Question

If \( \tan \theta = \frac{1}{2} \) and \( \tan \phi = \frac{1}{3} \), then the value of \( \tan(2\theta + \phi) \) is:

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{3}$
• D. $3$
• E. $\frac{1}{2}$

Hint:

Recognize that if $\tan A = 1/2$ and $\tan B = 1/3$, then $\tan(A+B) = 1$. This implies $(A+B) = 45^\circ$. Since $\tan(2\theta + \phi) = \tan(\theta + (\theta + \phi))$, you can sometimes use these small blocks to solve faster.

Answer 1

The Correct Option is D

Concept: This problem requires a multi-step application of tangent identities. First, we must find the value of $\tan 2\theta$ using the double-angle formula. Once we have that intermediate value, we can use the sum formula $\tan(A+B)$ where $A = 2\theta$ and $B = \phi$.

Step 1: Calculate the value of $\tan 2\theta$.
Using the double-angle identity for tangent: $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$. Substitute the given value $\tan \theta = 1/2$: \[ \tan 2\theta = \frac{2(\frac{1}{2})}{1 - (\frac{1}{2})^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

Step 2: Apply the tangent sum formula.
Now we need to find $\tan(2\theta + \phi)$. Let $A = 2\theta$ and $B = \phi$. The formula is: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] Substituting $\tan 2\theta = 4/3$ and the given $\tan \phi = 1/3$: \[ \tan(2\theta + \phi) = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \left(\frac{4}{3} \times \frac{1}{3}\right)} \]

Step 3: Simplify the arithmetic expression.
Combine the terms in the numerator and the denominator: \[ \text{Numerator: } \frac{4}{3} + \frac{1}{3} = \frac{5}{3} \] \[ \text{Denominator: } 1 - \frac{4}{9} = \frac{9 - 4}{9} = \frac{5}{9} \] Divide the numerator by the denominator: \[ \tan(2\theta + \phi) = \frac{5/3}{5/9} = \frac{5}{3} \times \frac{9}{5} = \frac{9}{3} = 3 \]