prove: tanθ = (1 - cos2θ) / (sin2θ)?

(1 - cos2θ)/sin2θ

...1 - [2cos^2θ - 1]...... note cos2θ = 2cos^2θ - 1, sin2θ = 2sinθcosθ

= -------------------

.....2sinθcosθ

...1 - 2cos^2θ + 1

=-----------------------

.....2sinθcosθ

...2 - 2cos^2θ

=--------------------

...2sinθcosθ

...2(1 - cos^2θ)

=--------------------

.....2sinθcosθ

......2sin^2θ

=----------------- note sin^2θ = 1 - cos^2θ

....2sinθcosθ

.....sinθ

=----------.... note tanθ = sinθ/cosθ

....cosθ

=tanθ prove//

cos2θ = 1 - 2sin²θ and sin2θ = 2sinθcosθ

so, (1 - cos2θ)/sin2θ => (1 - (1 - 2sin²θ))/2sinθcosθ  

=> 2sin²θ/2sinθcosθ  

i.e. sinθ/cosθ = tanθ

:)>  

For θ ≠ kπ/2 for any integer k:

RHS

= [1 - cos(2θ)] / sin(2θ)

= {1 - [1 - 2sin²(θ)]} / [2sin(θ)cos(θ)]

= 2sin²(θ) / [2sin(θ)cos(θ)]

= sin(θ) / cos(θ)

= tan(θ)

= LHS

Mind that exclusion I started with. The equation is not an identity without it, and in fact, the step where I divided above and below by 2sin(θ) would have been invalid. If you doubt that, try letting θ equal zero. In that case the left side would be zero while the right side is undefined, clearly not an identity.

tan(θ) = (1 - cos(2θ)) / (sin(2θ))

Multiply through by sin(2θ):

tan(θ)*(sin(2θ)) = 1 - cos(2θ)

Use double-angle formulas:

(sin(θ)/cos(θ))*2*sin(θ)*cos(θ) = 1 - (2*cos^2(θ) - 1)

Simplify:

2*sin^2(θ) = 1 - 2*cos^2(θ) + 1

2*sin^2(θ) = 2 - 2*cos^2(θ)

2*sin^2(θ) + 2*cos^2(θ) = 2

sin^2(θ) + cos^2(θ) = 1

That's Pythagoras.

I invite you to try your OWN homework and show your work.  Then, if you STILL can't get it, I am sure someone will help you.