Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If \(\tan\theta + \tan 2\theta + \sqrt{3}\tan\theta \tan 2\theta = \sqrt{3}\), then</p>
<p>(a) \(\theta = \frac{(6n+1)\pi}{18}, \forall n \in \mathbb{I}\)</p>
<p>(b) \(\theta = \frac{(6n+1)\pi}{9}, \forall n \in \mathbb{I}\)</p>
<p>(c) \(\theta = \frac{(3n+1)\pi}{9}, \forall n \in \mathbb{I}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize the pattern in the equation as the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
<p>The given equation matches the form $\tan A + \tan B + \sqrt{3}\tan A \tan B = \sqrt{3}$, which is equivalent to $\tan(A+B) = \sqrt{3}$ when rearranged using the tangent addition formula. With $A = \theta$ and $B = 2\theta$, we get $\tan 3\theta = \sqrt{3}$, so $3\theta = \frac{\pi}{3} + n\pi$, giving $\theta = \frac{(6n+1)\pi}{18}$.</p>
Correct Answer: a