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Question Number 105759 by bramlex last updated on 31/Jul/20
If α and β are the solution of equation a tan θ + b sec θ = c . find the value of tan (α+β).
$${If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{solution}\:{of} \\ $$$${equation}\:{a}\:\mathrm{tan}\:\theta\:+\:{b}\:\mathrm{sec}\:\theta\:=\:{c}\:.\: \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{tan}\:\left(\alpha+\beta\right). \\ $$
Commented by bramlex last updated on 31/Jul/20
thx both
$${thx}\:{both} \\ $$
Answered by john santu last updated on 31/Jul/20
⇒a tan θ + b sec θ = c ⇒ a sin θ + b = c cos θ using identity sin θ = ((2tan ((θ/2)))/(1+tan^2 ((θ/2)))) cos θ = ((1−tan^2 ((θ/2)))/(1+tan^2 ((θ/2)))) we have (b+c)tan^2 ((θ/2))+2a tan ((θ/2))+b−c=0 the equation is quadratic in tan ((θ/2)) and then tan ((α/2)) &tan ((β/2)) are the roots of this equation. by Vieta′s rule { ((tan ((α/2))+tan ((β/2))=−((2a)/(b+c)))),((tan ((α/2)).tan ((β/2))=((b−c)/(b+c)))) :} applied identity tan (((α+β)/2))=((−((2a)/(b+c)))/(1−((b−c)/(b+c)))) tan (((α+β)/2)) = −(a/c) using double angle formula tan (α+β) = ((2(−(a/c)))/(1−(a^2 /c^2 ))) = ((2ac)/(a^2 −c^2 )) ♠⧫
$$\Rightarrow{a}\:\mathrm{tan}\:\theta\:+\:{b}\:\mathrm{sec}\:\theta\:=\:{c}\: \\ $$$$\Rightarrow\:{a}\:\mathrm{sin}\:\theta\:+\:{b}\:=\:{c}\:\mathrm{cos}\:\theta\: \\ $$$${using}\:{identity}\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{2tan}\:\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\mathrm{cos}\:\theta\:=\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$${we}\:{have}\:\left({b}+{c}\right)\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{2}{a}\:\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\right)+{b}−{c}=\mathrm{0} \\ $$$${the}\:{equation}\:{is}\:{quadratic}\:{in}\: \\ $$$$\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\right)\:{and}\:{then}\:\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)\:\&\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right) \\ $$$${are}\:{the}\:{roots}\:{of}\:{this}\:{equation}. \\ $$$${by}\:{Vieta}'{s}\:{rule}\: \\ $$$$\begin{cases}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right)+\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)=−\frac{\mathrm{2}{a}}{{b}+{c}}}\\{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}\right).\mathrm{tan}\:\left(\frac{\beta}{\mathrm{2}}\right)=\frac{{b}−{c}}{{b}+{c}}}\end{cases} \\ $$$${applied}\:{identity}\:\mathrm{tan}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\right)=\frac{−\frac{\mathrm{2}{a}}{{b}+{c}}}{\mathrm{1}−\frac{{b}−{c}}{{b}+{c}}} \\ $$$$\mathrm{tan}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\right)\:=\:−\frac{{a}}{{c}} \\ $$$${using}\:{double}\:{angle}\:{formula} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{2}\left(−\frac{{a}}{{c}}\right)}{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }}\:=\:\frac{\mathrm{2}{ac}}{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$\spadesuit\blacklozenge \\ $$
Commented by Coronavirus last updated on 01/Aug/20
clear thanks you sir
Answered by bobhans last updated on 31/Jul/20
a tan θ + b sec θ = c →a tan θ−c = −b sec θ squaring both side (a tan θ−c)^2 = b^2 (1+tan^2 θ) (a^2 −b^2 )tan^2 θ −2ac tan θ + c^2 −b^2 =0 has the roots are tan α and tan β Vieta′s rule { ((tan α+tan β=((2ac)/(a^2 −b^2 )))),((tan α.tan β = ((c^2 −b^2 )/(a^2 −b^2 )))) :} therefore tan (α+β) = ((tan α+tan β)/(1−tan α.tan β)) = ((2ac)/(a^2 −c^2 )) ★
$${a}\:\mathrm{tan}\:\theta\:+\:{b}\:\mathrm{sec}\:\theta\:=\:{c}\: \\ $$$$\rightarrow{a}\:\mathrm{tan}\:\theta−{c}\:=\:−{b}\:\mathrm{sec}\:\theta \\ $$$${squaring}\:{both}\:{side}\: \\ $$$$\left({a}\:\mathrm{tan}\:\theta−{c}\right)^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{tan}\:^{\mathrm{2}} \theta\:−\mathrm{2}{ac}\:\mathrm{tan}\:\theta\:+\:{c}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0}\: \\ $$$${has}\:{the}\:{roots}\:{are}\:\mathrm{tan}\:\alpha\:{and}\:\mathrm{tan}\:\beta \\ $$$${Vieta}'{s}\:{rule}\:\begin{cases}{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta=\frac{\mathrm{2}{ac}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\\{\mathrm{tan}\:\alpha.\mathrm{tan}\:\beta\:=\:\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\end{cases} \\ $$$${therefore}\:\mathrm{tan}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha.\mathrm{tan}\:\beta} \\ $$$$\:=\:\frac{\mathrm{2}{ac}}{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }\:\bigstar \\ $$
Commented by Coronavirus last updated on 01/Aug/20
wouah very fast method

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