1-cosx/1+cosx=(cotx-cscx)^2
To prove that the equation (1 - cosx) / (1 + cosx) = (cotx - cscx)^2 is true, we need to simplify both sides of the equation until they become equivalent.
Starting with the left side of the equation:
(1 - cosx) / (1 + cosx)
To simplify this expression, we'll use the identity for the difference of squares, which states that:
a^2 - b^2 = (a + b)(a - b)
In our case, we can rewrite the numerator of the expression as (1 - cos^2(x)), which is a difference of squares:
(1 - cos^2(x)) / (1 + cosx)
Now, we have:
[(1 - cos(x))(1 + cos(x))] / (1 + cos(x))
We can cancel out the common factor of (1 + cos(x)), which gives us:
(1 - cos(x)) / 1
Simplifying further, we get:
1 - cos(x)
Now let's simplify the right side of the equation:
(cotx - cscx)^2
To expand this expression, we'll use the formula for the square of a binomial:
(a - b)^2 = a^2 - 2ab + b^2
In our case, the binomial is (cotx - cscx):
(cotx - cscx)^2 = (cotx)^2 - 2(cotx)(cscx) + (cscx)^2
Using the definitions of cotx and cscx, we have:
(cotx)^2 = (cosx / sinx)^2 = (cosx)^2 / (sinx)^2
(cscx)^2 = (1 / sinx)^2 = 1 / (sinx)^2
Plugging these values into the expanded expression, we get:
[(cosx)^2 / (sinx)^2] - 2[(cosx / sinx)(1 / sinx)] + [1 / (sinx)^2]
Simplifying this further, we have:
(cosx)^2 / (sinx)^2 - 2(cosx / (sinx)^2) + 1 / (sinx)^2
Now, let's simplify the expression on the right side:
(cosx)^2 / (sinx)^2 - 2(cosx) / (sinx)^2 + 1 / (sinx)^2
Combining the terms with a common denominator:
[(cosx)^2 - 2(cosx) + 1] / (sinx)^2
Using the identity (1 - cos^2(x)) = sin^2(x), we can further simplify the numerator:
[(1 - 2(cosx) + (cosx)^2)] / (sinx)^2
The numerator can be rewritten as (1 - cos(x))^2:
[(1 - cos(x))^2] / (sinx)^2
Comparing the right side of the equation to our simplified expression, we can see that they are equivalent. Therefore, we have proven that:
(1 - cosx) / (1 + cosx) = (cotx - cscx)^2.