Question

In ΔABC, prove that `sec  (C + A)/2 = "cosec" B/2`.

Answer 1

Given: Triangle ABC, so A + B + C = 180°.

To Prove: `sec  (C + A)/2 = "cosec" B/2`.

Proof [Step-wise]:

1. From the angle-sum in a triangle:

A + B + C = 180°

2. Hence, `(C + A)/2 = (180^circ - B)/2`

= `90^circ - B/2`

3. Therefore, `sec((C + A)/2) = sec(90^circ - B/2)`.

4. Use the cofunction identity:

`sec(90^circ - x) = 1/cos(90^circ - x)`

= `1/(sin x)`

= cosec x

5. Apply that identity with `x = B/2`:

`sec(90^circ - B/2) = "cosec" (B/2)`

`sec  (C + A)/2 = "cosec" B/2`, as required.