# Question-68191

Question Number 68191 by peter frank last updated on 06/Sep/19
Answered by mind is power last updated on 06/Sep/19
$${sin}\left({a}−{b}\right)={sin}\left({a}\right){cos}\left({b}\right)−{cos}\left({a}\right){sin}\left({b}\right) \\$$$$\frac{{sin}\left({a}−{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}={tg}\left({a}\right)−{tg}\left({b}\right) \\$$$$\Rightarrow\frac{{sin}\left({a}−{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}={tg}\left({b}\right)\left({k}−\mathrm{1}\right) \\$$$$\frac{{sin}\left({a}+{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}={tg}\left({a}\right)+{tg}\left({b}\right)={tg}\left({b}\right)\left({k}+\mathrm{1}\right) \\$$$$\Rightarrow\frac{{sin}\left({a}−{b}\right)}{{sin}\left({a}+{b}\right)}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\$$$$\Rightarrow{sin}\left({a}−{b}\right)=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}{sin}\left({a}+{b}\right)=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}{sin}\left({c}\right) \\$$
Commented by peter frank last updated on 07/Sep/19
$${thank}\:{you} \\$$
Answered by \$@ty@m123 last updated on 06/Sep/19
$$\frac{\mathrm{tan}\:{A}}{\mathrm{tan}\:{B}}=\frac{{k}}{\mathrm{1}} \\$$$${by}\:{componendo}\:\&\:{dividendo}, \\$$$$\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{B}}{\mathrm{tan}\:{A}−\mathrm{tan}\:{B}}\:=\:\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\$$$$\frac{\frac{\mathrm{sin}\:{A}}{\mathrm{cos}\:{A}}+\frac{\mathrm{sin}\:{B}}{\mathrm{cos}\:{B}}}{\frac{\mathrm{sin}\:{A}}{\mathrm{cos}\:{A}}−\frac{\mathrm{sin}\:{B}}{\mathrm{cos}\:{B}}}\:=\:\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\$$$$\frac{\mathrm{sin}\:{A}\mathrm{cos}\:{B}+\mathrm{cos}\:{A}\mathrm{sin}\:{B}}{\mathrm{sin}\:{A}\mathrm{cos}\:{B}−\mathrm{cos}\:{A}\mathrm{sin}\:{B}}\:=\:\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\$$$$\frac{\mathrm{sin}\:\left({A}+{B}\right)}{\mathrm{sin}\:\left({A}−{B}\:\right)}\:=\:\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\$$$$\frac{\mathrm{sin}\:{C}}{\mathrm{sin}\:\left({A}−{B}\:\right)}\:=\:\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\$$$${I}\:{think}\:{the}\:{question}\:{should}\:{be}: \\$$$$\:{prove}\:{that}:\:\mathrm{sin}\:\left({A}−{B}\right)=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\mathrm{sin}\:{C} \\$$
Commented by peter frank last updated on 07/Sep/19
$${thank}\:{you} \\$$