If $\alpha$ and $\beta$ are the roots of the equation $a \cos \ 2 \theta + b \sin \ 2 \theta = c$, then $\tan \ ( \alpha + \beta )$ is equal to:
- $\dfrac{a}{b}$
- $\dfrac{-b}{a}$
- $\dfrac{b}{a}$
- $\dfrac{-a}{b}$
The equation a cos(2theta) + b sin(2theta) = c can be written as a(1-tan^2(theta))/(1+tan^2(theta)) + b(2tan(theta))/(1+tan^2(theta)) = c. This is a quadratic in tan(theta). Let t = tan(theta). Then a(1-t^2) + 2bt = c(1+t^2), or (a+c)t^2 - 2bt + (c-a) = 0. The sum of roots tan(alpha)+tan(beta) = 2b/(a+c) and product tan(alpha)tan(beta) = (c-a)/(a+c). Using tan(alpha+beta) = (tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta)), we get (2b/(a+c)) / (1 - (c-a)/(a+c)) = (2b/(a+c)) / (2a/(a+c)) = b/a.
Using the double angle identities for cosine and sine, we rewrite the given equation as a(cos^2 x - sin^2 x) + b(2 sin x cos x) = c. Applying the substitution t = tan x transforms the terms into a(1 - t^2)/(1 + t^2) + b(2t)/(1 + t^2) = c. Multiplying by the denominator yields the quadratic (a + c)t^2 - 2bt + (c - a) = 0. Since alpha and beta are the given roots, their tangents t1 and t2 are the roots of this new quadratic equation. The formula for tan(alpha + beta) is (t1 + t2)/(1 - t1*t2), where the sum t1 + t2 = 2b/(a+c) and the product t1*t2 = (c-a)/(a+c). Substituting these values gives (2b/(a+c)) / (1 - (c-a)/(a+c)), which simplifies to 2b / ((a+c) - (c-a)) = b/a.