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Is just it a coincidence that $$\frac{0.5}{ \cos^2(30)} = \frac{\tan(30)}{\cos(30)} $$ However $$\frac{0.5}{ \cos^2(13) } \neq \frac{\tan(13)}{\cos(13)} $$ is not equal ? And if not does anyone know a reason why they just so happen to be equatable?

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It's simply because $\sin(30^\circ)=1/2$.

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$$\frac{\tan(30)}{\cos(30)}=\frac{\frac{\sin(30)}{\cos(30)}}{\cos(30)}=\frac{\sin(30)}{\cos^2(30)}=\frac{0.5}{\cos^2(30)}$$since $\sin(30^\circ)=0.5$. The other equation would work if instead of $0.5$, you chose $\sin(13)\approx0.225$ as the numerator on the LHS.

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Rewrite the right side: $$ \begin{align} \frac{0.5}{\cos^2(30^\circ)}&=\frac{\tan(30^\circ)}{\cos(30^\circ)}\\ &=\frac{\sin(30^\circ)}{\cos^2(30^\circ)}\\ &=\frac{0.5}{\cos^2(30^\circ)} \end{align} $$

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If we use $x$ as a variable, we get the following:

$$\begin{align} & \frac {0.5}{\cos^2x} = \frac {\tan x}{\cos x} \\ \implies& \frac {1}{2} \cos x = \cos^2 x \tan x\\ \implies& \cos x \tan x\ = \frac {1}{2} \\ \implies& \cos x \frac {\sin x}{\cos x} = \frac {1}{2}\\ \implies& \sin x\ = \frac {1}{2}\ \end{align}$$

Thus the only solutions to the above equation are $x = 30^\circ \pm 360^\circ k$ or $x = 150^\circ \pm 360^\circ k$ where $k$ is any natural number.

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