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