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Question-16000

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Question Number 16000 by tawa tawa last updated on 16/Jun/17
Answered by mrW1 last updated on 16/Jun/17
I assume you mean α and β are roots of a tan θ + b sec θ = c a tan θ + b sec θ = c ((a sin θ)/(cos θ)) + (b/(cos θ)) = c ((a sin θ + b)/(cos θ)) = c c cos θ − a sin θ = b (√(a^2 +c^2 ))[cos θ (c/( (√(a^2 +c^2 ))))− sin θ (a/( (√(a^2 +c^2 ))))]= b (√(a^2 +c^2 ))[cos θ cos ϕ− sin θ sin ϕ]= b (√(a^2 +c^2 ))cos (θ + ϕ) = b cos (θ + ϕ) = (b/( (√(a^2 +c^2 )))) ⇒θ+ϕ=±cos^(−1) (b/( (√(a^2 +c^2 )))) with ϕ=cos^(−1) (c/( (√(a^2 +c^2 )))) ⇒θ=±cos^(−1) (b/( (√(a^2 +c^2 )))) − ϕ ⇒θ=±cos^(−1) (b/( (√(a^2 +c^2 )))) − cos^(−1) (c/( (√(a^2 +c^2 )))) ∵ α and β are the roots ⇒α=cos^(−1) (b/( (√(a^2 +c^2 )))) − cos^(−1) (c/( (√(a^2 +c^2 )))) ⇒β=−cos^(−1) (b/( (√(a^2 +c^2 )))) − cos^(−1) (c/( (√(a^2 +c^2 )))) ∴ α+β=−2cos^(−1) (c/( (√(a^2 +c^2 ))))=−2ϕ cos ϕ=(c/( (√(a^2 +c^2 )))) tan ϕ=(a/c) tan (α+β)=−tan 2ϕ =−((2tan ϕ)/(1−tan^2 ϕ))=−((2×(a/c))/(1−(a^2 /c^2 )))=((2ac)/(a^2 −c^2 ))
$$\mathrm{I}\:\mathrm{assume}\:\mathrm{you}\:\mathrm{mean}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{tan}\:\theta\:+\:\mathrm{b}\:\mathrm{sec}\:\theta\:=\:\mathrm{c} \\ $$$$ \\ $$$$\mathrm{a}\:\mathrm{tan}\:\theta\:+\:\mathrm{b}\:\mathrm{sec}\:\theta\:=\:\mathrm{c} \\ $$$$\frac{\mathrm{a}\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:+\:\frac{\mathrm{b}}{\mathrm{cos}\:\theta}\:=\:\mathrm{c} \\ $$$$\frac{\mathrm{a}\:\mathrm{sin}\:\theta\:+\:\mathrm{b}}{\mathrm{cos}\:\theta}\:\:=\:\mathrm{c} \\ $$$$\mathrm{c}\:\mathrm{cos}\:\theta\:−\:\mathrm{a}\:\mathrm{sin}\:\theta\:=\:\mathrm{b} \\ $$$$\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \:}\left[\mathrm{cos}\:\theta\:\frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}−\:\mathrm{sin}\:\theta\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}\right]=\:\mathrm{b} \\ $$$$\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \:}\left[\mathrm{cos}\:\theta\:\mathrm{cos}\:\varphi−\:\mathrm{sin}\:\theta\:\mathrm{sin}\:\varphi\right]=\:\mathrm{b} \\ $$$$\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \:}\mathrm{cos}\:\left(\theta\:+\:\varphi\right)\:=\:\mathrm{b} \\ $$$$\mathrm{cos}\:\left(\theta\:+\:\varphi\right)\:=\:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\Rightarrow\theta+\varphi=\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\mathrm{with}\:\varphi=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\Rightarrow\theta=\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}\:−\:\varphi \\ $$$$\Rightarrow\theta=\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}\:−\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\because\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}\:−\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\Rightarrow\beta=−\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}\:−\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\therefore\:\alpha+\beta=−\mathrm{2cos}^{−\mathrm{1}} \frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }}=−\mathrm{2}\varphi \\ $$$$\mathrm{cos}\:\varphi=\frac{\mathrm{c}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{a}}{\mathrm{c}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=−\mathrm{tan}\:\mathrm{2}\varphi \\ $$$$=−\frac{\mathrm{2tan}\:\varphi}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\varphi}=−\frac{\mathrm{2}×\frac{\mathrm{a}}{\mathrm{c}}}{\mathrm{1}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }}=\frac{\mathrm{2ac}}{\mathrm{a}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} } \\ $$
Commented by tawa tawa last updated on 16/Jun/17
Wow God bless you sir.
$$\mathrm{Wow}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 17/Jun/17
atan θ+bsec θ=c let x=atan θ ⇒ tan θ=x/a c−x=bsec θ ⇒ (c−x)^2 =b^2 (1+tan^2 θ) c^2 −2cx+x^2 =b^2 (1+(x^2 /a^2 )) a^2 c^2 −2a^2 cx+a^2 x^2 =a^2 b^2 +b^2 x^2 (b^2 −a^2 )x^2 +2a^2 cx+(a^2 b^2 −a^2 c^2 )=0 x_1 +x_2 =atan α+atan β) = ((−2a^2 c)/( b^2 −a^2 )) , and x_1 x_2 = (atan α)(atan β) = ((a^2 b^2 −a^2 c^2 )/(b^2 −a^2 )) tan (α+β)= ((a(atan α+atan β))/(a^2 −a^2 tan αtan β)) = ((a(((−2a^2 c)/(b^2 −a^2 ))))/(a^2 −(((a^2 b^2 −a^2 c^2 )/(b^2 −a^2 ))))) =((−2a^3 c)/(a^4 −a^2 c^2 )) = ((2ac)/(c^2 −a^2 )) .
$${a}\mathrm{tan}\:\theta+{b}\mathrm{sec}\:\theta={c} \\ $$$${let}\:{x}={a}\mathrm{tan}\:\theta\:\:\:\:\Rightarrow\:\:\mathrm{tan}\:\theta={x}/{a} \\ $$$$\:\:\:\:\:\:{c}−{x}={b}\mathrm{sec}\:\theta \\ $$$$\:\:\Rightarrow\:\:\:\left({c}−{x}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} −\mathrm{2}{cx}+{x}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right) \\ $$$$\:\:\:{a}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {cx}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\:\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {cx}+\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left.\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} ={a}\mathrm{tan}\:\alpha+{a}\mathrm{tan}\:\beta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{−\mathrm{2}{a}^{\mathrm{2}} {c}}{\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\:,\:\:\:\:{and} \\ $$$$\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\:\left({a}\mathrm{tan}\:\alpha\right)\left({a}\mathrm{tan}\:\beta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\: \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\:\frac{{a}\left({a}\mathrm{tan}\:\alpha+{a}\mathrm{tan}\:\beta\right)}{{a}^{\mathrm{2}} −{a}^{\mathrm{2}} \mathrm{tan}\:\alpha\mathrm{tan}\:\beta} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{{a}\left(\frac{−\mathrm{2}{a}^{\mathrm{2}} {c}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} −\left(\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:=\frac{−\mathrm{2}{a}^{\mathrm{3}} {c}}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} }\:\:=\:\:\frac{\mathrm{2}\boldsymbol{{ac}}}{\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:\:. \\ $$
Commented by tawa tawa last updated on 17/Jun/17
I really appreciate. God bless you sir.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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