Question

If \( A, B, C \) 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:

Using complementary angle identity: \[ \csc(90^\circ - x) = \sec x, \quad \sec(90^\circ - x) = \tan x. \]

Answer 1

The Correct Option is B

Since \( A + B + C = 180^\circ \) in a triangle:
\[ \frac{A+B}{2} = \frac{180^\circ - C}{2} = 90^\circ - \frac{C}{2}. \] Taking cosecant: \[ \csc \left(\frac{A+B}{2}\right) = \csc (90^\circ - \frac{C}{2}). \] Using identity: \[ \csc (90^\circ - x) = \sec x. \] Thus: \[ \csc \left(\frac{A+B}{2}\right) = \sec \frac{C}{2}. \]