Verify the function: (1-cosΘ)/(1+cosΘ) = (cscΘ - cotΘ)²

Explain how to do this, I cannot figure it out!

(1-cosΘ)/(1+cosΘ)
= (1-cosΘ)^2/(1-cos^2Θ)
= [(1-cosΘ)/sinΘ]^2
= (cscΘ - cotΘ)^2

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To θ βяιαη θ,
TD my answer won't always help you, right?

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We can convert the right side into the left side by converting the right side in terms of sines and cosines only. Doing this gives:
RHS = (cscθ - cotθ)^2
= (1/sinθ - cos/sinθ)^2, since cscθ = 1/cosθ and cotθ = cosθ/sinθ
= [(1 - cosθ)/sinθ]^2, by combining fractions
= (1 - cosθ)^2/sin^2θ, by squaring
= (1 - cosθ)^2/(1 - cos^2θ), since sin^2θ = 1 - cos^2θ
= (1 - cosθ)^2/[(1 + cosθ)(1 - cosθ)], by difference of squares
= (1 - cosθ)/(1 + cosθ), by canceling 1 - cosθ
= LHS.

I hope this helps!

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