$$2-cos^2 heta = 1+1-cos^2 heta = 1+sin^2 heta$$
Dividing the LHS and RHS by $$cos^2 heta$$
$$1+sin^2 heta = sec^2 heta + tan^2 heta$$
$$3sin heta cos heta = 3 tan heta$$
$$sec ^2 heta + tan^2 heta = 3tan heta$$
$$sec ^2 heta = 1+tan^2 heta$$
$$1+tan^2 heta+tan^2 heta = 3tan heta$$
$$1+2tan^2 heta = 3tan heta$$
$$2tan^2 heta - 3tan heta + 1 =0$$
let x=$$tan heta$$
The equation becomes
$$2x^2 + 3x + 1=0$$
On solving for x we get x=$$1$$ and x=$$frac{1}{2}$$
Option A is the correct answer.