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I was remembering about Pythagorean trigonometric identities so I grabbed some of my old notebooks and I was having so much fun, solving some equalities, until I found these four equalities that I just can't find the way to solve them, probably they are easier than I thought but I really need some help, otherwise I wont sleep tonight!

I'll just show you how I tried one, because the rest of them are all pretty the same, but I'll post all of them in case someone can help me with all.

$$\frac{\sin (\theta)}{\cot(\theta)\csc(\theta)-\cos(\theta)}= \frac{\cos(\theta)-\sec(\theta)}{\sin(\theta)-\csc(\theta)}$$

So, I started with

$$LHC= \frac{\sin(\theta)}{\Big(\frac{\cos(\theta)}{\sin(\theta)}\Big) \Big(\frac{1}{\sin(\theta)}\Big)-\cos(\theta)} = \frac{\sin(\theta)}{\Big(\frac{\cos(\theta)}{\sin^2(\theta)}\Big)-\cos(\theta)}=\frac{\sin(\theta)}{\frac{\cos(\theta)-\cos(\theta)\sin(\theta)}{\sin^2(\theta)}}=\frac{\sin^3(\theta)}{\cos(\theta)-\cos(\theta)\sin^2(\theta)}=\frac{\sin^3(\theta)}{\cos(\theta)(1-\sin^2(\theta))}=\frac{\sin^3(\theta)}{\cos(\theta)\cos^2(\theta)}= \tan^3 (\theta)$$

$$RHC = \frac{\cos (\theta) - \frac{1}{\cos(\theta)}}{\sin(\theta)-\frac{1}{\sin(\theta)}} = \frac{\frac{\cos^2(\theta)-1}{\cos(\theta)}}{\frac{\sin^2(\theta)-1}{\sin(\theta)}}= \frac{\sin(\theta)(\cos^2-1)}{\cos(\theta)(\sin^2(\theta)-1)}= \frac{\sin(\theta)(-\sin(\theta)}{\cos(\theta)(-\cos(\theta)} = \tan^3(\theta).$$ And I seriously don't know how to manage to get to $\frac{\cos(\theta)-\sec(\theta)}{\sin(\theta)-\csc(\theta)}.$

And I have the same issue with

$$\frac{1+\cot^3 (\epsilon)}{1+\cot (\epsilon)}= \csc^2 (\epsilon) - \cot (\epsilon),$$ and $$\sin^2 (\lambda)+\tan(\lambda) = \frac{\tan^3(\lambda)-1}{\tan (\lambda)-1}.$$

Please guys, help me to manipulate these equations, I'll appreciate it!

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  • $\begingroup$ Hint: try starting with the RHS and deducing the LHS. $\endgroup$ – Will R Sep 5 '15 at 23:45
  • $\begingroup$ Off course, I already tried that and I got that same answer $tan^3 (\alpha)$ I'll edit the question to show you. $\endgroup$ – HelaHelheim Sep 5 '15 at 23:49
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    $\begingroup$ So you're saying you showed that both the LHS and the RHS were equal to the same thing? Doesn't that mean the LHS and the RHS are equal to each other? $\endgroup$ – Will R Sep 5 '15 at 23:52
  • $\begingroup$ Yes! But, I'm suppose to be able to manipulate just one side of the equation to get the other side. $\endgroup$ – HelaHelheim Sep 5 '15 at 23:58
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    $\begingroup$ Surely you can at least put the workings to turn LHS into $\tan^{3}{\theta}$, and then you can deduce the workings necessary to turn $\tan^{3}{\theta}$ into RHS by simply reversing the steps used to turn RHS into $\tan^{3}{\theta}$. $\endgroup$ – Will R Sep 6 '15 at 0:00
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We wish to prove that $$\frac{\sin{\theta}}{\cot{\theta}\csc{\theta}-\cos{\theta}} = \frac{\cos{\theta}-\sec{\theta}}{\sin{\theta}-\csc{\theta}}.$$

We begin from the right-hand side and show that this is equal to the left-hand side. We have,

\begin{eqnarray*} \frac{\cos{\theta}-\sec{\theta}}{\sin{\theta}-\csc{\theta}} & = & \frac{\cos{\theta}-\sec{\theta}}{\sin{\theta}-\csc{\theta}}\cdot\frac{\cos{\theta}}{\cos{\theta}}\\[6pt] & = & \frac{\cos^{2}{\theta}-1}{\sin{\theta}\cos{\theta}-\csc{\theta}\cos{\theta}}\\[6pt] & = &\frac{\sin^{2}{\theta}}{\csc{\theta}\cos{\theta}-\sin{\theta}\cos{\theta}}\\[6pt] & = &\frac{\sin{\theta}}{\csc{\theta}\cot{\theta}-\cos{\theta}}\\[6pt] & = &\frac{\sin{\theta}}{\cot{\theta}\csc{\theta}-\cos{\theta}} \end{eqnarray*}

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Lets start from the LHS, then multiply by $\frac{RHS}{RHS}$

$$\begin{array}{lll} \displaystyle\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}&=&\displaystyle\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot\frac{\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}}{\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}}\\ \displaystyle&=&\displaystyle\bigg(\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot \frac{\sin\theta-\csc\theta}{\cos\theta-\sec\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\ &=&\displaystyle\frac{\sin^2\theta\cos\theta}{\sin^2\theta\cos\theta}\cdot\displaystyle\bigg(\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot \frac{\sin\theta-\csc\theta}{\cos\theta-\sec\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\ &=&\displaystyle\displaystyle\bigg(\frac{\sin^2\theta\cos\theta}{\cos\theta-\cos\theta\sin^2\theta}\cdot \frac{\sin^2\theta-1}{\cos^2\theta-1}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\ &=&\displaystyle\displaystyle\bigg(\frac{\sin^2\theta\cos\theta}{\cos\theta(1-\sin^2\theta)}\cdot \frac{-\cos^2\theta}{-\sin^2\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\ &=&\displaystyle\displaystyle\bigg(\frac{\sin^2\theta\cos\theta}{\cos^3\theta}\cdot \frac{-\cos^2\theta}{-\sin^2\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\ &=&\displaystyle1\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}=\displaystyle \frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\ \end{array}$$

In general $$\frac{LHS}{RHS}\cdot RHS = RHS$$

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