# Proving $\frac{p}{\sin x+n\cos x}=\frac{p\csc(\arctan(n)+x)\sin(\arctan(n))}{ n}$

The following two expressions produce the same graph. $$\frac{p}{\sin(x)+n\cos(x)} = \frac {p\csc(\arctan(n)+x)\sin(\arctan(n))}{n}.$$ How to prove that they are equal?

If drawn with polar coordinates, $x$ is $\theta$, the $p$ is the the point of intersection with the $y$-axis and $-n$ is the slope.

Hint:

$$\arctan(n) = y \implies n = \tan(y) \implies \sin(y) = \frac{n}{\sqrt{n^2+1}}, \text{and }\cos(y) = \frac{1}{\sqrt{n^2+1}}.$$ Also, $$\csc(\arctan(n)+x) = \frac{1}{\sin(\arctan(n)+x)} = \frac{1}{\sin(\arctan(n))\cos(x)+\cos(\arctan(n))\sin(x)}$$

Whenever you are dealing with expressions like $\sin(\arctan(n))$, the trick is to draw a triangle like below:

You can construct this by noting that for the angle to be $\arctan(n)$ the opposite and adjacent sides must be $n$ and $1$ respecitvely, in which case the Pythagoras rule gives the hypotenuse.

From this you can see: $$\sin(\arctan(n))=\frac{n}{\sqrt{1+n^2}},$$ $$\cos(\arctan(n))=\frac{1}{\sqrt{1+n^2}}.$$ You will also want to expand $\sin(\arctan(n)+x)$, which you can do using the identity $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$.

• It's not $\cos(x)$; it's $\csc(x)$. – Michael McGovern Jul 29 '18 at 1:38

Hint:

$$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$$

and

$$\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$$

• Just letting you know, your two equations here are showing on the same line. – Ruvi Lecamwasam Jul 29 '18 at 1:41