Question:
Let \(A, B, C\) be the angles of a plane triangle. If
\[
\tan \frac{A}{2} = \frac{1}{3} \quad \text{and} \quad \tan \frac{B}{2} = \frac{2}{3},
\]
then \(\tan \frac{C}{2}\) is equal to:
Let \(A, B, C\) be the angles of a plane triangle. If
\[ \tan \frac{A}{2} = \frac{1}{3} \quad \text{and} \quad \tan \frac{B}{2} = \frac{2}{3}, \]
then \(\tan \frac{C}{2}\) is equal to:
\[ \tan \frac{A}{2} = \frac{1}{3} \quad \text{and} \quad \tan \frac{B}{2} = \frac{2}{3}, \]
then \(\tan \frac{C}{2}\) is equal to:
Show Hint
Use the half-angle tangent identity for triangle angle problems.
Updated On: Mar 23, 2026
- \( \dfrac{7}{9} \)
- \( \dfrac{2}{9} \)
- \( \dfrac{1}{3} \)
- (2)/(3)
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The Correct Option is A
Solution and Explanation
Step 1: In a triangle,
\[ \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 \]
Step 2: Substitute values:
\[ \frac{1}{3} \cdot \frac{2}{3} + \frac{2}{3} x + \frac{1}{3} x = 1 \]
Step 3: Solve to get:
\[ x = \frac{7}{9} \]
\[ \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 \]
Step 2: Substitute values:
\[ \frac{1}{3} \cdot \frac{2}{3} + \frac{2}{3} x + \frac{1}{3} x = 1 \]
Step 3: Solve to get:
\[ x = \frac{7}{9} \]
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