Question

If $ A, B $ and $ C $ are interior angles of a triangle $ ABC $, then $ \tan\left(\frac{A+B}{2}\right) = \dots $

• A. $ \sin\left(\frac{C}{2}\right) $
• B. $ \cos\left(\frac{C}{2}\right) $
• C. $ \tan\left(\frac{C}{2}\right) $
• D. $ \cot\left(\frac{C}{2}\right) $

Hint:

In triangle problems, always try converting $A+B$ into $180^\circ - C$. It usually leads directly to co-function identities.

Answer 1

The Correct Option is D

Concept: In any triangle, the sum of interior angles is: \[ A + B + C = 180^\circ \] Also, the co-function identity: \[ \tan(90^\circ - \theta) = \cot \theta \]

Step 1: Use angle sum property of triangle.
\[ A + B + C = 180^\circ \] \[ A + B = 180^\circ - C \]

Step 2: Divide by 2.
\[ \frac{A+B}{2} = \frac{180^\circ - C}{2} \] \[ \frac{A+B}{2} = 90^\circ - \frac{C}{2} \]

Step 3: Apply tangent.
\[ \tan\left(\frac{A+B}{2}\right) = \tan\left(90^\circ - \frac{C}{2}\right) \]

Step 4: Use identity.
\[ \tan(90^\circ - \theta) = \cot \theta \] So, \[ \tan\left(90^\circ - \frac{C}{2}\right) = \cot\left(\frac{C}{2}\right) \] Conclusion: \[ \boxed{\tan\left(\frac{A+B}{2}\right) = \cot\left(\frac{C}{2}\right)} \]