Question

In triangle abc prove that c/a+b = 1-tana/2 tanb/2 / 1+tana/2 tanb/2. solution

Answer 1

We know that in a triangle $a=k\sin A,b=k\sin B,c=k\sin C$.

Hence

[tex]LHS=\frac{c}{a+b} =\frac{\sin C}{\sin A+\sin B} \\ LHS=\frac{\sin (\pi - A-B)}{\sin A+\sin B} \\ LHS=\frac{\sin ( A+B)}{\sin A+\sin B} \\ LHS=\frac{2\sin ( \frac{A+B}{2} )\cos ( \frac{A+B}{2} )}{2\sin ( \frac{A+B}{2} )\cos ( \frac{A-B}{2} )} \\[/tex]

Further simplifying,

[tex]LHS=\frac{\cos ( \frac{A+B}{2} )}{\cos ( \frac{A-B}{2} )} \\ LHS=\frac{\cos ( \frac{A}{2} )\cos ( \frac{B}{2} )-\sin ( \frac{A}{2} )\sin ( \frac{B}{2} )}{\cos ( \frac{A}{2} )\cos ( \frac{B}{2} )+\sin ( \frac{A}{2} )\sin ( \frac{B}{2} )} \\[/tex]

Divide numerator and denominator by $\cos ( \frac{A}{2} )\cos ( \frac{B}{2} )}$. Then

[tex]LHS=\frac{1-\frac{\sin ( \frac{A}{2} )\sin ( \frac{B}{2} )}{\cos ( \frac{A}{2} )\cos ( \frac{B}{2} )}}{1+\frac{\sin ( \frac{A}{2} )\sin ( \frac{B}{2} )}{\cos ( \frac{A}{2} )\cos ( \frac{B}{2} )}} \\ LHS=\frac{1-\tan ( \frac{A}{2} )\tan ( \frac{B}{2} )}{1+\tan ( \frac{A}{2} )\tan ( \frac{B}{2} )} =RHS[/tex].

The proof is complete.