Question

If $\sec \theta +\tan \theta =x$ then show that: $\sec \theta =\frac{1}{2}[x+\frac{1}{x}]$ and $\tan \theta =\frac{1}{2}[x-\frac{1}{x}]$

Answer 1

We have,

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

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

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

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

Thus,

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

Adding and subtracting these two equations.

we get,

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

Therefore,$sec\theta =\frac{1}{2}(x+\frac{1}{x})andtan\theta =\frac{1}{2}(x-\frac{1}{x})$