1-cos/1+cos= 2(cosec-cot)
I presume you want to solve this equation for something. It would be easier if you actually wrote it with variables, coefficients, all that cool stuff.
Or is it just an identity you want to prove? If you want help, don't just toss trash over the transom.
Ahh. an identity
(1-cos)/(1+cos) * (1-cos)/(1-cos)
= (1-cos)^2/(1-cos^2)
= (1 - 2cos + cos^2)/sin^2
= (1 - 2cos + 1 - sin^2)/sin^2
= ***
Hmmm. mistake?
Take x=π/4. Things don't match.
To solve the equation 1 - cos(x) / 1 + cos(x) = 2(cosec(x) - cot(x)), we need to simplify both sides of the equation.
Starting with the left side:
1 - cos(x) / 1 + cos(x)
To simplify this, we can use the identity a^2 - b^2 = (a + b)(a - b). The denominator 1 + cos(x) can be regarded as a^2 while the numerator 1 - cos(x) can be seen as b^2. Applying the identity, we get:
(1 - cos(x))(1 + cos(x)) / (1 + cos(x))
Now, the common factor (1 + cos(x)) can be canceled out, resulting in:
1 - cos^2(x) / 1
As per the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can substitute sin^2(x) for 1 - cos^2(x). Doing this gives us:
sin^2(x) / 1
which is just sin^2(x).
Now, moving on to the right side:
2(cosec(x) - cot(x))
Using the reciprocal identities, we can express cosec(x) and cot(x) as follows:
2((1/sin(x)) - (cos(x)/sin(x)))
Combining the two terms gives us:
2(1 - cos(x)) / sin(x)
Now that we have simplified both sides of the equation, we can set them equal to each other:
sin^2(x) = 2(1 - cos(x)) / sin(x)
To proceed further, we can multiply both sides by sin(x) to eliminate the denominator:
sin^3(x) = 2(1 - cos(x))
Expanding the right side, we get:
sin^3(x) = 2 - 2cos(x)
Finally, we have solved the equation 1 - cos(x) / 1 + cos(x) = 2(cosec(x) - cot(x)) and simplified it to sin^3(x) = 2 - 2cos(x).