Question

In a triangle ABC, prove that Tan(A+B)/2 + CotC/2 =0

Answer 1

Answer:

Step-by-step explanation:

In ABC, using angle sum property

A+B+C=180

⇒(A+B+C)/2=180/2

⇒(A+B)/2=90−C/2

⇒tan((A+B)/2)= tan(90−C/2)    

⇒tan((A+B)/2)= cot(C/2)                               (using tan(90−θ)=cotθ)                                                                          

Hence proved