Question Number 38288 by ajfour last updated on 23/Jun/18
$$\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta\:=\:\mathrm{2tan}\:\theta \\ $$$${a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\:{l}\mathrm{sin}\:\theta \\ $$$${express}\:\mathrm{sin}\:\alpha,\:\mathrm{sin}\:\beta\:\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
$${t}_{\mathrm{1}} ={tan}\frac{\alpha}{\mathrm{2}}\:\:\:{t}_{\mathrm{2}} ={tan}\frac{\beta}{\mathrm{2}}\:\:\:{p}={tan}\frac{\theta}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{2}{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{\mathrm{4}{p}}{\mathrm{1}−{p}^{\mathrm{2}} } \\ $$$${a}\frac{\mathrm{2}{t}_{\mathrm{1}} }{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }−{b}\frac{\mathrm{2}{t}_{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }={l}\frac{\mathrm{2}{p}}{\mathrm{1}+{p}^{\mathrm{2}} } \\ $$$${wait} \\ $$$$ \\ $$