Question

If $\sec \theta -\tan \theta =x$, then show that
$\sec \theta =\frac{1}{2}\left(x+\frac{1}{x}\right)$ and $\tan \theta =\frac{1}{2}\left(\frac{1}{x}-x\right)$

Answer 1

$\sec \theta -\tan \theta =x$ (1)

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

$\left(2\right)÷\left(1\right)$

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

$\sec \theta +\tan \theta =\frac{1}{x}$ (3)

Adding (1) & (3)

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

$\sec \theta =\frac{1}{2}\left(x+\frac{1}{x}\right)$

Substract (3) – (1)

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

$\tan \theta =\frac{1}{2}\left(\frac{1}{x}-x\right)$