Question

In any $\Delta ABC,\frac{a+b}{c}=$

• A. $\frac{cos\left(\frac{A+B}{2}\right)}{sin\left(\frac{C}{2}\right)}$
  
• B. $\frac{sin\left(\frac{A+B}{2}\right)}{cos\left(\frac{C}{2}\right)}$
  
• C. $\frac{cos\left(\frac{A-B}{2}\right)}{sin\left(\frac{C}{2}\right)}$
  
• D. $\frac{sin\left(\frac{A-B}{2}\right)}{cos\left(\frac{C}{2}\right)}$
  
  

Answer 1

The correct option is C $\frac{cos\left(\frac{A-B}{2}\right)}{sin\left(\frac{C}{2}\right)}$



We know $\frac{a}{sin\left(A\right)}\frac{b}{sin\left(B\right)}=\frac{c}{sin\left(C\right)}$ (From Sine rule)
$\frac{a}{sin\left(A\right)}\frac{b}{sin\left(B\right)}=\frac{c}{sin\left(C\right)}=k$ (Let's say)
a = k sin(A) , b = k sin(B) and c = k sin(C)
So, $\frac{a+b}{c}=\frac{k(sinA+sinB)}{ksinC}$
$=\frac{sin\left(\frac{A+B}{2}\right).cos\left(\frac{A-B}{2}\right)}{sin\left(\frac{C}{2}\right).cos\frac{c}{2}}$
Since, $A+B=\pi -C$
$sin\left(\frac{A+B}{2}\right)=cos\left(\frac{c}{2}\right)$
$=\frac{cos\left(\frac{C}{2}\right).cos\left(\frac{A-B}{2}\right)}{sin\left(\frac{c}{2}\right).cos\frac{c}{2}}$
$=\frac{cos\left(\frac{A-B}{2}\right)}{sin\left(\frac{C}{2}\right)}$