Question

If $\tan \theta -\sec \theta =p$ then $\tan \theta +\sec \theta =......$

Answer 1

Given : $tan\theta -sec\theta =P$

To find : $tan\theta +sec\theta$

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

$\Rightarrow sec\theta =\sqrt{1+ta{n}^2\theta }$

$tan\theta -\sqrt{1+ta{n}^2\theta }=P$

$tan\theta -P=\sqrt{1+ta{n}^2\theta }$

Squaring both sides

$ta{n}^2\theta +P^2-2Ptan\theta =1+ta{n}^2\theta$

$P^2-2Ptan\theta -1=0$

Using quadratic formula,

$P=\frac{2tan\theta \pm \sqrt{4ta{n}^2\theta +4}}{2}$

$=\frac{2tan\theta \pm 2\sqrt{ta{n}^2\theta +1}}{2}$

$=\frac{2+tan\theta \pm 2sec\theta }{2}$

$tan\theta +sec\theta =P$