Prove cscθ - sinθ = cotθ*cosθ
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Prove cscθ - sinθ = cotθ*cosθ

[From: ] [author: ] [Date: 12-07-04] [Hit: ]
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This was from my math exam and I couldn't get it, no matter what I did (plus I was limited for time). It's been driving my crazy and I want to know if my attempts were going in the right direction.

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Strategy: If you're trying to show that when one trig function has another added or subtracted to it is equal to two multiplied trig functions, re-write it so you can factor it and then play around with it until you get the result you're looking for.

cscθ - sinθ
= (1/sinθ) - sinθ
= (1/sinθ) - [(sinθ)^2 / sinθ]
= [1 - (sinθ)^2] / sinθ
= (cosθ)^2 / sinθ
= cotθ * cosθ

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k = csc θ - sin θ

k = 1/sin θ - sin θ

k = (1 - sin² θ)/sin θ

k = cos² θ/sin θ

k = (cos θ/sin θ) cos θ

k = cot θ cos θ

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cscθ - sinθ

=( 1 /sinθ ) - sinθ

= ( 1 - sin^2(θ) / sinθ

= ( cos^2(θ) / sinθ

= cos θ ( cos θ / sinθ)

= cotθ cosθ
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