Jaydan Aguirre

Answered

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}(\alpha -\beta )=(1-n)\mathrm{tan}\alpha $

Answer & Explanation

kawiarkahh

Expert

2022-07-13Added 15 answers

As we need to eliminate $\beta ,$, write $\mathrm{tan}\beta =\mathrm{tan}\{\alpha -(\alpha -\beta )\}$ and expand.

For the RHS,

$\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha}{1-n{\mathrm{sin}}^{2}\alpha}={\displaystyle \frac{{\displaystyle \frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha}{{\mathrm{cos}}^{2}\alpha}}}{{\displaystyle \frac{1-n{\mathrm{sin}}^{2}\alpha}{{\mathrm{cos}}^{2}\alpha}}}}={\displaystyle \frac{n{\mathrm{tan}}^{2}\alpha}{1+(1-n){\mathrm{tan}}^{2}\alpha}}$

For the RHS,

$\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha}{1-n{\mathrm{sin}}^{2}\alpha}={\displaystyle \frac{{\displaystyle \frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha}{{\mathrm{cos}}^{2}\alpha}}}{{\displaystyle \frac{1-n{\mathrm{sin}}^{2}\alpha}{{\mathrm{cos}}^{2}\alpha}}}}={\displaystyle \frac{n{\mathrm{tan}}^{2}\alpha}{1+(1-n){\mathrm{tan}}^{2}\alpha}}$

Kaeden Hoffman

Expert

2022-07-14Added 3 answers

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

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

$\mathrm{tan}\beta =\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha}{1-n{\mathrm{sin}}^{2}\alpha}$

we write

$n=\frac{\mathrm{tan}\beta}{\mathrm{sin}\alpha \mathrm{cos}\alpha +\mathrm{tan}\beta {\mathrm{sin}}^{2}\alpha}$

so

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

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