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
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