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I have to show that the left equation simplifies to $\tan\theta$:

Show that: $$\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$$

I do have prior knowledge that: $$\tan \theta = \frac{\sin\theta}{\cos \theta}$$

But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far.

Community
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Hatmix5
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4 Answers4

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Using the double angle formulae:

$$\cos(2\theta)=1-2\sin^2\theta$$ $$\sin(2\theta)=2\sin\theta\cos\theta$$

$$\frac{1-\cos(2\theta)}{\sin(2\theta)}=\frac{1-1+2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

rae306
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you need the double angle formulae $ cos2\theta=2cos^2\theta-1$ and $ sin2\theta=2sin\theta cos\theta $

VigneshM
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use $cos2\theta$=$1-2{sin^2{\theta}}$ and $sin2\theta$=$2{sin\theta}{cos\theta}$ and you will get your required answer.

adember
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Here is a detailed answer.
Let's go! $$ \require{cancel} \begin{align} \frac{1-\cos2\theta}{\sin2\theta}&=\frac{1-\left(\cos^2\theta-\sin^2\theta\right)}{2\sin\theta\cos\theta}\\ &=\frac{1-\cos^2\theta+\sin^2\theta}{2\sin\theta\cos\theta}\\ &=\frac{\sin^2\theta\cancel{+\cos^2\theta}\cancel{-\cos^2\theta}+\sin^2\theta}{2\sin\theta\cos\theta}\\ &=\frac{\bcancel2\cancelto{\sin\theta}{\sin^2\theta}}{\bcancel2\cancel{\sin\theta}\cos\theta}\\ &=\frac{\sin\theta}{\cos\theta}\\ &=\tan\theta \end{align} $$ I hope this helps.

Kang Oedin
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