How does $\frac{\sin\theta}{\cos\theta}$ become $\frac{y}{x}$

Question

I ended up in the wrong math class (trigonometry) for my level but am trying to survive by catching up on some more basic principles. I'm wondering if the same principle (and if so, what is it) is applying in the second and third part of each equation. In the first equation, how does it get reduced to $_y^x$ if, in the middle part, x was on the bottom and y was on the top and what happened to r

$$ \frac{\sin\theta}{\cos\theta} = \frac{\frac yr}{\frac xr} = \frac yx$$

In the bottom equation, I understand that the equation in the middle (fraction divided by a fraction) is rewritten as multiplication by making the denominator the numerator etc. Is that's what's happening in the top equation?

$$ {\tan\theta} = \frac{\frac{\sqrt{15}}{4}}{\frac 14} = \frac{\sqrt{15}}{4} \times \frac 41 $$

Answer 1

When dividing fractions you can make use of the fact that:$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$$

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In your case, your first equation is most likely more like:$$\require{cancel}\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{y}{r}}{\frac{x}{r}}=\frac{y}{r}\div\frac{x}{r}=\frac{y}{\cancel{r}}\times\frac{\cancel{r}}{x}=\frac{y}{x}$$

and your second equation is:$$\require{cancel}\tan{\theta}=\frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}}=\frac{\sqrt{15}}{4}\div\frac{1}{4}=\frac{\sqrt{15}}{\cancel{4}}\times\frac{\cancel{4}}{1}=\sqrt{15}$$

Answer 2

1) $\require {cancel}\frac {\frac y r}{\frac x r} = \frac {r*\frac y r}{r*\frac x r} =\frac {\cancel r*\frac y {\cancel r}}{\cancel r*\frac x {\cancel r}} \frac y x$

2) $\frac {\frac y r}{\frac x r} = \frac {\frac 1 r y}{\frac 1 r x} = \frac{ \cancel {\frac 1 r} y}{ \cancel {\frac 1 r} x}= \frac y x$

3) $\frac {\frac y r}{\frac x r} = \frac y r \cdot \frac r x = \frac y {\cancel r} \cdot \frac{\cancel r} x = \frac y x$

Take your pick. Ultimately these should become very quick and automatic