Question

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

Options

• \[\operatorname{cosec} \frac{B}{2}\]

• \[\sec \frac{B}{2}\]

• \[\operatorname{cosec} \frac{B + A}{2}\]

• none of these

Answer 1

In a triangle ABC, \[ \sec \frac{C + A}{2}\] is equal to \[\underline{\boldsymbol{\operatorname{cosec} \frac{B}{2}}}\]. 

Explanation:

In triangle ABC,

By angle sum property of triangle,

⇒ A + B + C = 180°

⇒ A + C = 180° − B

\[\Rightarrow \frac{A + C}{2} = \frac{180^\circ - B}{2} \; \text{........(1)}\]

Substituting value of \[\frac{A + C}{2}\] in \[\sec \frac{C + A}{2}\], we get:

\[\Rightarrow \sec \frac{C + A}{2}\]

\[\Rightarrow \sec \frac{180^\circ - B}{2}\]

\[\Rightarrow \sec\left(90^\circ - \frac{B}{2}\right)\]

\[\Rightarrow \operatorname{cosec} \frac{B}{2}\]