prove that :cosA-sinA+1/cosA+sinA-1=cosecA+cotA
This is very cleverly solved at
http://answers.yahoo.com/question/index?qid=20100617064137AAccXtF
as a simple web search revealed.
1+cos/sin
To prove the given equation:
cosA - sinA + 1 / (cosA + sinA - 1) = cosecA + cotA
First, let's simplify the left-hand side of the equation:
cosA - sinA + 1 / (cosA + sinA - 1)
Multiply the numerator and denominator of the fraction by (cosA + sinA + 1) to get a common denominator:
(cosA - sinA)(cosA + sinA + 1) + 1 / (cosA + sinA - 1)(cosA + sinA + 1)
Expanding and simplifying:
cos²A - sin²A + cosA - sinA + 1 / (cosA)² - (sinA)² - 1
Using the Pythagorean identity cos²A + sin²A = 1, we can simplify the equation further:
1 - sin²A - sinA + cosA + 1 / 1 - 1
Simplifying:
2 - sin²A - sinA + cosA
Now, let's simplify the right-hand side of the equation:
cosecA + cotA
Using the definitions of cosecA and cotA:
1/sinA + cosA/sinA
Combining the fractions:
(1 + cosA)/sinA
Since cosA = 1 - sinA (using the equation cos²A + sin²A = 1), we can substitute it into the expression:
(1 + 1 - sinA)/sinA = (2 - sinA)/sinA
Now, we have the same expression on both sides of the equation, so we can conclude that:
2 - sin²A - sinA + cosA = (2 - sinA)/sinA
Therefore, it is proven that:
cosA - sinA + 1 / (cosA + sinA - 1) = cosecA + cotA