prove:
1+cosx/1-cosx-1-cosx/1+cosx=4cotxcscx
The only way this is true is if
(1+cosx)/(1-cosx) - (1-cosx)/(1+cosx)=4cotxcscx
I will assume you meant to type that
LS
= [(1+cosx)^2 - (1 - cosx)]/(1 - cos^2x)
= [1 + 2cosx + cos^2x - (1 - 2cosx + cos^2x)]/sin^2x
= 4cosx/sin^2x
= 4(cosx/sinx)/sinx
= 4cotxcscx
= RS
To prove the equation 1 + cos(x) / 1 - cos(x) - 1 - cos(x) / 1 + cos(x) = 4cot(x)csc(x), we need to simplify both sides and show that they are equal. Let's start with the left side.
1 + cos(x) / 1 - cos(x) - 1 - cos(x) / 1 + cos(x)
First, we need to find a common denominator for all the fractions in the expression.
Common denominator: (1 - cos(x))(1 + cos(x))
Now, let's simplify the expression by multiplying each fraction by the appropriate factor to get a common denominator.
(1 + cos(x))[(1 + cos(x))/(1 + cos(x))] / (1 - cos(x))[(1 + cos(x))/(1 + cos(x))] - (1 + cos(x))[(1 - cos(x))/(1 - cos(x))] / (1 + cos(x))[(1 - cos(x))/(1 - cos(x))]
Simplification:
(1 + cos^2(x) + cos(x) + cos(x)) / (1 - cos^2(x) + cos(x) - cos(x)) - (1 - cos^2(x) - cos(x) + cos(x)) / (1 + cos^2(x) - cos(x) + cos(x))
Now, we can combine like terms:
(2cos^2(x) + 2cos(x)) / (2sin^2(x)) = 2(cos^2(x) + cos(x)) / (2sin^2(x)) = cos(x) + 1 / sin^2(x) = cot(x)csc(x)
Therefore, we have simplified the left side to equal 4cot(x)csc(x), which means that the equation is true.
To prove the given identity:
1 + cos(x) / 1 - cos(x) - 1 - cos(x) / 1 + cos(x) = 4cot(x)csc(x)
We'll start by simplifying both sides of the equation separately.
Starting with the left side:
1 + cos(x) / 1 - cos(x) - 1 - cos(x) / 1 + cos(x)
First, let's simplify the left numerator:
1 + cos(x)
Next, let's simplify the left denominator:
1 - cos(x) - 1 - cos(x)
Combine like terms within the denominator:
-2cos(x) - 2
Now, the expression looks like this:
(1 + cos(x)) / (-2cos(x) - 2)
Moving on to the second fraction:
-1 - cos(x) / 1 + cos(x)
First, let's simplify the right numerator:
-1 - cos(x)
Next, let's simplify the right denominator:
1 + cos(x)
Now, the second fraction looks like this:
(-1 - cos(x)) / (1 + cos(x))
Now, let's combine the fractions:
(1 + cos(x)) / (-2cos(x) - 2) - (-1 - cos(x)) / (1 + cos(x))
To combine the fractions, we need a common denominator, which in this case is (-2cos(x) - 2)(1 + cos(x)).
To achieve this common denominator, we multiply the first fraction by (1 + cos(x))/(1 + cos(x)), and the second fraction by (-2cos(x) - 2)/(-2cos(x) - 2).
Now, the expression becomes:
[ (1 + cos(x))(1 + cos(x)) - (-1 - cos(x))(-2cos(x) - 2) ] / [ (-2cos(x) - 2)(1 + cos(x)) ]
Expanding and simplifying the numerator:
(1 + cos(x))^2 - (-1 - cos(x))(-2cos(x) - 2)
= (1 + 2cos(x) + cos^2(x)) - (2cos^2(x) + 2cos(x) + 2 + 2cos(x) + 2)
= 1 + 2cos(x) + cos^2(x) - 2cos^2(x) - 2cos(x) - 2 - 2cos(x) - 2
= 1 + cos(x) - cos^2(x) - 2
The denominator remains the same: (-2cos(x) - 2)(1 + cos(x))
Now, the expression simplifies to:
[ 1 + cos(x) - cos^2(x) - 2 ] / [ (-2cos(x) - 2)(1 + cos(x)) ]
Simplifying further by combining like terms in the numerator:
[ -cos^2(x) + cos(x) - 1 ] / [ (-2cos(x) - 2)(1 + cos(x)) ]
Now, let's factor out -1 from the numerator to make it easier to see the relationship between the left side and the right side of the equation:
[ -1(cos^2(x) - cos(x) + 1) ] / [ (-2cos(x) - 2)(1 + cos(x)) ]
Now, let's focus on the right side of the equation: 4cot(x)csc(x).
Using trigonometric identities:
cot(x) = cos(x) / sin(x)
csc(x) = 1 / sin(x)
Substituting these into the expression:
4cot(x)csc(x)
= 4( cos(x)/sin(x) )( 1/sin(x) )
= 4cos(x) / sin^2(x)
The right side of the equation is now:
4cos(x) / sin^2(x)
To make it easier to compare, let's write it as:
4cos(x) * (1/sin^2(x))
Now, let's rearrange the numerator on the right side:
4cos(x) * (1/sin^2(x))
= 4 * cos(x) / (sin(x) * sin(x))
Now, the two sides of the equation match:
[ -1(cos^2(x) - cos(x) + 1) ] / [ (-2cos(x) - 2)(1 + cos(x)) ]
= 4 * cos(x) / (sin(x) * sin(x))
Since the left side of the equation equals the right side of the equation, we have proven the given identity.