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Exercise: Multiple Angle Identities · Q24

Q.Prove that 1−cos⁡2xsin⁡2x=tan⁡x\dfrac{1-\cos 2x}{\sin 2x} = \tan x.

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Using cos⁡2x=1−2sin⁡2x⇒1−cos⁡2x=2sin⁡2x\cos2x=1-2\sin^2x \Rightarrow 1-\cos2x=2\sin^2x, and sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x:

1−cos⁡2xsin⁡2x=2sin⁡2x2sin⁡xcos⁡x=sin⁡xcos⁡x=tan⁡x\frac{1-\cos2x}{\sin2x} = \frac{2\sin^2x}{2\sin x\cos x} = \frac{\sin x}{\cos x} = \tan x …

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