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







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$endgroup$












  • $begingroup$
    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






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edited Feb 3 '14 at 0:58









user127.0.0.1

5,95162139




5,95162139










asked Feb 3 '14 at 0:43









LearnerLearner

34941022




34941022












  • $begingroup$
    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











      $endgroup$













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        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

            8,06121738




            8,06121738























                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

                    6,19611526




                    6,19611526























                        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.

                            25k22539




                            25k22539






























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