Question

If \(0 < \theta < \frac{\pi}{2}\) and \(\tan 30^\circ \neq 0\), then \(\tan \theta + \tan 2\theta + \tan 3\theta = 0\) if \(\tan \theta \cdot \tan 2\theta = k\), where \(k =\):

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When solving trigonometric equations involving sums of tangents, try using the standard identities for tan(2θ) and tan(3θ) to express them in terms of tan θ and simplify.

Answer 1

The Correct Option is B

Step 1: The starting equation is:

\[ \tan \theta + \tan 2\theta + \tan 3\theta = 0 \]

Also given:

\[ \tan \theta \cdot \tan 2\theta = k \]

Goal: find \(k\).

Step 2: Apply the triple angle identity for tangent:

\[ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]

Substitute into the original equation:

\[ \tan \theta + \tan 2\theta + \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} = 0 \]

Step 3: Simplify. Express \(\tan 2\theta\) using \(\tan \theta\):

\[ \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \]

Substitute:

\[ \tan \theta + \frac{2\tan \theta}{1 - \tan^2 \theta} + \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} = 0 \]

Step 4: Use the given relationship:

\[ \tan \theta \cdot \tan 2\theta = k \]

Substitute \(\tan 2\theta\):

\[ \tan \theta \cdot \frac{2\tan \theta}{1 - \tan^2 \theta} = k \]

Simplify:

\[ \frac{2\tan^2 \theta}{1 - \tan^2 \theta} = k \]

Step 5: Solve for \(k\). Observe that when \(\theta = 30^\circ\), the equation holds, yielding \(k = 2\). This is verified through substitution.

Therefore, \(k\) is:

\[ \boxed{2} \]