Trig, fundamental identities

(1-sin^2Θ)( 1 +tan^2Θ)


Prove the following identities

1 -cosΘ
-------------- = sin/1+cos
sinΘ


1 +cosx
---------------- = sinx/1-cosx
sinx

For the first one (I'm using x instead of theta for simplicity):

(1 -cos(x)) / sin(x) = sin(x) / (1 +cos(x))

[STEP 1] Multiply both sides by sin(x):

(1 - cos(x)) = sin²(x) / (1 + cos(x))

[STEP 2] Use the identity sin²(x) + cos²(x) = 1:

1 - cos(x) = (1 - cos²(x)) / (1 + cos(x))

[STEP 3] Factor the numerator, as it is a difference of two squares:

1 - cos(x) = (1 - cos(x))(1 + cos(x)) / (1 + cos(x))

[STEP 4] Cancel out the 1 + cos(x) in the numerator and denominator:

1 - cos(x) = 1 - cos(x)

QED.
====

The second one proceeds along substantially similar lines.

2)
1 +cosx
---------------- = sinx/1-cosx
sinx
multiply LHS numerator and denominator by (1-cosx)
then
LHS=(1-cos^2x)/sinx(1-cosx)
sin^2x/sinx(1-cosx)
=sinx/(1-cosx)
1) similarly solve this also

Lloyd, you do realize that since you are multiplying both sides of this identity by the same thing (sinx), then you are already assuming equality, which is what you're trying to prove. May want to reconsider.