Question

In an triangle ABC prove that (a+b+c) (tanA/2+tanB/2)= 2c cot C/2

Answer 1

ANSWER

2s = a+b+c

We know tan (A/2)

= √(s-b)(s-c)/s(s-a)

And tan (B/2) = √(s-a) (s-c)/s(s-b)

Now, (a+b+c) (tanA/2) + tan(B/2) = (a+b+c)(√(s-b) (s-c)/s(s-a)

√(s-a) (s-c)/s(s-b) = 2s √s-c/s (√s-b/s-a + √s-a/s-b

= 2s√(s-a) (s-c) (s-b+s-a/√(s-b) (s-a)

= 2c√s(s-c)/(s-a)(s-b)

2ctan(c/2)