Question

prove that cotA + cot B + cotc =
a² + b²+c²/4∆​

Answer 1

Question

Show that cotA+cotB+cotC= $\bold{\dfrac{a²+b²+c²}{4∆}}$

Solution:

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:

$\bold{\dfrac{1}{2}}$ ab sinC = K

$\bold{\dfrac{1}{2}}$ ac sinB = K

$\bold{\dfrac{1}{2}}$ bc sinA = K

That is given two sides and an included angle the area is $\bold{\dfrac{1}{2}}$ 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 $\bold{\frac{1}{2}}$ 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 = $\bold{\dfrac{1}{2}}$ bc cosA + $\bold{\dfrac{1}{2}}$ ac cosB + $\bold{\dfrac{1}{2}}$ ab cosC

But $\bold{\dfrac{1}{2}}$ bc = $\bold{\dfrac{k}{sinA}}$; $\bold{\dfrac{1}{2}}$ ac = $\bold{\dfrac{k}{sinB}}$ ; $\bold{\dfrac{1}{2}}$ ab = $\bold{\dfrac{k}{sinC}}$

(these from the earlier mentioned formulas for area)

Substitute into I giving on the right side: ($\bold{\dfrac{k}{sinA}}$) cosA + ($\bold{\dfrac{k}{sinB}}$) cosB + ($\bold{\dfrac{k}{sinC}}$) 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. $\bold{\dfrac{a²+b²+c²}{4∆}}$ = cotA + cotB + cotC

Hence, Proved

hope it helps ☺️