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10 questions from this quiz
sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1sin2θ+cos2θ=1
1+tan2θ=sec2θ1 + \tan^2 \theta = \sec^2 \theta1+tan2θ=sec2θ
cscθ=1sinθ\csc \theta = \frac{1}{\sin \theta}cscθ=sinθ1
tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}tanθ=cosθsinθ
cosθ\cos \thetacosθ
−sinθ-\sin \theta−sinθ
sin2θ=2sinθcosθ\sin 2\theta = 2 \sin \theta \cos \thetasin2θ=2sinθcosθ
Identical to; true for all values
It's not an identity; only true for specific θ\thetaθ
Dividing loses the solution where sinθ=0\sin \theta = 0sinθ=0
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