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Question-68191

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

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