Question

If $\tan\theta$ and $\cot\theta$ are two distinct roots of the equation $ax^2+bx+c=0, a\neq0, b\neq0$, then

• A. $\cos 2\theta = -\frac{2b}{c}$
• B. $\sin 2\theta = -\frac{2c}{b}$
• C. $\tan 2\theta = \frac{2b}{c}$
• D. $\cot 2\theta = \frac{2c}{a}$

Hint:

When the roots of a quadratic equation are trigonometric functions of the same angle, use Vieta's formulas (sum and product of roots) and then simplify using trigonometric identities to find the relationship between the coefficients.

Answer 1

The Correct Option is B

Given that $\tan\theta$ and $\cot\theta$ are the roots of the quadratic equation $ax^2+bx+c=0$.
From the properties of quadratic equations, we can write the sum and product of the roots.
Sum of the roots: $\tan\theta + \cot\theta = -\frac{b}{a}$.
Product of the roots: $(\tan\theta)(\cot\theta) = \frac{c}{a}$.
Since $\tan\theta \cdot \cot\theta = 1$, we have $1 = \frac{c}{a}$, which implies $a=c$.
Now, let's work with the sum of the roots.
$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}$.
Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:
$\frac{1}{\sin\theta\cos\theta} = -\frac{b}{a}$.
Rearranging gives $\sin\theta\cos\theta = -\frac{a}{b}$.
We know the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$.
Multiplying both sides by 2: $2\sin\theta\cos\theta = -2\frac{a}{b}$.
Therefore, $\sin(2\theta) = -\frac{2a}{b}$.
Since we established that $a=c$, we can substitute $c$ for $a$.
$\sin(2\theta) = -\frac{2c}{b}$.