If A, B and C be the angles of a triangle, then of the following the incorrect relation is :

$cot left(frac{mathrm{A}+mathrm{B}}{2} ight)= an frac{mathrm{C}}{2}$

$cos left(frac{mathrm{A}+mathrm{B}}{2} ight)=sin frac{mathrm{C}}{2}$

$an left(frac{mathrm{A}+mathrm{B}}{2} ight)=sec frac{mathrm{C}}{2}$

$sin frac{A+B}{2}=cos frac{C}{2}$

Answer : Option C

Explanation :

A + B + C = π

$Rightarrow frac{mathrm{A}+mathrm{B}}{2}=frac{pi}{2}-frac{mathrm{C}}{2}$

$Rightarrow sin left(frac{mathrm{A}+mathrm{B}}{2} ight)$

$=sin left(frac{pi}{2}-frac{mathrm{C}}{2} ight)=cos frac{mathrm{C}}{2}$

Similarly,

$cos left(frac{A+B}{2} ight)=sin frac{C}{2}$

$cot left(frac{A+B}{2} ight)= an frac{C}{2}$

$an left(frac{A+B}{2} ight)=cot frac{C}{2}$