In a $\triangle A B C, \cot A+\cot B+\cot C=$

• A

  $\frac{a^2+b^2+c^2}{\Delta}$
• B

  $\frac{a+b+c}{4 \Delta}$
• Correct answer
  C

  $\frac{a^2+b^2+c^2}{4 \Delta}$
• D

  $\frac{a^2+b^2+c^2}{2 \Delta}$

Solution

Let a triangle $A B C$ of sides $a, b, c$, having area $\Delta$
Area $=\Delta$ $\frac{1}{2} b c \sin A=\Delta$ $\ldots(\mathrm{i})$ and $\quad \frac{1}{2} a c \sin B=\Delta$ $\ldots(\mathrm{ii})$ and $\quad \frac{1}{2} a b \sin c=\Delta$ $\ldots(\mathrm{iii})$ Using cosine rule $a^2=b^2+c^2-2 b c \cos A$ and $\quad b^2=a^2+c^2-2 a c \cos B$ and $\quad c^2=a^2+b^2-2 a b \cos C$ On adding, we get $a^2+b^2+c^2=2 a^2+2 b^2+2 c^2$ $-2 a b \cos C-2 a c \cos B-2 b c \cos A$ or $a^2+b^2+c^2=2(a b \cos C+a c \cos B+b c \cos A)$ $\ldots(\mathrm{iv})$ Now, from Eq. (i) $b c=\frac{2 \Delta}{\sin A}$ Eq. (ii), $a c=\frac{2 \Delta}{\sin B}$ Eq. (iii), $\quad a b=\frac{2 \Delta}{\sin C}$ Putting these values in Eq. (iv), we get $a^2+b^2+c^2=2$ $\left(\frac{2 \Delta}{\sin C} \cos C+\frac{2 \Delta}{\sin B} \cos B+\frac{2 \Delta}{\sin A} \cos A\right)$ $a^2+b^2+c^2=4 \Delta(\cot C+\cot B+\cot A)$ or $\cot C+\cot B+\cot A=\frac{a^2+b^2+c^2}{4 \Delta}$ or $\cot A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4 \Delta}$

Hence, the correct answer is option: 'C'.