# solve this problem using Pythagorean identities

$$\frac{\sin(\theta)\cdot\tan(\theta)}{1-\cos(\theta)}= \sec(\theta)+1$$ 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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Convert both tangent and secant to sines and cosines, and simplify. You should get an expression equivalent to $\sin^2+\cos^2=1$.
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?
On the lhs, convert $\tan(\theta)$ to $\frac{\sin\theta}{\cos\theta}$ and replace $\sin^2(\theta)$ with $1-\cos^2(\theta)$. Simplify.
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.
Use $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and then $\sin^2\theta=1-\cos^2\theta$