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
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
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
=( 1 /sinθ ) - sinθ
= ( 1 - sin^2(θ) / sinθ
= ( cos^2(θ) / sinθ
= cos θ ( cos θ / sinθ)
= cotθ cosθ