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: Recognize the algebraic pattern.
The given expression is \[ \tan \frac{\pi}{12}+\tan \frac{\pi}{6}+\tan \frac{\pi}{12}\tan \frac{\pi}{6} \] This resembles the numerator form in the tangent addition formula: \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \] But here we can also use known exact values directly.

Step 2: Write the exact values of the required tangents.
We know: \[ \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \] Also, \[ \tan \frac{\pi}{12}=\tan 15^\circ=2-\sqrt{3} \]

Step 3: Substitute these values into the expression.
So the expression becomes \[ (2-\sqrt{3})+\frac{1}{\sqrt{3}}+(2-\sqrt{3})\cdot\frac{1}{\sqrt{3}} \]

Step 4: Simplify the product term.
First compute \[ (2-\sqrt{3})\cdot\frac{1}{\sqrt{3}} = \frac{2-\sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3}}-1 \] Therefore the whole expression becomes \[ (2-\sqrt{3})+\frac{1}{\sqrt{3}}+\left(\frac{2}{\sqrt{3}}-1\right) \]

Step 5: Combine like terms.
Now combine constants and radical terms: \[ (2-1)-\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}} \] \[ =1-\sqrt{3}+\frac{3}{\sqrt{3}} \] But \[ \frac{3}{\sqrt{3}}=\sqrt{3} \] Hence, \[ 1-\sqrt{3}+\sqrt{3}=1 \]

Step 6: Alternative identity check.
Let \[ a=\tan \frac{\pi}{12},\qquad b=\tan \frac{\pi}{6} \] Then the expression is \[ a+b+ab \] Since \[ \frac{\pi}{12}+\frac{\pi}{6}=\frac{\pi}{12}+\frac{2\pi}{12}=\frac{3\pi}{12}=\frac{\pi}{4} \] we know \[ \tan\left(\frac{\pi}{12}+\frac{\pi}{6}\right)=\tan\frac{\pi}{4}=1 \] Using \[ 1=\frac{a+b}{1-ab} \] we get \[ a+b=1-ab \] So \[ a+b+ab=1 \] This confirms the result.

Step 7: Final conclusion.
Therefore, the required value is \[ \boxed{1} \] Hence, the correct option is \[ \boxed{(1)\ 1} \]