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$$\frac{\sin(\theta)\cdot\tan(\theta)}{1-\cos(\theta)}= \sec(\theta)+1$$ I need to prove that both sides of the equation are equal to each other, we are supposed to be using Pythagorean identities to help us solve.

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  • $\begingroup$ Welcome to MSE! It is very helpful to show the work you have done as it helps responders get to the specific issue as opposed to rehashing items you already know. Regards $\endgroup$ – Amzoti Jun 25 '13 at 14:28
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Convert both tangent and secant to sines and cosines, and simplify. You should get an expression equivalent to $\sin^2+\cos^2=1$.

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Hints: Recall that $\tan\theta =\frac{\sin\theta}{\cos\theta}.$ Given the Pythagorean identity, how can we rewrite $\sin^2\theta$? What is the factored form of a difference of squares?

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On the lhs, convert $\tan(\theta)$ to $\frac{\sin\theta}{\cos\theta}$ and replace $\sin^2(\theta)$ with $1-\cos^2(\theta)$. Simplify.

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Note that your equation is equivalent to $$ \sin(\theta)\tan(\theta) = (\sec(\theta) + 1)(1 - \cos(\theta)) $$ This is equivalent to $$ \sin(\theta)\tan(\theta) = \sec(\theta) - 1 + 1 - \cos(\theta) $$ Now write the $\tan(\theta)$ on the left hand side as $\sin(\theta)$ over $\cos(\theta)$. Try to multiply both sides by $\cos(\theta)$. Then simplify.

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Use $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and then $\sin^2\theta=1-\cos^2\theta$

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