Trigonometry: Solve for Cosec²(theta)

We are given the trigonometric equation:

$ \tan\theta + 3\cot\theta = 2\sqrt{3} $

And the condition $0^\circ < \theta < 90^\circ$. We need to find the value of $\text{cosec}^2\theta$.

Equation Transformation

Rewrite $\cot\theta$ as $\frac{1}{\tan\theta}$:

$ \tan\theta + \frac{3}{\tan\theta} = 2\sqrt{3} $

Quadratic Formulation

Let $x = \tan\theta$. Substitute $x$ into the equation:

$ x + \frac{3}{x} = 2\sqrt{3} $

Multiply the entire equation by $x$ (since $0^\circ < \theta < 90^\circ$, $\tan\theta \neq 0$):

$ x^2 + 3 = 2\sqrt{3}x $

Rearrange into a standard quadratic form:

$ x^2 - 2\sqrt{3}x + 3 = 0 $

Solving for Tan(theta)

This quadratic equation is a perfect square:

$ (x - \sqrt{3})^2 = 0 $

Therefore, the solution is:

$ x = \sqrt{3} $

This means $\tan\theta = \sqrt{3}$.

Cosec²(theta) Calculation

We can find $\text{cosec}^2\theta$ using the identity $\text{cosec}^2\theta = 1 + \cot^2\theta$.

First, find $\cot\theta$:

$ \cot\theta = \frac{1}{\tan\theta} = \frac{1}{\sqrt{3}} $

Now, calculate $\cot^2\theta$:

$ \cot^2\theta = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} $

Substitute this value into the identity:

$ \text{cosec}^2\theta = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} $

Final Answer

The value of $\text{cosec}^2\theta$ is $\frac{4}{3}$.