Jaydan Aguirre

2022-07-12

Given $\mathrm{tan}\beta =\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha }{1-n{\mathrm{sin}}^{2}\alpha }$, show that $\mathrm{tan}\left(\alpha -\beta \right)=\left(1-n\right)\mathrm{tan}\alpha$

kawiarkahh

Expert

As we need to eliminate $\beta ,$, write $\mathrm{tan}\beta =\mathrm{tan}\left\{\alpha -\left(\alpha -\beta \right)\right\}$ and expand.
For the RHS,
$\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha }{1-n{\mathrm{sin}}^{2}\alpha }=\frac{\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha }{{\mathrm{cos}}^{2}\alpha }}{\frac{1-n{\mathrm{sin}}^{2}\alpha }{{\mathrm{cos}}^{2}\alpha }}=\frac{n{\mathrm{tan}}^{2}\alpha }{1+\left(1-n\right){\mathrm{tan}}^{2}\alpha }$

Kaeden Hoffman

Expert

With
$\mathrm{tan}\beta =\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha }{1-n{\mathrm{sin}}^{2}\alpha }$
we write
$n=\frac{\mathrm{tan}\beta }{\mathrm{sin}\alpha \mathrm{cos}\alpha +\mathrm{tan}\beta {\mathrm{sin}}^{2}\alpha }$
so
$\begin{array}{rcl}1-n& =& \frac{\mathrm{sin}\alpha \mathrm{cos}\alpha +\mathrm{tan}\beta {\mathrm{sin}}^{2}\alpha -\mathrm{tan}\beta }{\mathrm{sin}\alpha \mathrm{cos}\alpha +\mathrm{tan}\beta {\mathrm{sin}}^{2}\alpha }\\ & =& \frac{\mathrm{sin}\alpha \mathrm{cos}\alpha -{\mathrm{cos}}^{2}\alpha \mathrm{tan}\beta }{\mathrm{sin}\alpha \mathrm{cos}\alpha +\mathrm{tan}\beta {\mathrm{sin}}^{2}\alpha }\\ & =& \frac{\mathrm{tan}\alpha -\mathrm{tan}\beta }{\mathrm{tan}\alpha +\mathrm{tan}\beta {\mathrm{tan}}^{2}\alpha }\\ & =& \frac{1}{\mathrm{tan}\alpha }\mathrm{tan}\left(\alpha -\beta \right)\end{array}$

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