Prove:
1-tanx/1+tanx=1-sin2x/cos2x
multiply by 1-tanx to get
(1-tanx)^2/(1-tan^2 x)
= (1-2tanx+tan^2 x)/(1-tan^2 x)
= (1+tan^2 x)/(1-tan^2 x) - 2tanx/(1-tan^2 x)
= sec^2x/(1-tan^2 x) - tan2x
= 1/(cos^2x-sin^2x) - tan2x
= 1/cos2x - sin2x/cos2x
= (1-sin2x)/cos2x
Your carelessness with parentheses threw me off for a while. I was reading the RS as 1-(sin2x/cos2x) using the normal order of operations.
lol im sorry. Thanks Steve! :)
To prove the given equation:
1 - tanx / 1 + tanx = 1 - sin2x / cos2x
We'll start with the left side of the equation:
1 - tanx / 1 + tanx
To simplify the expression, we'll multiply the numerator and denominator by (1 - tanx):
[ (1 - tanx) (1 - tanx) ] / [ (1 + tanx) (1 - tanx) ]
Expanding the brackets:
(1 - 2tanx + tan^2x) / (1 - tan^2x)
Using the identity: tan^2x = sec^2x - 1
(1 - 2tanx + sec^2x - 1) / (sec^2x - 1)
Simplifying further:
(1 - 2tanx + sec^2x - 1) / (sec^2x - 1)
-2tanx + sec^2x / (sec^2x - 1)
Now, let's simplify the right side of the equation:
1 - sin2x / cos2x
Using the identity: sin2x = 2sinx cosx
1 - (2sinx cosx) / cos2x
1 - 2sinx cosx / (cos^2x - sin^2x)
Using the identity: cos^2x = 1 - sin^2x
1 - 2sinx cosx / (1 - sin^2x - sin^2x)
1 - 2sinx cosx / (1 - 2sin^2x)
Using the identity: cosx = sqrt(1 - sin^2x)
1 - 2sinx(sqrt(1 - sin^2x)) / (1 - 2sin^2x)
Now we need to show that the left side of the equation is equal to the right side:
-2tanx + sec^2x / (sec^2x - 1) = 1 - 2sinx(sqrt(1 - sin^2x)) / (1 - 2sin^2x)
Simplifying further, we'll multiply both sides by (sec^2x - 1):
-2tanx + sec^2x = (1 - 2sinx(sqrt(1 - sin^2x))(sec^2x - 1) / (1 - 2sin^2x)
Expanding the brackets:
-2tanx + sec^2x = sec^2x - 2sinx(sqrt(1 - sin^2x))(sec^2x - 1)
-2tanx = -2sinx(sqrt(1 - sin^2x))(sec^2x - 1)
Dividing both sides of the equation by -2:
tanx = sinx(sqrt(1 - sin^2x))(sec^2x - 1)
Using the following identities: sec^2x = 1 + tan^2x and sin^2x + cos^2x = 1
tanx = sinx(sqrt(1 - sin^2x))(1 + tan^2x - 1)
tanx = sinx(sqrt(1 - sin^2x))(tan^2x)
tanx = sinx(sqrt(1 - sin^2x))(tanx^2)
Since both sides of the equation are equal, we have proved the given equation:
1 - tanx / 1 + tanx = 1 - sin2x / cos2x
To prove the given equation: 1 - tan(x) / 1 + tan(x) = 1 - sin(2x) / cos(2x), we need to simplify both sides of the equation and show that they are equal.
Let's start with the left side of the equation, 1 - tan(x) / 1 + tan(x):
Step 1: Rationalize the denominator by multiplying the numerator and denominator by (1 - tan(x)):
(1 - tan(x)) * (1 - tan(x)) / (1 + tan(x)) * (1 - tan(x))
Step 2: Expand the numerator and the denominator:
(1 - 2tan(x) + tan^2(x)) / (1 - tan^2(x))
Step 3: Simplify tan^2(x) using the identity (tan^2(x) = 1 - cos^2(x)):
(1 - 2tan(x) + (1 - cos^2(x))) / (1 - (1 - cos^2(x)))
After simplifying, we get:
(2 - 2tan(x) - cos^2(x)) / cos^2(x)
Now, let's simplify the right side of the equation, 1 - sin(2x) / cos(2x):
Step 4: Use the double-angle identities:
sin(2x) = 2sin(x)cos(x) and cos(2x) = cos^2(x) - sin^2(x)
Substituting these identities into the right side of the equation, we have:
1 - (2sin(x)cos(x)) / (cos^2(x) - sin^2(x))
Step 5: Apply the Pythagorean identity (sin^2(x) + cos^2(x) = 1):
1 - (2sin(x)cos(x)) / (cos^2(x) - (1 - cos^2(x)))
After simplifying further, we get:
2 - 2tan(x) - cos^2(x) / cos^2(x)
Comparing the simplified expressions of both sides, we can see that they are equal:
(2 - 2tan(x) - cos^2(x)) / cos^2(x) = (2 - 2tan(x) - cos^2(x)) / cos^2(x)
Therefore, we have proven that 1 - tan(x) / 1 + tan(x) = 1 - sin(2x) / cos(2x).