Question

Sec^4theta-sec^2theta=3,then tan^4theta+tan^2theta

Answer 1

Answer: $4-\sqrt{13}$

Step-by-step explanation:

(For typing convenience, I'll use α in place of theta)

The given equation is

$(sec\alpha)^4-(sec\alpha)^2-3=0$ ......(1)

Let

$x=(sec\alpha)^2$

Then, equation (1) is reduced to a quadratic form

$x^2-x-3=0$

Using quadratic formula, we obtain the solution

$x=\frac{1+\sqrt{13} }{2}$

I have discared the negative solution, ∵

$x=(sec\alpha)^2$⇒$x\geq 0$ ∀α∈R

Now, we have

$(sec\alpha)^2=\frac{1+\sqrt{13} }{2}$

⇒$1+(tan\alpha)^2 = \frac{1+\sqrt{13} }{2}$

⇒$(tan\alpha)^2=\frac{-1+\sqrt{13} }{2}$

Using the above value, the required expression can be computed numerically.

$(tan\alpha)^4+(tan\alpha)^2=[(tan\alpha)^2]^2+(tan\alpha)^2$

Substitute the value of (tanα)^2 to get the answer.