Question

Show that $\cot A+cotB+\mathrm{cotC}=\frac{a^2+b^2+c^2}{4\Delta }$

Answer 1

Assume triangle ABC with a opposite A, b opposite B, and c opposite C.

Also assume you have already proved the formula for the area of a triangle given 2 sides and an included angle. Let K = area of triangle ABC. Then by this formula the following are true:

1/2 ab sinC = K

1/2 ac sinB = K

1/2 bc sinA = K

That is given two sides and an included angle the area is 1/2 the product of the two sides times the sine of the included angle.

Using the Law of Cosines we also have:

a² = b²+c²-2bccosA

b² = a²+c²-2accosB

c² = a²+b²-2abcosC

The trick is to add these three equations together, simplify, divide by 4, and then replace 1/2 ac with K/sinB (for example).

a²+b²+c²=2a²+2b²+2c²-2bccosA-2accosB-2abcosC

This can be rewritten as:

a²+b²+c²=2bccosA + 2accosB + 2abcosC

Divide every term by 4:

I. (a²+b²+c²)/4 = 1/2 bc cosA + 1/2 ac cosB + 1/2 ab cosC

But 1/2 bc = K/sinA; 1/2 ac = K/sinB ; 1/2 ab = K/sinC (these from the earlier mentioned formulas for area)

Substitute into I giving on the right side: (K/sinA) cosA + (K/sinB) cosB + (K/sinC) cosC

Replace cos X / sin X with cot X and on right side we have:

= KcotA + KcotB + KcotC

Now divide by K (which is the area of the triangle)

II. (a²+b²+c²)/4K = cotA + cotB + cotC

Hence, Proved