Prove that 1 + cot 60 degree upon 1 - Cos 60 degree ka whole square equal 1 + cos 30 degree upon 1 - cos 30 degree

Answer 1

To prove the given equation,
1 + cot 60° / (1 - cos 60°)^2 = 1 + cos 30° / (1 - cos 30°)

First, let's evaluate the values on both sides:

cot 60° = 1 / tan 60°
tan 60° = √3
cot 60° = 1 / √3
cot 60° = √3 / 3

cos 60° = 1/2

(1 - cos 60°)^2 = (1 - 1/2)^2 = (1/2)^2 = 1/4

cos 30° = √3/2

(1 - cos 30°) = (1 - √3/2) = (2 - √3) / 2

Now, substituting the values in the equation,

LHS = 1 + (√3 / 3) / (1 - 1/2)^2
= 1 + (√3 / 3) / (1/4)
= 1 + (√3 / 3) * (4/1)
= 1 + 4√3 / 3

RHS = 1 + (√3/2) / (2 - √3 / 2)^2
= 1 + (√3/2) / ((4 - √3) / 2)^2
= 1 + (√3/2) / ((4 - √3)^2 / 4)
= 1 + (√3/2) * (4/(4 - √3)^2)
= 1 + (4√3/2) * (4/(4 - √3)^2)
= 1 + 4√3 / 3

Since LHS = RHS, we have proven that:

1 + cot 60° / (1 - cos 60°)^2 = 1 + cos 30° / (1 - cos 30°)