Question

for triangle ABC prove that (AB+BC)/AC=COT B/2

Answer 1

Answer:

Explanation:

By sin rule

a /sin A = b /sin B = c/ sin C = 2 R

(AB+BC)/AC= c + a / b = sin a' + sin A / sin B

= 2 sin (A+C /2) cos ( A - C /2) / 2 sin (B/2) cos (B/2)

= {2 cos B/2 cos A- C /2 } / {2 sin (B/2) cos (B/2)}

= {cos A - C /2} / sin B/2

= cos B/2 / sin B/2

= cot B /2.