Question

If $a \cos 2\theta + b \sin 2\theta = c$ has $\alpha$ and $\beta$ as its roots, then the value of $\tan \alpha + \tan \beta$ is

• A. $\frac{2b}{c+a}$
• B. $\frac{2c}{b+c}$
• C. $\frac{2b}{a+c}$
• D. $\frac{2a}{b+c}$

Hint:

For trigonometric equations where the roots are angles, convert trigonometric functions into algebraic expressions involving $\tan \theta$ (or $t = \tan \theta$) using double-angle formulas. This often transforms the equation into a polynomial, whose roots directly correspond to $\tan$ of the original roots.

Answer 1

The Correct Option is A

Step 1: Express $\cos 2\theta$ and $\sin 2\theta$ in terms of $\tan \theta$. Let $t = \tan \theta$. Recall the double angle identities: \[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - t^2}{1 + t^2} \] \[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2t}{1 + t^2} \]
Step 2: Substitute these expressions into the given equation $a \cos 2\theta + b \sin 2\theta = c$. \[ a\left(\frac{1 - t^2}{1 + t^2}\right) + b\left(\frac{2t}{1 + t^2}\right) = c \]
Step 3: Clear the denominator by multiplying by $(1 + t^2)$. \[ a(1 - t^2) + b(2t) = c(1 + t^2) \] \[ a - at^2 + 2bt = c + ct^2 \]
Step 4: Rearrange the terms to form a quadratic equation in $t$. \[ ct^2 + at^2 - 2bt + c - a = 0 \] \[ (c+a)t^2 - 2bt + (c-a) = 0 \]
Step 5: Identify the roots of this quadratic equation. If $\alpha$ and $\beta$ are the roots of the original trigonometric equation, then $\tan \alpha$ and $\tan \beta$ are the roots of this quadratic equation in $t$. Let $t_1 = \tan \alpha$ and $t_2 = \tan \beta$.
Step 6: Use Vieta's formulas for the sum of the roots of a quadratic equation $Ax^2 + Bx + C = 0$, which is $x_1 + x_2 = -\frac{B}{A}$. Here, $A = (c+a)$, $B = (-2b)$, and $C = (c-a)$. \[ \tan \alpha + \tan \beta = t_1 + t_2 = -\frac{(-2b)}{c+a} \] \[ \tan \alpha + \tan \beta = \frac{2b}{c+a} \]