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Q.
In a $\Delta ABC , \cot A +\cot B +\cot C =$
TS EAMCET 2020
Solution:
Let a triangle $A B C$ of sides $a, b, c$, having area Delta
Area $=\Delta$
$\frac{1}{2} b c \sin A=\Delta \ldots$ (i)
and $\frac{1}{2} a c \sin B=\Delta \ldots$ (ii)
and $\frac{1}{2} a b \sin c=\Delta \ldots$ (iii)
Using cosine rule
$a^{2}=b^{2}+c^{2}-2 b c \cos A$
and $b^{2}=a^{2}+c^{2}-2 a c \cos B$
and $c^{2}=a^{2}+b^{2}-2 a b \cos C$
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 $a^{2}+b^{2}+c^{2}=2(a b \cos C +a c \cos B +b c \cos A) \ldots$ (iv)
Now, from Eq. (i) $b c=\frac{2 \Delta}{\sin A}$
Eq. (ii), $a c=\frac{2 \Delta}{\sin B}$
Eq. (iii), $a b=\frac{2 \Delta}{\sin C}$
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 $\cot A+\cot B+\cot C=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta}$
