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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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    @WillJagy: I merely require proficiency with a mace.2012-11-28

2 Answers 2

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