Patatiniuh

2022-07-09

Find $\mathrm{tan}\left(\alpha +\beta \right)$ when $\mathrm{tan}\beta =\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha }{1-n{\mathrm{cos}}^{2}\alpha }$

SweallySnicles3

Expert

I think it will be easier if one first simplifies $\mathrm{tan}\beta$,
$\mathrm{tan}\beta =\frac{n\mathrm{sin}\alpha \mathrm{cos}\alpha }{1-n{\mathrm{cos}}^{2}\alpha }=\frac{n\mathrm{tan}\alpha }{1/{\mathrm{cos}}^{2}\alpha -n}=\frac{n\mathrm{tan}\alpha }{{\mathrm{tan}}^{2}\alpha +1-n}.$
Next, use the addition formula for $\mathrm{tan}\left(\alpha +\beta \right)$ and insert the expression above everywhere you encounter tanβ. Simplify, and you got your result.

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