prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx
To prove that the expression 2sin(x)cos(x) - cos(x) / (1 - sin(x) + sin^2(x) - cos^2(x)) is equivalent to cot(x), we can simplify the expression step by step and show that it simplifies to cot(x).
We'll start by simplifying the numerator and denominator of the expression separately.
Numerator:
The numerator is 2sin(x)cos(x) - cos(x). We can factor out a common term of cos(x) and rewrite it as cos(x)(2sin(x) - 1).
Denominator:
The denominator is 1 - sin(x) + sin^2(x) - cos^2(x). We can rewrite it as 1 - sin(x) + sin^2(x) - (1 - sin^2(x)), which simplifies to 1 - sin(x) + sin^2(x) - 1 + sin^2(x). Simplifying further, we get -sin(x) + 2sin^2(x).
Now that we have simplified the numerator and denominator, we can substitute these simplified forms back into the original expression:
cos(x)(2sin(x) - 1) / (-sin(x) + 2sin^2(x))
Next, we'll simplify the expression further by factoring out common terms in the numerator and denominator:
cos(x)(2sin(x) - 1) / (-sin(x)(1 - 2sin(x)))
Now we can cancel out the common factor of -sin(x) from the numerator and denominator:
cos(x)(2sin(x) - 1) / (1 - 2sin(x))
Finally, we can simplify the expression even further by factoring out -1 from the denominator:
cos(x)(2sin(x) - 1) / -(2sin(x) - 1)
Now, we can cancel out the common factor of (2sin(x) - 1):
cos(x) / -1
The simplified expression is equal to -cos(x).
To prove that -cos(x) is equivalent to cot(x), we can recall the trigonometric identity for cotangent:
cot(x) = cos(x) / sin(x)
By dividing -cos(x) by sin(x), we get:
-cos(x) / sin(x)
Since -cos(x) / sin(x) is equal to cot(x), we have proven that the original expression simplifies to cot(x).
To prove the given statement, we will start with the left-hand side (LHS) and simplify it until we obtain the right-hand side (RHS).
LHS = (2sin(x)cos(x) - cos(x))/(1 - sin(x) + sin^2(x) - cos^2(x))
First, we will factor out -1 from the denominator:
LHS = (2sin(x)cos(x) - cos(x))/(1 + (sin^2(x) - sin(x)) - cos^2(x))
Next, we can simplify the denominator further by using the trigonometric identity sin^2(x) + cos^2(x) = 1:
LHS = (2sin(x)cos(x) - cos(x))/(1 + sin^2(x) - sin(x) - (1 - sin^2(x)))
Combining like terms:
LHS = (2sin(x)cos(x) - cos(x))/(1 - sin(x)^2 + sin(x) + sin(x)^2)
LHS = (2sin(x)cos(x) - cos(x))/(2sin(x) + sin(x))
Factoring out the common term sin(x) from the numerator:
LHS = cos(x)(2sin(x) - 1)/(sin(x)(2 + 1))
Simplifying further:
LHS = cos(x)(2sin(x) - 1)/(3sin(x))
Using the identity cot(x) = cos(x)/sin(x):
LHS = cos(x)(2sin(x) - 1)/(3sin(x)) = (2sin(x) - 1)/(3sin(x)) * cos(x)/sin(x)
LHS = (2sin(x) - 1)/(3sin(x)) * cot(x)
Therefore, we have shown that LHS = cot(x), which is the RHS of the given equation.
LHS = (2sinxcosx-cosx)/(1-sinx-cos^2x+sin^2x)
= (2sinxcosx-cosx)/(sin^2x+cos^2x-sinx-cos^2x+sin^2x)
= (2sinxcosx-cosx)/(2sin^2x+cos^2x)
= cosx(2sinx-1)/sinx(2sinx-1)
= cosx/sinx
= cotx
=RHS