrt{x-{\left(\frac{1}{4x}\right)}^2}$

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

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

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