prove:(cscx-cotx)^4 x (cscx+cotx)^4 = 1
LS
= [(cscx - cotx)(cscx + cotx)]^4
= [csc^2x - cot^2x]^4
= [1 + cot^2x - cot^2x]^4
= [1]^4
= 1
= RS
To prove the given equation, we need to demonstrate that both sides of the equation are equal. Let's start by simplifying each side step by step.
On the left side of the equation, we have:
(csc(x) - cot(x))^4 * (csc(x) + cot(x))^4
To simplify this expression, we can use the identity: (a - b)^4 = (a^2 - 2ab + b^2)^2
Applying this identity to the left side, we get:
((csc(x))^2 - 2(csc(x) * cot(x)) + (cot(x))^2)^2 * (csc(x) + cot(x))^4
Now, let's expand the square:
((csc^2(x) - 2(csc(x) * cot(x)) + cot^2(x))^2 * (csc(x) + cot(x))^4
Now, let's simplify further by expanding the fourth power:
(csc^4(x) - 4(csc^3(x) * cot(x)) + 6(csc^2(x) * cot^2(x)) - 4(csc(x) * cot^3(x)) + cot^4(x)) * (csc(x) + cot(x))^4
We can simplify this expression by distributing the power of (csc(x) + cot(x))^4 to each term individually:
(csc^4(x) - 4(csc^3(x) * cot(x)) + 6(csc^2(x) * cot^2(x)) - 4(csc(x) * cot^3(x)) + cot^4(x)) * (csc^4(x) + 4(csc^3(x) * cot(x)) + 6(csc^2(x) * cot^2(x)) + 4(csc(x) * cot^3(x)) + cot^4(x))
Now, if we distribute and simplify the expression further, we get:
csc^8(x) + 4csc^7(x)cot(x) + 6csc^6(x)cot^2(x) + 4csc^5(x)cot^3(x) + cot^8(x)
Now, let's simplify the right side of the equation:
1
Since the right side is just a constant value of 1, we can see that the left side does not simplify to the right side. Therefore, the given equation is not true.
Hence, we have proven that (csc(x) - cot(x))^4 * (csc(x) + cot(x))^4 ≠ 1.