Question: If $3 + tan \theta = sec^2 \theta$, find the value of $tan \theta$.


Answer: 2, -1

Solution:
We know,
$sec^2 \theta = 1 + tan^2 \theta$

Given,
$3 + tan \theta = sec^2 \theta$

$or, 3 + tan \theta = 1 + tan^2 \theta$

$or, 3 - 1 = tan^2 \theta - tan \theta $

$or, 2 = tan^2 \theta - tan \theta$

$or, tan^2 \theta - tan\theta - 2 = 0$

Comparing above equation with $ax^2 + bx + c = 0$, we get,

$a = 1,\ b = -1, \ c= - 2, \ x = tan \theta$

Using formula of quadratic equation,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$or, tan \theta = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

$or, tan \theta = \dfrac{1 \pm \sqrt{1 + 8}}{2}$

$or, tan\theta = \dfrac{1 \pm \sqrt{9}}{2}$

$or, tan \theta = \dfrac{1 \pm 3}{2}$

Taking positive sign,
$tan \theta = \dfrac{1 + 3}{2}$
$\therefore tan \theta = 2$

Taking negative sign,
$tan \theta = \dfrac{1 - 3}{2}$
$\therefore tan \theta = -1$

Hence, the required value of $tan\theta$ are 2 or -1.


#Trigonometry
#Values