5. In a triangle ABC, prove that
A B
С
cot — + cot
+ cot
(a+b+c)2
2 2 2
Prove that
a? +62 +62 Cot A + cot B+Cot C
Answers
Answer:
Step-by-step explanation:
There are few key formula used here around Properties of triangle:
Let A , B , C be the angles of the triangle and a , b , c be the length of sides opposite to respective angles.
Sine Rule:
a/SinA = b/SinB = c/Sinc = 2R
Cosine Rule: a² = b² + c² - 2bcCosA = > CosA = b² + c² - a² / 2bc
Similarly, CosB = c² + a² - b² / 2ac, CosC = a² + b² - c² / 2ab.
Area of triangle in terms of Sine is
Δ = (1/2)bcSinA = (1/2)caSinB = (1/2)abSinC
Let us now calculate the values:
CotA = CosA/SinA
= 2bcCosA/2bcSinA (∵ Multiply numerator and denominator by 2bc)
= b² + c² - a² / 4Δ ( ∵ CosA = b² + c² - a² / 2bc => 2bcCosA = b² + c² - a²
∵ Δ = (1/2)bcSinA => bcSinA = 2Δ)
Similarly we can calculate CotB and CotC,
=> CotB = c² + a² - b² / 4Δ
=> CotC = a² + b² - c² / 4Δ
let us calculate the half angles
CotA/2 = CosA/2 / SinA/2
= 2Cos²A/2 / 2SinA/2CosA/2 (∵ Multiply numerator and
denominator by 2CosA/2)
= 1 + CosA/SinA (∵ Cos2A = 2Cos²A - 1 and Sin2A = 2SinACosA)
= [1 + b² + c² - a²/2bc] / 2Δ/bc
= [2bc + b² + c² - a² / 2bc ] * [bc/2Δ]
= [2bc + b² + c² - a²] /4Δ
Similarly we can calculate CotB/2 and CotC/2,
=> CotB/2 = [2ca + c² + a² - b²] /4Δ
=> CotC/2 = [2ab + a² + b² - c²] /4Δ
R.H.S
CotA/2 + CotB/2 + CotC/2 / CotA + CotB + CotC
Numerator => CotA/2 + CotB/2 + CotC/2
= [2bc + b² + c² - a²] /4Δ + [2ca + c² + a² - b²] /4Δ + [2ab + a² + b² - c²] /4Δ
= 1/4Δ [2bc + b² + c² - a² + 2ca + c² + a² - b² + 2ab + a² + b² - c²]
= (a + b + c )²/4Δ --------- (1)
Denominator => CotA + CotB + CotC
= b² + c² - a² / 4Δ + c² + a² - b² / 4Δ + a² + b² - c² / 4Δ
= 1/4Δ [b² + c² - a² + c² + a² - b² + a² + b² - c²]
= a² + b² + c²/4Δ ---------- (2)
Now (1) / (2)
[(a + b + c )²/4Δ] / [a² + b² + c²/4Δ]
= (a + b + c )² / a² + b² + c²
= L.H.S
Hence Proved