To prove that
(1 + cot 60°) / (1 - cos 60°)^2 = (1 + cos 30°) / (1 - cos 30°),
we'll start by simplifying each side individually:
Left-hand side (LHS):
(1 + cot 60°) / (1 - cos 60°)^2
Now, cot 60° can be simplified as 1 / tan 60°. And, cos 60° is equal to 1/2.
So, cot 60° = 1 / tan 60° = 1 / √3
Now, let's substitute these values:
LHS = (1 + 1/√3) / (1 - 1/2)^2
= (√3 + 1) / (1/2)^2
= (√3 + 1) / (1/4)
= 4(√3 + 1)
Right-hand side (RHS):
(1 + cos 30°) / (1 - cos 30°)
Now, cos 30° is equal to √3/2.
So, RHS = (1 + √3/2) / (1 - √3/2)
= (2 + √3) / (2 - √3)
= (2 + √3) * (2 + √3) / (2 - √3) * (2 + √3) [Rationalizing the denominator]
= (4 + 2√3 + 2√3 + 3) / (4 - 3)
= (7 + 4√3) / 1
= 7 + 4√3
Now, we can observe that LHS = RHS:
4(√3 + 1) = 7 + 4√3
Multiplying the entire equation by √3:
4√3 + 4 = 7√3 + 4√3
Combining like terms:
4 = 7√3
This equation is not true, which means our initial statement is false. Therefore, we have shown that
(1 + cot 60°) / (1 - cos 60°)^2 ≠ (1 + cos 30°) / (1 - cos 30°).