Question

In a triangle ABC, sec \[\frac{A + C}{2}\] is equal to ______.

Options

• 0

• `sec  B/2`

• cosec B

• `cosec  B/2`

Answer 1

In a triangle ABC, sec \[\frac{A + C}{2}\] is equal to `bbunderline(cosec  B/2)`.

Explanation:

Since A + B + C = 180°, we have A + C = 180° − B. Dividing by 2 yields \[\frac{A + C}{2} = 90^\circ - \frac{B}{2}\]. Substituting this into the expression gives \[\sec\left(90^\circ - \frac{B}{2}\right)\], which simplifies to \[\csc \frac{B}{2}\] by complementary angle identities.