Proving identities: (tanA sinA)/(1 - cosA) = 1 + (1/cosA)

Please refrain from providing complete solution in the first post. Let the OP discover the solution
The equation is the same as tanasina=(1-cosa)(1+(1/cosa))=(1-cosa)(1+cosa)/cosa=(1-cos^2a)/cosa=sin^2a/cosa=sina/cosasina=tanasina
 
PROVE THE IDENTITY:
(tanAsinA)/(1-cosA) = (1+(1/cosA))
\(\displaystyle \begin{align*}\frac{\tan(A)\sin(A)}{1-\cos(A)}&= \frac{\sin^2(A)}{\cos(A)(1-\cos(A)}\\&=\frac{1-\cos^2(A)}{\cos(A)*(1-\cos(A))}\\&=\frac{(1-\cos(A))*(1+\cos(A))}{\cos(A)*(1-\cos(A))}\\&=\frac{(1+\cos(A))}{\cos(A)}\\&=~? \end{align*}\)

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What sometimes helps with trig identities is to start off by reducing everything to just sine and cosine.

[MATH]\dfrac{tan(a) * sin(a)}{1 - cos(a)} = \dfrac{\dfrac{sin(a)}{cos(a)} * sin(a)}{1 - cos(a)} = \dfrac{\dfrac{sin^2(a)}{cos(a)}}{\dfrac{1 - cos(a)}{1}} =[/MATH]
[MATH]\dfrac{sin^2(a)}{cos(a)\{1 - cos(a)\}}.[/MATH]
And whenever you see sine squared, you should ask yourself whether the basic identities of

[MATH]sin^2(a) = 1 - cos^2(a) = \{1 - cos(a)\}\{1 + cos(a)\}[/MATH]
might be helpful.

These are not the only generic tricks useful for identities, but they are ones to keep in mind. And in this problem, they are virtually all you need except basic algebra.
 
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