−tanθ
−cosθ
tanθ
cosθ
Correct answer is A
cos2θ−1sin2θ
cos(x+y)=cosxcosy−sinxsiny⟹cos2θ=cos2θ−sin2θ
cos2θ=1−sin2θ⟹cos2θ=1−2sin2θ
sin2θ=2sinθcosθ
∴
= \frac{-2 \sin^{2} \theta}{2\sin \theta \cos \theta} = \frac{- \sin \theta}{\cos \theta}
= -\tan \theta
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