Question

In any triangle ABC, prove that :-

Answer 1

Hi ....
here is your answer..
sorry for bad handwriting

Answer 2

Let a = 2K sinA, b = 2K sinB, c = 2K sinC

1st we shall calculate: $\frac{b - c}{a}$

$= \ \textgreater \ \frac{b - c}{a} = \frac{2KsinB - 2ksinC}{2KsinA}$

$= \ \textgreater \ \frac{sinB - sinC}{sin \pi - [B + C]}$

$= \ \textgreater \ \frac{2 cos \frac{B + C}{2}sin \frac{B - C}{2} }{sin(B + C)}$

$= \ \textgreater \ \frac{2 cos \frac{B + C}{2} sin \frac{B - C}{2} }{2 sin \frac{B + C}{2} cos \frac{B + C}{2} }$

$= \ \textgreater \ \frac{2sin \frac{B - C}{2} }{2 sin \frac{B + C}{2} }$

$= \ \textgreater \ \frac{sin \frac{B - C}{2} }{sin( \frac{ \pi - A}{2} )}$

$= \ \textgreater \ \frac{sin \frac{B - C}{2} }{cos \frac{A}{2} }$

Therefore:

$= \ \textgreater \ \frac{b - c}{a} = \frac{sin \frac{B - C}{2} }{cos \frac{A}{2} }$


(or) 

$\frac{b - c}{a} cos \frac{A}{2} = \frac{sin(B - C)}{2}$



Hope this helps!