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.

• Clearing denominators is apt to simplify this particular problem, but there is more than one approach to simplification. – hardmath Feb 28 '16 at 19:13
• For the RHS, how do you get a common denominator? Hint: find $$\frac{3}{5}-\frac{3}{7}$$ – John Joy Feb 29 '16 at 3:38

\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}

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 $$\frac{6\sin \phi}{\cos^2 \phi}=\frac{6\sin \phi}{(\cos \phi)(\cos \phi)}$$ $$6\sec \phi \tan \phi$$

• but why would you cross multiply to get the numerator of 3+3.... – Esma Feb 28 '16 at 19:48
• @Esma - to add two fractions together, they need to have a common denominator. The smallest common denominator in this case is their product. So $$\frac 3{1 - \sin \phi} = \frac 3{1 - \sin \phi}\frac {1 + \sin \phi}{1 + \sin \phi} = \frac {3(1+\sin \phi)}{1 - \sin^2 \phi}$$ and $$\frac 3{1 + \sin \phi} = \frac 3{1 + \sin \phi}\frac {1 - \sin \phi}{1 - \sin \phi} = \frac {3(1-\sin \phi)}{1 - \sin^2 \phi}$$ – Paul Sinclair Feb 28 '16 at 20:02
• @Esma : You start with one term on the left and two terms on the right. At some point, you're going to have to cancel or merge the terms on the right. One way to do this is putting them over a common denominator. (Nothing was cross-multiplied. That term only applies to multiplication across an equals sign. The two fractions were put over a common denominator.) – Eric Towers Feb 28 '16 at 20:03
• thank you so much. I understand now. This is what I did at first, but got lost. Once again, thank you so much for your help @PaulSinclair – Esma Feb 28 '16 at 20:06

We have: $$\textbf{RHS} = \dfrac{3(1+\sin \phi)-3(1-\sin \phi)}{(1-\sin \phi)(1+\sin \phi)}= \dfrac{3+3\sin \phi-3+3\sin \phi}{1-\sin^2\phi}\\ =\dfrac{6\sin \phi}{\cos^2\phi}= 6\cdot \dfrac{\sin \phi}{\cos \phi}\cdot \dfrac{1}{\cos \phi}= 6\tan\phi\sec\phi= \textbf{LHS}$$

\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*}

• Just for notice - you can use align environment without adding "" tags. – Galc127 Feb 28 '16 at 18:59 • How do I align left? In my browser my equation tags are overlapping with hot network posts. – SplitInfinity Feb 28 '16 at 19:00 • You can "add" negative horizontal space before the aligned environment. Iv'e edited your post, adding negative horizontal space of 3 cm. You might edit it to your satisfaction. – Galc127 Feb 28 '16 at 19:05 • This is not cross multiplication. Cross multiplication is multiplying across an equals sign, which is not what is being demonstrated. What you have done is put your fractions over a common denominator. – Eric Towers Feb 28 '16 at 20:05 • @Eric Towers Fixed, thanks. – SplitInfinity Feb 28 '16 at 20:07 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