In any ∆ ABC, Prove that :\(\frac{a+b}{c}=\frac{cos(\frac{A-B}{2})}{

1.	In any ∆ ABC, Prove that :\(\frac{a+b}{c}=\frac{cos(\frac{A-B}{2})}{sin\frac{c}{2}}\)
Answer» \(\frac{a}{sin\,A}=\frac{b}{sin\,B}=\frac{c}{sin\,C}=k\) ⇒ k sin A, b = k sin B and c = k sin C LHS = \(\frac{a+b}{c}=\frac{k\,sin\,A+k\,sin\,B}{k\,sin\,C}\) \(=\frac{k(sin\,A+sin\,B)}{k\,sin\,C}\) \(=\frac{sin\,A+sin\,B}{sin\,C}\) \(=\frac{2sin\,\frac{A+B}{2}\,cos\frac{A-B}{2}}{2\,sin\frac{C}{2}cos\frac{C}{2}}\) \(=\frac{sin\,(90°-\frac{C}{2})cos\frac{A-B}{2}}{\,sin\frac{C}{2}cos\frac{C}{2}}\) [∴ \(\frac{A+B}{2}\)= 90° − \(\frac{C}{2}\)] \(=\frac{cos\frac{C}{2}cos\frac{A-B}{2}}{\,sin\frac{C}{2}cos\frac{C}{2}}\) \(=\frac{cos\frac{A-B}{2}}{\,sin\frac{C}{2}}\) = RHS ∴ LHS = RHS

In any ∆ ABC, Prove that :\(\frac{a+b}{c}=\frac{cos(\frac{A-B}{2})}{sin\frac{c}{2}}\)

Answer»

\(\frac{a}{sin\,A}=\frac{b}{sin\,B}=\frac{c}{sin\,C}=k\)

⇒ k sin A, b = k sin B and c = k sin C

LHS = \(\frac{a+b}{c}=\frac{k\,sin\,A+k\,sin\,B}{k\,sin\,C}\)

\(=\frac{k(sin\,A+sin\,B)}{k\,sin\,C}\)

\(=\frac{sin\,A+sin\,B}{sin\,C}\)

\(=\frac{2sin\,\frac{A+B}{2}\,cos\frac{A-B}{2}}{2\,sin\frac{C}{2}cos\frac{C}{2}}\)

\(=\frac{sin\,(90°-\frac{C}{2})cos\frac{A-B}{2}}{\,sin\frac{C}{2}cos\frac{C}{2}}\)

[∴ \(\frac{A+B}{2}\)= 90° − \(\frac{C}{2}\)]

\(=\frac{cos\frac{C}{2}cos\frac{A-B}{2}}{\,sin\frac{C}{2}cos\frac{C}{2}}\)

\(=\frac{cos\frac{A-B}{2}}{\,sin\frac{C}{2}}\) = RHS

∴ LHS = RHS