Menu Close

Given-x-x-2-1-y-y-2-4-9-find-the-value-of-x-y-2-4-y-x-2-1-

Tinku Tara 0 Comments
Pin




Question Number 138150 by liberty last updated on 10/Apr/21
Given (x+(√(x^2 +1)))(y+(√(y^2 +4)))=9 find the value of x(√(y^2 +4)) + y(√(x^2 +1)) .
$${Given}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{9} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\:\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:+\:{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:. \\ $$
Answered by EDWIN88 last updated on 10/Apr/21
set { ((x=cot 2α)),((y=2cot 2β)) :} with 0<α,β<π/2 we get determinant (((x+(√(x^2 +1)) = ((cos 2α)/(sin 2α))+(1/(sin 2α))=cot α)),((y+(√(y^2 +4)) = ((2cos 2β)/(sin 2β))+(2/(sin 2β)) = 2cot β))) then from the given equation 2cot α.cot β = a ((cos α cos β)/(sin α sin β)) = (a/2) we want to compute x(√(y^2 +4)) +y(√(x^2 +1)) = cot 2α ((2/(sin 2β)))+ 2cot 2β ((1/(sin 2α)))= ((2(cos 2α+cos 2β))/(sin 2α sin 2β)) = ((cos 2α +cos 2β)/(2(cos α cos β)(sin α sin β))) = ((2−2sin^2 α−2sin^2 β )/(2((a/2))sin^2 α sin^2 β)) =(2/a) (((1−sin^2 α−sin^2 β)/(sin^2 α sin^2 β))) = (2/a)((1+cot^2 α)(1+cot^2 β)−1−cot^2 α−1−cot^2 β) =(2/a)(cot^2 α cot^2 β−1) =(2/a)((a^2 /4)−1) = (2/a)(((a^2 −4)/4))= ((a^2 −4)/(2a)) in this case a=9 ; we get ((81−4)/(18))=((77)/(18))
$${set}\:\begin{cases}{{x}=\mathrm{cot}\:\mathrm{2}\alpha}\\{{y}=\mathrm{2cot}\:\mathrm{2}\beta}\end{cases}\:{with}\:\mathrm{0}<\alpha,\beta<\pi/\mathrm{2} \\ $$$${we}\:{get}\:\begin{array}{|c|c|}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{sin}\:\mathrm{2}\alpha}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha}=\mathrm{cot}\:\alpha}\\{{y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:=\:\frac{\mathrm{2cos}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\beta}+\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}\beta}\:=\:\mathrm{2cot}\:\beta}\\\hline\end{array} \\ $$$${then}\:{from}\:{the}\:{given}\:{equation} \\ $$$$\mathrm{2cot}\:\alpha.\mathrm{cot}\:\beta\:=\:{a}\: \:\frac{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}\:=\:\frac{{a}}{\mathrm{2}} \\ $$$${we}\:{want}\:{to}\:{compute}\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:= \\ $$$$\mathrm{cot}\:\mathrm{2}\alpha\:\left(\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}\beta}\right)+\:\mathrm{2cot}\:\mathrm{2}\beta\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha}\right)= \\ $$$$\frac{\mathrm{2}\left(\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\beta\right)}{\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{sin}\:\mathrm{2}\beta}\:=\:\frac{\mathrm{cos}\:\mathrm{2}\alpha\:+\mathrm{cos}\:\mathrm{2}\beta}{\mathrm{2}\left(\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\right)\left(\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\right)} \\ $$$$=\:\frac{\mathrm{2}−\mathrm{2sin}^{\mathrm{2}} \alpha−\mathrm{2sin}\:^{\mathrm{2}} \beta\:}{\mathrm{2}\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{sin}\:^{\mathrm{2}} \alpha\:\mathrm{sin}\:^{\mathrm{2}} \beta}\: \\ $$$$=\frac{\mathrm{2}}{{a}}\:\left(\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \alpha−\mathrm{sin}\:^{\mathrm{2}} \beta}{\mathrm{sin}\:^{\mathrm{2}} \alpha\:\mathrm{sin}\:^{\mathrm{2}} \beta}\right) \\ $$$$=\:\frac{\mathrm{2}}{{a}}\left(\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \beta\right)−\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} \alpha−\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} \beta\right) \\ $$$$=\frac{\mathrm{2}}{{a}}\left(\mathrm{cot}\:^{\mathrm{2}} \alpha\:\mathrm{cot}\:^{\mathrm{2}} \beta−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}}{{a}}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\right)\:=\:\frac{\mathrm{2}}{{a}}\left(\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}\right)=\:\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}{a}} \\ $$$${in}\:{this}\:{case}\:{a}=\mathrm{9}\:;\:{we}\:{get}\:\frac{\mathrm{81}−\mathrm{4}}{\mathrm{18}}=\frac{\mathrm{77}}{\mathrm{18}} \\ $$
Commented by liberty last updated on 11/Apr/21
nice
$${nice} \\ $$
Answered by MJS_new last updated on 10/Apr/21
u=x+(√(x^2 +1)) ⇔ x=((u^2 −1)/(2u)) v=y+(√(y^2 +4)) ⇔ y=((v^2 −4)/(2v)) v=(9/u) ⇒ y=((81−4u^2 )/(18u)) x(√(y^2 +4))+y(√(x^2 +1))=...=((77)/(18))
$${u}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow\:{x}=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}} \\ $$$${v}={y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:\Leftrightarrow\:{y}=\frac{{v}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}{v}} \\ $$$${v}=\frac{\mathrm{9}}{{u}}\:\Rightarrow\:{y}=\frac{\mathrm{81}−\mathrm{4}{u}^{\mathrm{2}} }{\mathrm{18}{u}} \\ $$$${x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=…=\frac{\mathrm{77}}{\mathrm{18}} \\ $$
Commented by liberty last updated on 11/Apr/21
short cut?
$${short}\:{cut}? \\ $$
Commented by MJS_new last updated on 11/Apr/21
well I guess it′s not necessary to type all the algebraic steps... btw. we can also solve the given equation for y: X(y+(√(y^2 +4)))=9 ⇒ y=((81−4X^2 )/(18X))=((−85x+77(√(x^2 +1)))/(18)) ⇒ x(√(y^2 +4))+y(√(x^2 +1))=...=((77)/(18))
$$\mathrm{well}\:\mathrm{I}\:\mathrm{guess}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{type}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{algebraic}\:\mathrm{steps}… \\ $$$$\mathrm{btw}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\mathrm{for}\:{y}: \\ $$$${X}\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{9}\:\Rightarrow\:{y}=\frac{\mathrm{81}−\mathrm{4}{X}^{\mathrm{2}} }{\mathrm{18}{X}}=\frac{−\mathrm{85}{x}+\mathrm{77}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{18}} \\ $$$$\Rightarrow\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=…=\frac{\mathrm{77}}{\mathrm{18}} \\ $$
Commented by SLVR last updated on 11/Apr/21
this part of solution is marvoles
$${this}\:{part}\:{of}\:{solution}\:{is}\:{marvoles} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *