Question

The value of \( \tan \dfrac{\pi}{12}+\tan \dfrac{\pi}{6}+\left(\tan \dfrac{\pi}{12}\tan \dfrac{\pi}{6}\right) \) is equal to

• A. \( 1 \)
• B. \( 2 \)
• C. \( \sqrt{3} \)
• D. \( -\sqrt{3} \)
• E. \( -1 \)

Hint:

When tangent values involve standard angles like \( \frac{\pi}{12}, \frac{\pi}{6}, \frac{\pi}{4} \), always check whether their sum or difference becomes a familiar angle. That often gives a much shorter solution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The expression strongly resembles the expansion of \(\tan(A+B)\). Let's analyze the identity for \(\tan(A+B)\).
Step 2: Key Formula or Approach:
The tangent addition formula is:
\[ \tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \] Rearranging this formula gives:
\[ \tan(A+B)(1 - \tan(A)\tan(B)) = \tan(A) + \tan(B) \] \[ \tan(A+B) - \tan(A+B)\tan(A)\tan(B) = \tan(A) + \tan(B) \] \[ \tan(A) + \tan(B) + \tan(A+B)\tan(A)\tan(B) = \tan(A+B) \] This doesn't quite match the given expression. Let's look at the original expression again: \(\tan A + \tan B + \tan A \tan B\), where \(A=\pi/12\) and \(B=\pi/6\).
This structure often appears when \(A+B\) is a special angle, like \(\pi/4\).
Let's check the sum of the angles: \(A+B = \frac{\pi}{12} + \frac{\pi}{6} = \frac{\pi}{12} + \frac{2\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}\).
Step 3: Detailed Explanation:
Since \(A+B = \frac{\pi}{4}\), we have \(\tan(A+B) = \tan(\frac{\pi}{4}) = 1\).
Now, let's use the tangent addition formula:
\[ \tan(A+B) = 1 \] \[ \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} = 1 \] \[ \tan(A) + \tan(B) = 1 - \tan(A)\tan(B) \] Rearranging the terms to match the expression in the question:
\[ \tan(A) + \tan(B) + \tan(A)\tan(B) = 1 \] Substituting \(A = \pi/12\) and \(B = \pi/6\):
\[ \tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} = 1 \] Step 4: Final Answer:
The value of the expression is 1. Therefore, option (A) is correct.