prove:(cscx-cotx)^4 x (cscx+cotx)^4 = 1
To prove the given equation, we need to simplify both sides of the equation separately and show that they are equal to each other. Let's start by simplifying the left-hand side (LHS) of the equation:
LHS: (cscx - cotx)^4 * (cscx + cotx)^4
First, we can expand both binomial expressions using the fourth power:
LHS: (cscx - cotx) * (cscx - cotx) * (cscx - cotx) * (cscx - cotx) * (cscx + cotx) * (cscx + cotx) * (cscx + cotx) * (cscx + cotx)
To simplify this expression, we can use the formula for expanding the product of two binomials:
(a - b)(c - d) = ac - ad - bc + bd
Using this formula, we can expand the first two terms:
LHS: (cscx - cotx)^2 * (cscx + cotx)^2 * (cscx + cotx) * (cscx + cotx)
Expanding the first two terms again:
LHS: (cscx - cotx)(cscx - cotx) * (cscx - cotx)(cscx - cotx) * (cscx + cotx)(cscx + cotx) * (cscx + cotx)
Expanding further:
LHS: (csc^2(x) - 2csc(x)cot(x) + cot^2(x)) * (csc^2(x) - 2csc(x)cot(x) + cot^2(x)) * (csc^2(x) + 2csc(x)cot(x) + cot^2(x))
Now, we can simplify this expression by recognizing some trigonometric identities. The Pythagorean identities state:
csc^2(x) = 1 + cot^2(x)
cot^2(x) = csc^2(x) - 1
Using these identities, we can substitute and simplify further:
LHS: (1 + cot^2(x) - 2csc(x)cot(x) + cot^2(x)) * (1 + cot^2(x) - 2csc(x)cot(x) + cot^2(x)) * (1 + cot^2(x) + 2csc(x)cot(x) + cot^2(x))
Simplifying terms:
LHS: (2 + 2cot^2(x) - 2csc(x)cot(x)) * (2 + 2cot^2(x) + 2csc(x)cot(x))
Multiplying further:
LHS: 4 + 4cot^2(x) - 4csc(x)cot(x) + 4cot^2(x) + 4cot^4(x) - 4csc(x)cot(x) + 4csc(x)cot(x) + 4csc(x)cot^3(x))
Combining like terms:
LHS: 4 + 8cot^2(x) - 8csc(x)cot(x) + 4cot^4(x) + 4csc(x)cot^3(x)
Now, let's simplify the right-hand side (RHS) of the equation, which is equal to 1:
RHS: 1
Since RHS is already simplified, we can compare the LHS and RHS to see if they are equal:
4 + 8cot^2(x) - 8csc(x)cot(x) + 4cot^4(x) + 4csc(x)cot^3(x) = 1
We can see that the LHS and RHS are not equal to each other, so the given equation is not true for all values of x.
Therefore, we have proved that (cscx - cotx)^4 * (cscx + cotx)^4 ≠ 1.