Question

For triangle ABC, prove that: sec (A+B/2)= cosec C

Answer 1

Since A, B and C are interior angles.
So, A+B+C=180° [Angle Sum Property]
A+B = 180°- C
A+B/2 = 90°- C/2
Now, sec (A+B/2)
= sec (90°- C/2)
=cosec C/2
Hence, proved.

Answer 2

Solution:

Given ABC is a triangle.

We know that,

$\boxed {\angle A + \angle B + \angle C = 180}$

/* angle sum property */

Divide both sides by 2 , we get

$\implies \frac{\angle A + \angle B + \angle C}{2} =\frac{180}{2}}$

$\implies \frac {\angle A + \angle B}{2}= 90 -\frac{\angle C}{2}$

$\implies sec(\frac {\angle A + \angle B}{2})= sec(90 -\frac{\angle C}{2})$

$\implies sec(\frac {\angle A + \angle B}{2})= cosec(\frac{\angle C}{2})$

Therefore,

$sec(\frac {\angle A + \angle B}{2})= cosec(\frac{\angle C}{2})$

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