Question

in triangle ABC show that (tan A/2) (tan B/2)=a+b-c/a+b+c

Answer 1

Answer:

Step-by-step explanation:

L.H.S -

= (tan A/2) (tan B/2)

√(s-b) √(s-c) √(s-a) √(s-c)

= _________ • _________

√s√(s-a) √s√(s-b)

= (s-c)/s

= [(a+b+c)/2 - c] / [(a+b+c)/2]

= [(a+b+c-2c)/2] / [(a+b+c)/2]

= (a+b-c)/(a+b+c)

= R.H.S