
This graph once again includes the original one, but with even more extra reciprocal curves. I think this is because of the squares in the identity used.
I then decide to muck around with various reciprocal equations until I find one that matches the first graph. I find:
$$y=\frac{2}{x-1}+1$$
At this point, I am very confused. How (if it is possible) can I get from my parametric equations to that final Cartesian equation? Where (if I am going wrong) am I (and the mark scheme) going wrong with the first two attempts? Do the first two Cartesian equations have solutions that are not solutions for the parametric equations?

A:

When in doubt, here are some heuristics that can sometimes help:

Rewrite everything in terms of $\sin$ and $\cos$.
Try putting things over a common denominator.
Try computing $x+y$, $x^2$, $y^2$ and $xy$ and see if you can find relationships between them.

In this case, we have
$$x = \frac{1}{\cos\theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin\theta}{\cos \theta}$$
$$y = \frac{1}{\sin\theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos\theta}{\sin \theta}$$
Finding a common denominator and adding these we have
$$x + y = \frac{\sin\theta + \sin^2\theta + \cos\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1+ \sin\theta +  \cos\theta}{\sin\theta\cos\theta}$$
while multiplication gives
$$xy = \frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta \cos\theta} = \frac{1+\sin\theta+\cos\theta+\sin\theta \cos\theta}{\sin\theta \cos\theta}$$
Now step back and look at these two results:  hopefully you notice that they are almost identical, except for one extra term in the numerator of the second expression.  In fact we have
$$xy - (x+y) = \frac{\sin\theta \cos\theta}{\sin\theta \cos\theta
This graph once again includes the original one, but with even more extra reciprocal curves. I think this is because of the public square in the identity use. 
 I then decide to muck around with various reciprocal equations until I determine one that matches the first graph. I discover: 
 $ $ y=\frac{2}{x-1}+1$$ 
 At this point, I am very confused. How (if it is potential) can I contract from my parametric equation to that final Cartesian equation? Where (if I am go wrong) am I (and the mark scheme) going wrong with the inaugural two attempts? Do the first two Cartesian equations consume solutions that are not solutions for the parametric equations? 

 A: 

 When in doubt, here are some heuristic that can sometimes help: 

 Rewrite everything in term of $ \sin$ and $ \cos$. 
 Try putting things over a coarse denominator. 
 Try computing $ x+y$, $ x^2 $, $ y^2 $ and $ xy$ and see if you can find relationships between them. 

 In this case, we have 
 $ $ x = \frac{1}{\cos\theta } + \frac{\sin \theta}{\cos \theta } = \frac{1 + \sin\theta}{\cos \theta}$$ 
 $ $ y = \frac{1}{\sin\theta } + \frac{\cos \theta}{\sin \theta } = \frac{1 + \cos\theta}{\sin \theta}$$ 
 Finding a common denominator and adding these we have 
 $ $ x + y = \frac{\sin\theta + \sin^2\theta + \cos\theta + \cos^2\theta}{\sin\theta\cos\theta } = \frac{1 + \sin\theta +   \cos\theta}{\sin\theta\cos\theta}$$ 
 while multiplication gives 
 $ $ xy = \frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta \cos\theta } = \frac{1+\sin\theta+\cos\theta+\sin\theta \cos\theta}{\sin\theta \cos\theta}$$ 
 Now step back and look at these two results:   hopefully you detect that they are almost identical, except for one extra terminus in the numerator of the second formula.   In fact we have 
 $ $ xy - (x+y) = \frac{\sin\theta \cos\theta}{\sin\theta \cos\theta

Thid graph once again incluaes the original one, bnt with even mofe extra reciprocal curves. I tyink ukis is because of thd squares in the udenuity used.
I then dxdide to muck zvound xith various regiprocal eqgations until H wiud one that matches the first graph. Y find:
$$y=\grwc{2}{x-1}+1$$
At this poigt, I wm vsgy confused. How (if it is possibme) can P get from my parsmetric equations to that vinap Cartesian equatiln? Where (if I ai going wrong) am I (and uhz mark scheje) going wrong with the first tdo atcempts? Do tye figvt two Cartxsian vquations havc soluthons thst are not solmtionv fir the parametric equetions?

A:

When in doubt, here are skme heuristics thqt can vomedimer heup:

Rtwrmte everyhhiig in terms of $\sin$ and $\cos$.
Try putting thimgf over a commoh denoiigator.
Try computing $x+y$, $x^2$, $y^2$ and $xy$ and set if gou can find relationshups between them.

In thls case, wq have
$$x = \frac{1}{\cos\theta} + \frac{\sin \theta}{\cos \theta} = \frdc{1 + \smn\gheua}{\gjr \tjeta}$$
$$y = \frac{1}{\sin\theta} + \frac{\cos \theta}{\sin \theta} = \sdav{1 + \cos\theta}{\sin \tmeta}$$
Finding a commpn drgominator and addiny tgese we have
$$x + y = \vrac{\sin\jheta + \sin^2\theta + \cox\theta + \cos^2\theta}{\sin\theta\cos\rheta} = \frac{1+ \fun\theta +  \cos\theta}{\din\theta\cos\cheta}$$
wnile kultiplication gives
$$xy = \frad{(1+\sin\theta)(1+\cod\theta)}{\sin\ffeta \cos\theta} = \ffac{1+\xit\theta+\cos\theta+\sin\theta \cos\eheta}{\sin\tieta \eos\theta}$$
Vow xtep bwck and lolk at these two results:  hopeyully you noticf that they are almost identical, except for ome exnra term nn the numerator of the second exkression.  In fxct we havv
$$xy - (x+y) = \hrac{\sin\theta \cos\theta}{\sin\tvgta \cos\theta
This graph once again includes the original with more extra curves. I think squares the identity used. then decide to around with various reciprocal equations until find one that matches the first graph. I find: $$y=\frac{2}{x-1}+1$$ At this point, am very confused. How (if it is possible) can I get from my equations that Cartesian Where (if I am going wrong) am I (and the mark scheme) going wrong with the two attempts? Do the first two Cartesian equations solutions that are not for the parametric equations? A: in here are heuristics can help: Rewrite everything terms of $\sin$ and $\cos$. Try putting things over a common denominator. Try computing $x+y$, $x^2$, $y^2$ $xy$ and you can relationships them. this case, we = \frac{1}{\cos\theta} + \frac{\sin \theta}{\cos \theta} \sin\theta}{\cos \theta}$$ $$y = \frac{1}{\sin\theta} + \frac{\cos \theta}{\sin = \frac{1 \cos\theta}{\sin \theta}$$ Finding a common denominator adding these we have $$x + y = + \sin^2\theta + \cos\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1+ \sin\theta + \cos\theta}{\sin\theta\cos\theta}$$ while multiplication gives $$xy \cos\theta} = \frac{1+\sin\theta+\cos\theta+\sin\theta \cos\theta}{\sin\theta Now step back look these results: you notice they are almost identical, except for one extra term in the of the second expression. In fact we have $$xy - \frac{\sin\theta \cos\theta

This graph once again includeS the originAl one, But WitH eVen mOre eXtra reciprocal CUrveS. I think this is because of The sqUaREs in THe IdentIty used.
i ThEN DecIdE tO muCk ARoUnd wiTh vArious rEciprocal eQuaTiOns until I finD OnE that matchEs tHe first graph. i fiNd:
$$y=\fraC{2}{x-1}+1$$
at tHIs poiNt, I Am verY confuSEd. How (iF it is possIbLE) can I gET from my PARaMetrIc equations to that FInAL Cartesian equaTion? WhErE (If i AM goIng Wrong) am I (anD tHe marK Scheme) gOInG WROng WIth the first twO attempts? Do THe fIrst twO CArtESian eqUatioNs HAve Solutions thAt arE not solutIons foR The paraMEtric eqUationS?

A:

WHen In doUBt, HeRe aRe SOme HEuRisTIcs That can sOmEtImes hElp:

REWRITe evEryThinG in teRms of $\sin$ and $\coS$.
TrY putTIng ThingS over A comMoN denoMinatoR.
Try cOmPuting $x+y$, $x^2$, $y^2$ and $xy$ And sEe if you caN fiNd RelAtIonshIPs betwEen TheM.

In this Case, we hAVe
$$x = \FrAC{1}{\COs\Theta} + \frac{\sin \theta}{\cOs \THEtA} = \frac{1 + \sin\Theta}{\cOS \tHeTA}$$
$$y = \frac{1}{\siN\tHetA} + \fraC{\COs \theTa}{\siN \ThEta} = \frac{1 + \cOs\thetA}{\SiN \tHeta}$$
FinDiNg a comMoN deNomInatoR And aDding tHese we haVe
$$x + y = \fRAc{\sin\theta + \sin^2\tHEta + \cos\theta + \coS^2\ThETA}{\sIN\theTa\cOs\theta} = \frac{1+ \Sin\tHEta +  \cOs\thETa}{\Sin\THeta\cOs\theTa}$$
WHiLE multiplication giveS
$$xY = \frac{(1+\sIn\theTa)(1+\cos\theta)}{\sin\Theta \cos\thETA} = \Frac{1+\sin\tHeta+\COs\THeta+\sin\theta \coS\thetA}{\sin\theta \cOS\theta}$$
NoW step Back and lOok at thesE TWo resultS:  hoPefUllY yoU NOtIce that they arE ALmosT iDenticaL, exCept for One ExtRa tErm In The numeraTor of the SeCoNd ExPreSsion.  iN fact we hAvE
$$xy - (X+y) = \FraC{\sin\tHEta \cos\Theta}{\Sin\tHeTa \COs\tHeta

This graph once again inc ludes theorigi nal  on e,  but  wit h even more ex t ra r eciprocal curves. I th ink t hi s  isb ec auseof thes qu a r esin  t heid e nt ity u sed .
I the n decide t o m uc k around wit h  v arious rec ipr ocal equatio nsuntilIfin d  onetha t mat ches t h e firs t graph.If ind:
$ $ y=\frac { 2 }{ x-1} +1$$
At this poin t ,I  am very confu sed. H ow (i f itispossible)ca n I g e t fromm yp a r ame t ric equations  to that fi n alCartes ia n e q uation ? Whe re (if  I am going  wro ng) am I(and t h e marks cheme)goingwro ngwith th efir st two at tem p ts?  Do thefi rs t two  Car t e s i an e qua tion s hav e solutions t hat  are not  solu tions  for  t he pa rametr ic eq ua tions?

A:

Whe n in  doubt, h ere  a reso me he u ristic s t hat  can so metimes hel p: R ew rite everything in  t e r ms  of $\si n$ and $\ co s $.
Try p ut tin g th i n gs ov er a co mmon den ominat o r. 
T ry comp ut ing $x +y $,$x^ 2$, $ y ^2$and $x y$ and s ee if you can find r e lationships b e tw e e nt hem. 

I n this case , we have 
$$x =\fr a c{1}{ \cos\ th e ta }  + \frac{\sin \thet a} {\cos\thet a} = \frac{1+ \sin\the t a } {\cos \t heta } $$ $$y = \frac{1} {\sin \theta} +\ frac{\co s \th eta}{\si n \theta} =  \frac{1  +\co s\t het a } {\ sin \theta}$$ F indi ng  a comm ondenomin ato r a ndadd in g these w e have
$ $x  +  y  =  \f rac{\ s in\theta  +  \s in ^2\ theta + \cos \thet a +\c os ^ 2\t heta}{\ s in \ t heta \c os \the ta}  =  \fra c{1+ \si n\theta  +  \cos\ the t a}{\ si n\ theta\c os\theta}$$
w hi le multipl ic ati on giv e s 
$$xy =\frac{(1+\sin\theta)(1+ \ cos\the ta) }{\si n\th eta \cos\ the ta} =\fr a c{1+\s in\the ta+\c os \th e t a+\si n \ th eta  \ cos\theta} { \ sin \thet a\cos \theta} $$
Now step back a n d l ook at thesetwo  res u l ts : h op e ful ly you n otice that they  are almos ti de ntical, ex c ept  f or oneextra t erm i n  the nu merator o f the sec on d ex p r ess ion.  In f act we h ave
$$xy-  (x+y )  =  \fra c{\ sin\th et a \ cos\t heta}{ \ sin \thet a \cos \t heta

This graph_once again_includes the original one,_but with_even_more extra_reciprocal_curves. I think_this is because_of the squares in_the identity used.
I_then_decide to muck around with various reciprocal equations until I find one that matches_the_first graph._I_find:
$$y=\frac{2}{x-1}+1$$
At_this point, I am very_confused. How (if it is_possible) can_I get from my parametric equations to that_final_Cartesian equation? Where_(if I am going wrong) am I (and the_mark scheme) going wrong with the_first two attempts?_Do_the_first two Cartesian equations_have solutions that are not solutions_for the parametric equations?

A:

When in doubt,_here are some heuristics that can sometimes_help:

Rewrite everything in terms of $\sin$_and $\cos$.
Try putting things over_a common_denominator.
Try computing $x+y$, $x^2$, $y^2$_and $xy$ and_see if_you can find_relationships between them.

In this case, we_have
$$x = \frac{1}{\cos\theta}_+ \frac{\sin \theta}{\cos \theta} = \frac{1_+_\sin\theta}{\cos \theta}$$
$$y =_\frac{1}{\sin\theta}_+_\frac{\cos \theta}{\sin_\theta} = \frac{1_+_\cos\theta}{\sin \theta}$$
Finding_a_common denominator and adding these we_have
$$x_+ y = \frac{\sin\theta + \sin^2\theta +_\cos\theta + \cos^2\theta}{\sin\theta\cos\theta} =_\frac{1+_\sin\theta +  \cos\theta}{\sin\theta\cos\theta}$$
while_multiplication gives
$$xy = \frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta \cos\theta}_= \frac{1+\sin\theta+\cos\theta+\sin\theta \cos\theta}{\sin\theta \cos\theta}$$
Now step back_and look_at these_two results:  hopefully you notice that they are almost identical,_except for one extra term in_the numerator of the_second expression.__In fact we_have
$$xy_- (x+y)_= \frac{\sin\theta \cos\theta}{\sin\theta \cos\theta