Question

If \( A \), \( B \), and \( C \) are angles of a triangle ABC, then the value of \( \csc \left( \frac{A + B}{2} \right) \) is:

• A. \( \tan \frac{C}{2} \)
• B. \( \sec \frac{C}{2} \)
• C. \( \cot \frac{C}{2} \)
• D. \( \sin \frac{C}{2} \)

Hint:

Use the identity \( \csc (90^\circ - x) = \sec x \) for angles in a triangle to simplify trigonometric expressions.

Answer 1

The Correct Option is B

In any triangle \( A + B + C = 180^\circ \). Therefore, \( A + B = 180^\circ - C \). Now, we have: \[ \frac{A + B}{2} = \frac{180^\circ - C}{2} = 90^\circ - \frac{C}{2}. \] Thus: \[ \csc \left( \frac{A + B}{2} \right) = \csc \left( 90^\circ - \frac{C}{2} \right). \] Using the identity \( \csc (90^\circ - x) = \sec x \), we get: \[ \csc \left( \frac{A + B}{2} \right) = \sec \frac{C}{2}. \] Thus, the correct answer is \( \boxed{\sec \frac{C}{2}} \).