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This is part of practice, and I am clueless.

I have to prove: $\dfrac{1}{\cos θ} - \cos θ = \sin θ \tan θ$

Using the identities: $\sin^2 θ + \cos^2 θ = 1$, and/or $\tan θ = \dfrac{\sin θ}{\cos θ}$.

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I would have expected the Lord of the Nazgul to be very good at mathematics. Something of a job requirement, really. –  Will Jagy Nov 28 '12 at 21:21
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@WillJagy: I merely require proficiency with a mace. –  Wk_of_Angmar Nov 28 '12 at 21:35
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2 Answers

up vote 2 down vote accepted

Using the identity:

$$\tan{\theta}\equiv\frac{\sin{\theta}}{\cos{\theta}}$$

And multiplying both sides by $\sin{\theta}$:

$$\sin{\theta}\tan{\theta}\equiv\frac{\sin^{2}{\theta}}{\cos{\theta}}\equiv\frac{1-\cos^{2}{\theta}}{\cos{\theta}}\equiv\frac{1}{\cos{\theta}}-\cos{\theta} \qquad\blacksquare$$

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Hint: $$\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}= \dots.$$

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