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

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.

Blue
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Esma
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  • 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 03:38

6 Answers6

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\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}

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

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

Galc127
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DeepSea
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$\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*} $

SplitInfinity
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As $(1-\sin\phi)(1+\sin\phi)=\cos^2\phi,$

$$\dfrac1{1\pm\sin\phi}=\dfrac{1\mp\sin\phi}{\cos^2\phi}=\sec^2\phi\mp\sec\phi\tan\phi$$

lab bhattacharjee
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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.

Kang Oedin
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