Prove trig identity: $tan + cot = sec csc$












7












$begingroup$


I appreciate the help.



My attempt:



$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$






trigonometry solution-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
    $endgroup$
    – colormegone
    Feb 3 '14 at 0:48








  • 1




    $begingroup$
    yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
    $endgroup$
    – Eleven-Eleven
    Feb 3 '14 at 0:49
















7












$begingroup$


I appreciate the help.



My attempt:



$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$






trigonometry solution-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
    $endgroup$
    – colormegone
    Feb 3 '14 at 0:48








  • 1




    $begingroup$
    yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
    $endgroup$
    – Eleven-Eleven
    Feb 3 '14 at 0:49














7












7








7


1



$begingroup$


I appreciate the help.



My attempt:



$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$






trigonometry solution-verification







share|cite|improve this question











$endgroup$




I appreciate the help.



My attempt:



$$
begin{align}
tan + cot &= frac{sin}{cos} + frac{cos}{sin} \
&= frac{sin^2}{cos sin}+frac{cos^2}{cos sin} \
&= frac{sin^2+cos^2}{cos sin}\
&= frac{1}{cos sin}\
&= frac{1}{frac{1}{sec}frac{1}{csc}}\
&=frac{1}{frac{1}{sec csc}}\
&=frac{1}{1}cdot frac{sec csc}{1}\
&= sec csc
end{align}
$$






trigonometry solution-verification




trigonometry solution-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 '14 at 0:58









user127.0.0.1

5,99162139




5,99162139










asked Feb 3 '14 at 0:43









LearnerLearner

35941022




35941022












  • $begingroup$
    OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
    $endgroup$
    – colormegone
    Feb 3 '14 at 0:48








  • 1




    $begingroup$
    yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
    $endgroup$
    – Eleven-Eleven
    Feb 3 '14 at 0:49


















  • $begingroup$
    OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
    $endgroup$
    – colormegone
    Feb 3 '14 at 0:48








  • 1




    $begingroup$
    yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
    $endgroup$
    – Eleven-Eleven
    Feb 3 '14 at 0:49
















$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48






$begingroup$
OK! If you have to do this for an exam, however, I suggest you write in all of the " $ theta $ "s (or whatever symbol you are using for angles). A grader may take points off for not writing the functions properly. (What you did is fine for your own "scrap work", of course.)
$endgroup$
– colormegone
Feb 3 '14 at 0:48






1




1




$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49




$begingroup$
yup. It's quicker to go from $frac1{coscdot{sin}}$ to $frac1{cos}frac1{sin}=seccdot{csc}$.
$endgroup$
– Eleven-Eleven
Feb 3 '14 at 0:49










3 Answers
3






active

oldest

votes


















5












$begingroup$

That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.



Finally, you could save time on your proof by noticing on the fourth step that
$$
frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.



    Note that
    $$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
    that
    $$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
    and that
    $$sec^2theta=tan^2+1$$
    then
    $$begin{array}{lll}
    tantheta+cottheta&=&1cdot(tantheta+cottheta)\
    &=&(cotthetatantheta)(tantheta+cottheta)\
    &=&(cottheta)(tantheta(tantheta+cottheta))\
    &=&frac{csctheta}{sectheta}(tan^2theta+1)\
    &=&frac{csctheta}{sectheta}sec^2theta\
    &=&secthetacsctheta
    end{array}$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$






      share|cite|improve this answer











      $endgroup$














        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f661439%2fprove-trig-identity-tan-cot-sec-csc%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.



        Finally, you could save time on your proof by noticing on the fourth step that
        $$
        frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
        $$






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.



          Finally, you could save time on your proof by noticing on the fourth step that
          $$
          frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
          $$






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.



            Finally, you could save time on your proof by noticing on the fourth step that
            $$
            frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
            $$






            share|cite|improve this answer









            $endgroup$



            That is exactly correct! Just two things: First, $tan,sin,cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $tan x$, $cos x$ etc rather than just $tan$ or $cos$.



            Finally, you could save time on your proof by noticing on the fourth step that
            $$
            frac{1}{cos xsin x}=frac{1}{cos x}frac{1}{sin x}=sec x csc x
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 3 '14 at 0:50









            mathematics2x2lifemathematics2x2life

            8,10621739




            8,10621739























                1












                $begingroup$

                Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.



                Note that
                $$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
                that
                $$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
                and that
                $$sec^2theta=tan^2+1$$
                then
                $$begin{array}{lll}
                tantheta+cottheta&=&1cdot(tantheta+cottheta)\
                &=&(cotthetatantheta)(tantheta+cottheta)\
                &=&(cottheta)(tantheta(tantheta+cottheta))\
                &=&frac{csctheta}{sectheta}(tan^2theta+1)\
                &=&frac{csctheta}{sectheta}sec^2theta\
                &=&secthetacsctheta
                end{array}$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.



                  Note that
                  $$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
                  that
                  $$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
                  and that
                  $$sec^2theta=tan^2+1$$
                  then
                  $$begin{array}{lll}
                  tantheta+cottheta&=&1cdot(tantheta+cottheta)\
                  &=&(cotthetatantheta)(tantheta+cottheta)\
                  &=&(cottheta)(tantheta(tantheta+cottheta))\
                  &=&frac{csctheta}{sectheta}(tan^2theta+1)\
                  &=&frac{csctheta}{sectheta}sec^2theta\
                  &=&secthetacsctheta
                  end{array}$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.



                    Note that
                    $$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
                    that
                    $$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
                    and that
                    $$sec^2theta=tan^2+1$$
                    then
                    $$begin{array}{lll}
                    tantheta+cottheta&=&1cdot(tantheta+cottheta)\
                    &=&(cotthetatantheta)(tantheta+cottheta)\
                    &=&(cottheta)(tantheta(tantheta+cottheta))\
                    &=&frac{csctheta}{sectheta}(tan^2theta+1)\
                    &=&frac{csctheta}{sectheta}sec^2theta\
                    &=&secthetacsctheta
                    end{array}$$






                    share|cite|improve this answer









                    $endgroup$



                    Your steps are correct, but just keep in mind that robotically converting everying into $sin$s and $cos$s isn't the only option available to you.



                    Note that
                    $$cottheta = frac{costheta}{sintheta}=frac{frac{1}{sintheta}}{frac{1}{costheta}}=frac{csctheta}{sectheta}$$
                    that
                    $$cotthetatantheta=frac{1}{tantheta}cdottantheta=1$$
                    and that
                    $$sec^2theta=tan^2+1$$
                    then
                    $$begin{array}{lll}
                    tantheta+cottheta&=&1cdot(tantheta+cottheta)\
                    &=&(cotthetatantheta)(tantheta+cottheta)\
                    &=&(cottheta)(tantheta(tantheta+cottheta))\
                    &=&frac{csctheta}{sectheta}(tan^2theta+1)\
                    &=&frac{csctheta}{sectheta}sec^2theta\
                    &=&secthetacsctheta
                    end{array}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 11 '14 at 20:51









                    John JoyJohn Joy

                    6,29911827




                    6,29911827























                        0












                        $begingroup$

                        An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$






                            share|cite|improve this answer











                            $endgroup$



                            An alternative approach writes $t=tan x/2$ so $$tan x+cot x=frac{2t}{1-t^2}+frac{1-t^2}{2t}=frac{1+t^2}{1-t^2}frac{1+t^2}{2t}=sec xcsc x.$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 19 '18 at 17:24

























                            answered Dec 19 '18 at 17:17









                            J.G.J.G.

                            32.6k23250




                            32.6k23250






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f661439%2fprove-trig-identity-tan-cot-sec-csc%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Plaza Victoria

                                Image
                                Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                                Read more

                                Puebla de Zaragoza

                                Image
                                Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                                Read more

                                Musa

                                Image
                                Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                                Read more