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I am working on a problem and obtained the following equation:

$$ \tan \beta = \frac{\lambda - \cos \theta}{\sin \theta} \Leftrightarrow \tan \beta = \lambda\cdot \csc \theta-\cot \theta$$

Where $\lambda$ is a constant.

Is it possible to simplify the right hand side term into a single trig function? My goal is to have an equation of the form $$\theta = f(\beta)$$

I have tried different trig identities but I only end up making it worse.

Any help is appreciated.

Thank you!

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Note that $$\sin \theta\sin\beta+\cos\theta\cos\beta=\lambda\cos\beta$$ and the LHS equivalent to $\cos(\theta-\beta).$

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  • $\begingroup$ Ah! Thank you so much :) $\endgroup$ – Marco Nunez Oct 31 '18 at 20:33
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The answer to your question is that the inverse equation, $\theta=f(\beta)$ has three values:

$\theta=arccos(\lambda * cos(\beta))$

$\theta=arcsin(\lambda *cos(\beta))+\beta+\frac{3 \pi}{2}$

$\theta=\beta-arccos(\lambda* cos(\beta))$

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