Question

a+ b /2 = sin A-B B /2 cosec C/2

Answer 1

Answer:

Step-by-step explanation:

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})