Question

In a $\Delta ABC,\cot A+\cot B+\cot C=$

• A. $\frac{a^2+b^2+c^2}{\Delta }$
• B. $\frac{a+b+c}{4\Delta }$
• C. $\frac{a^2+b^2+c^2}{4\Delta }$
• D. $\frac{a^2+b^2+c^2}{2\Delta }$

Answer 1

Answer: (C)

Let a triangle $ABC$ of sides $a,b,c$, having area Delta

Area $=\Delta$
$\frac{1}{2}bc\sin A=\Delta \dots$ (i)
and $\frac{1}{2}ac\sin B=\Delta \dots$ (ii)
and $\frac{1}{2}ab\sin c=\Delta \dots$ (iii)
Using cosine rule
$a^2=b^2+c^2-2bc\cos A$
and $b^2=a^2+c^2-2ac\cos B$
and $c^2=a^2+b^2-2ab\cos C$
On adding, we get
$a^2+b^2+c^2=2a^2+2b^2+2c^2-2ab\cos C-2ac\cos B-2bc\cos A$
or $a^2+b^2+c^2=2(ab\cos C+ac\cos B+bc\cos A)\dots$ (iv)
Now, from Eq. (i) $bc=\frac{2\Delta }{\sin A}$
Eq. (ii), $ac=\frac{2\Delta }{\sin B}$
Eq. (iii), $ab=\frac{2\Delta }{\sin C}$
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 $\cot A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4\Delta }$