Help simplify Trig identities ?

[bchang57]

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(sin²θ+cos²&#952😉(sin²θ-cos²&#952😉/sin2θ

I know sin2θ = 2sinθcosθ...but i am stuck -.-

i'm trying to get to -cos2θ/sin2θ to get the final answer choice of -cot2θ

 

[Neil45]

(sin²θ+cos²&#952😉(sin²θ-cos²&#952😉/sin2θ

I know sin2θ = 2sinθcosθ...but i am stuck -.-

i'm trying to get to -cos2θ/sin2θ to get the final answer choice of -cot2θ

We know the following:

sin²θ+cos²θ = 1

and

cos 2θ = cos²θ - sin²θ


So to solve the problem:

(sin²θ+cos²&#952😉(sin²θ-cos²&#952😉 / sin2θ

(1)(sin²θ-cos²&#952😉 / sin2θ

- (1)(cos²θ-sin²&#952😉 / sin2θ

- cos2θ /sin2θ

= -cot2θ

 

[bchang57]

cos 2θ = cos²θ - sin²θ

ahh thank u...a trig identity missing from my brain!