Given \(\sin \theta = \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd form.

Assessment: WAEC SSCE - Further Mathematics - 2009 (Objective) Subject: Further Mathematics

Question 1 Report

Given \(\sin \theta =  \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd form.

Answer Details

We know that:

\(\sin \theta = \frac{{\text{{opposite}}}}{{\text{{hypotenuse}}}}\)

We can draw a right triangle with an angle of \(\theta\), opposite side of \(\sqrt{3}\), and hypotenuse of 2. Using the Pythagorean theorem, we can find the adjacent side to be 1. Therefore, the triangle looks like:

         /|
        / |
   1  /  |sqrt(3)
     /   |
    /____|
       2

Using the double-angle formula for tangent, we have:

\(\tan 2\theta = \frac{{2 \tan \theta}}{{1 - \tan^{2} \theta}}\)

Using the formula for tangent:

\(\tan \theta = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}} = \sqrt{3}\)

Substituting in the formula for double-angle tangent, we have:

\(\tan 2\theta = \frac{{2 \sqrt{3}}}{{1 - (\sqrt{3})^{2}}} = \frac{{2 \sqrt{3}}}{{1 - 3}}\)

Simplifying, we get:

\(\tan 2\theta = \frac{{2 \sqrt{3}}}{{-2}} = -\sqrt{3}\)

Therefore, the correct answer is.

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