Question

prove that cotA+cotB+cotC=a²+b²+c²/4∆​

Answer 1

Step-by-step explanation:

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}}

2

1

ab sinC = K

\bold{\dfrac{1}{2}}

2

1

ac sinB = K

\bold{\dfrac{1}{2}}

2

1

bc sinA = K

That is given two sides and an included angle the area is \bold{\dfrac{1}{2}}

2

1

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}}

2

1

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}}

2

1

bc cosA + \bold{\dfrac{1}{2}}

2

1

ac cosB + \bold{\dfrac{1}{2}}

2

1

ab cosC

But \bold{\dfrac{1}{2}}

2

1

bc = \bold{\dfrac{k}{sinA}}

sinA

k

; \bold{\dfrac{1}{2}}

2

1

ac = \bold{\dfrac{k}{sinB}}

sinB

k

; \bold{\dfrac{1}{2}}

2

1

ab = \bold{\dfrac{k}{sinC}}

sinC

k

(these from the earlier mentioned formulas for area)

Substitute into I giving on the right side: (\bold{\dfrac{k}{sinA}}

sinA

k

) cosA + (\bold{\dfrac{k}{sinB}}

sinB

k

) cosB + (\bold{\dfrac{k}{sinC}}

sinC

k

) 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∆}}

4∆

a²+b²+c²

= cotA + cotB + cotC

Hence, Proved

hope it helps ☺️