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Question-196886

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Question Number 196886 by Amidip last updated on 02/Sep/23
Answered by MM42 last updated on 02/Sep/23
tan(α+β)=((tanα+tanβ)/(1−tanαtanβ))=(p/(q−1)) ⇒(p/(q−1))=((sin^2 (α+β))/(sin(α+β)cos(α+β))) ⇒psin(α+β)cos(α+β)=(q−1)sin^2 (α+β)) ⇒sin^2 (α+β)+psin(α+β)cos(α+β)=qsin^2 (α+β) ✓
$${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{{p}}{{q}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{p}}{{q}−\mathrm{1}}=\frac{{sin}^{\mathrm{2}} \left(\alpha+\beta\right)}{{sin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)} \\ $$$$\left.\Rightarrow{psin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)=\left({q}−\mathrm{1}\right){sin}^{\mathrm{2}} \left(\alpha+\beta\right)\right) \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \left(\alpha+\beta\right)+{psin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)={qsin}^{\mathrm{2}} \left(\alpha+\beta\right)\:\checkmark \\ $$
Answered by HeferH last updated on 02/Sep/23
(x−tan a)(x−tan b) =0 x^2 −x(tan a+tan b)+tan atan b = 0 p = −(tan a +tan b) q = tan a tan b sin^2 (a+b) + p sin (a+b)cos (a+b) + qcos^2 (a+b)=q sin^2 (a+b) + p sin (a+b)cos (a+b) = qsin^2 (a+b) 1+ p cot (a+b) = q cot (a+b) = ((q−1)/p) ((q−1)/p) = ((1−tan a tan b)/((tan a+tan b))) = tan^(−1) (a+b) =cot (a+b)
$$\:\left({x}−\mathrm{tan}\:{a}\right)\left({x}−\mathrm{tan}\:{b}\right)\:=\mathrm{0} \\ $$$$\:{x}^{\mathrm{2}} −{x}\left(\mathrm{tan}\:{a}+\mathrm{tan}\:{b}\right)+\mathrm{tan}\:{a}\mathrm{tan}\:{b}\:=\:\mathrm{0} \\ $$$$\:{p}\:=\:−\left(\mathrm{tan}\:{a}\:+\mathrm{tan}\:{b}\right) \\ $$$$\:{q}\:=\:\mathrm{tan}\:{a}\:\mathrm{tan}\:{b}\: \\ $$$$\:\mathrm{sin}^{\mathrm{2}} \:\left({a}+{b}\right)\:+\:{p}\:\mathrm{sin}\:\left({a}+{b}\right)\mathrm{cos}\:\left({a}+{b}\right)\:+\:{q}\mathrm{cos}\:^{\mathrm{2}} \left({a}+{b}\right)={q} \\ $$$$\:\mathrm{sin}^{\mathrm{2}} \:\left({a}+{b}\right)\:+\:{p}\:\mathrm{sin}\:\left({a}+{b}\right)\mathrm{cos}\:\left({a}+{b}\right)\:=\:{q}\mathrm{sin}\:^{\mathrm{2}} \left({a}+{b}\right) \\ $$$$\:\mathrm{1}+\:{p}\:\mathrm{cot}\:\left({a}+{b}\right)\:=\:{q} \\ $$$$\:\mathrm{cot}\:\left({a}+{b}\right)\:=\:\frac{{q}−\mathrm{1}}{{p}} \\ $$$$\:\frac{{q}−\mathrm{1}}{{p}}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:{a}\:\mathrm{tan}\:{b}}{\left(\mathrm{tan}\:{a}+\mathrm{tan}\:{b}\right)}\:=\:\:\mathrm{tan}\:^{−\mathrm{1}} \left({a}+{b}\right)\:=\mathrm{cot}\:\left({a}+{b}\right)\: \\ $$

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