# $\arctan$ of constant multiplied $\tan \varphi$

Helllo, I'm new here. I will try this great place to get my answer. I have a next problem:

$$\vartheta = \arctan(K \tan \varphi)$$

$K > 0$

Where $K$ is a constant. If $K = 1$ then $\vartheta = \varphi$, but I can't find more simple answer for $\vartheta$ if $K \neq 1$. Maybe someone will be able to help me out, if it is even possbile.

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So what is your question? What do you need to derive? Do you have to find $\theta$ ? – Kirthi Raman Feb 23 '12 at 16:00
A solution to $\vartheta = \arctan(K \tan \varphi)$ is an intersection of the graphs of $\tan x$ and $K\tan x$ in the interval $(-\pi/2,\pi/2)$. But there is only one such intersection when $K \neq 1$: the intersection at $x = 0$. See this plot. – Antonio Vargas Feb 23 '12 at 16:06
@Antonio: The question asks for $\vartheta=\arctan(K\tan(\varphi))$ not $\varphi=\arctan(K\tan(\varphi))$. – robjohn Feb 23 '12 at 16:13
@AntonioVargas it looks like you're finding values of x for which $\tan(x)$ and $K\tan(x)$ are equal, but the OP is seeking the relationship between $x$ and $y$ where $\tan(x)=K\tan(y)$. For example, $\theta = \pi/3$,$\phi = \pi/6$, $K=3$ works. – Aru Ray Feb 23 '12 at 16:15
Thanks for answers. I would like to get simpler form for $\vartheta$ if $K \neq 1$. I was wondering if there exists any derivation of that form. – Grega Feb 23 '12 at 16:16