Question

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

Answer 1

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

We know that $ta{n}^2\theta =se{c}^2\theta -1$

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

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

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

$=x^2+\frac{1}{16x^2}-2\cdot x\cdot \frac{1}{4x}$

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

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

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

$\Rightarrow sec\theta +tan\theta =2x$