Proving $6 \sec\phi \tan\phi = \frac{3}{1-\sin\phi} - \frac{3}{1+\sin \phi}$ 
$$6 \sec\phi \tan\phi = \frac{3}{1-\sin\phi} - \frac{3}{1+\sin \phi}$$

I can't seem to figure out how to prove this. 
Whenever I try to prove the left side, I end up with $\frac{6\sin\theta}{\cos\theta}$, which I think might be right. 
As for the right side, I get confused with the denominators and what to do with them. I know if I square root  $1-\sin\phi$, I'll get a Pythagorean identity, but then I don't know where to go from there.  
Please help me with a step-by-step guide. I really want to learn how to do this. 
 A: \begin{align}
\sec \phi\tan\phi &= \frac{\sin \phi}{\cos^2\phi}\\
&=\frac{\sin\phi}{1-\sin^2\phi}\\
&=\frac{\sin\phi}{(1+\sin\phi)(1-\sin\phi)}\\
&=\sin\phi\cdot \frac{1}{2\sin \phi} \left(\frac{1}{1-\sin\phi}-\frac{1}{1+\sin\phi}\right)\\
&=\frac{1}{2} \left(\frac{1}{1-\sin\phi}-\frac{1}{1+\sin\phi}\right)
\end{align}
A: Notice that 
$$\frac{3}{1-\sin\phi} - \frac{3}{1+\sin \phi}=\frac{3+3\sin\phi-3+3\sin\phi}{1-\sin^{2}\phi}$$
using $1-\sin^2\phi=\cos^2 \phi$,
We get 
$$\frac{6\sin \phi}{\cos^2 \phi}=\frac{6\sin \phi}{(\cos \phi)(\cos \phi)}$$
$$6\sec \phi \tan \phi$$
A: We have: $$\textbf{RHS} = \dfrac{3(1+\sin \phi)-3(1-\sin \phi)}{(1-\sin \phi)(1+\sin \phi)}= \dfrac{3+3\sin \phi-3+3\sin \phi}{1-\sin^2\phi}\\ =\dfrac{6\sin \phi}{\cos^2\phi}= 6\cdot \dfrac{\sin \phi}{\cos \phi}\cdot \dfrac{1}{\cos \phi}= 6\tan\phi\sec\phi= \textbf{LHS}$$
A: $\hspace{-3cm}\begin{align*}
\frac{3}{1-\sin\phi} - \frac{3}{1+\sin \phi} &= \frac{(3+3 \sin \phi) - (3 - 3 \sin \phi)}{1-\sin ^2 \phi} \tag{common denominator}\\
&= \frac{6 \sin \phi}{\cos^2 \phi} \tag{Pythagorean identity} \\
&= 6 \left ( \frac{\sin \phi}{\cos \phi} \right ) \left ( \frac{1}{\cos \phi} \right ) \\
&= 6 \sec \phi \tan \phi
\end{align*}
$
A: As $(1-\sin\phi)(1+\sin\phi)=\cos^2\phi,$
$$\dfrac1{1\pm\sin\phi}=\dfrac{1\mp\sin\phi}{\cos^2\phi}=\sec^2\phi\mp\sec\phi\tan\phi$$
A: You can do this either from LHS to RHS or from RHS to LHS.
Solution 1: LHS $\rightarrow$ RHS
$$\require{cancel}\begin{aligned}6\sec\phi\tan\phi&=6\frac{1}{\cos\phi}\frac{\sin\phi}{\cos\phi}\\&=\frac{6\sin\phi}{\cos^2\phi}\\&=\frac{3\sin\phi+3\sin\phi}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{3\left(1+\sin\phi\right)-3\left(1-\sin\phi\right)}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{3\cancel{\left(1+\sin\phi\right)}}{\left(1-\sin\phi\right)\cancel{\left(1+\sin\phi\right)}}-\frac{3\cancel{\left(1-\sin\phi\right)}}{\cancel{\left(1-\sin\phi\right)}\left(1+\sin\phi\right)}\\&=\frac{3}{1-\sin\phi}-\frac{3}{1+\sin\phi}\end{aligned}$$
Solution 2: RHS $\rightarrow$ LHS
$$\begin{aligned}\frac{3}{1-\sin\phi}-\frac{3}{1+\sin\phi}&=\frac{3\left(1+\sin\phi\right)-3\left(1-\sin\phi\right)}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{\cancel3+3\sin\phi\cancel{-3}+3\sin\phi}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{6\sin\phi}{1-\sin^2\phi}\\&=\frac{6\sin\phi}{\cos^2\phi}\\&=\frac{6\sin\phi}{\cos\phi\cos\phi}\\&=6\sec\phi\tan\phi\end{aligned}$$
I hope this helps.
