Proving $ frac{csc x + cot x}{tan x + sin x} = cot xcsc x $












4












$begingroup$


I am currently working on understanding trig identities.
A question has me stumped, and no matter how I look at it, it never leads to the proof. I believe I am making a mistake when dividing multiple fractions.



$$ frac{csc x + cot x}{tan x + sin x} = cot xcsc x $$



For my first step I break up the $csc x$ and $cot x$ in the numerator and add them together to make:



$$frac{frac{1+cos x}{sin xcos x}}{tan x+sin x}$$



I then simplify further and end up at:



$$ frac{cos x+cos^2 x}{sin^2 xcos^2 x} $$



From here on I don't see any identities, or possible ways to decompose this further.






trigonometry problem-solving







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  • 2




    $begingroup$
    Note that $csc x + cot x = dfrac{1}{sin x} + dfrac{cos x}{sin x} = dfrac{1 + cos x}{sin x}$
    $endgroup$
    – Chaitanya Tappu
    Nov 30 '18 at 2:29
















4












$begingroup$


I am currently working on understanding trig identities.
A question has me stumped, and no matter how I look at it, it never leads to the proof. I believe I am making a mistake when dividing multiple fractions.



$$ frac{csc x + cot x}{tan x + sin x} = cot xcsc x $$



For my first step I break up the $csc x$ and $cot x$ in the numerator and add them together to make:



$$frac{frac{1+cos x}{sin xcos x}}{tan x+sin x}$$



I then simplify further and end up at:



$$ frac{cos x+cos^2 x}{sin^2 xcos^2 x} $$



From here on I don't see any identities, or possible ways to decompose this further.






trigonometry problem-solving







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that $csc x + cot x = dfrac{1}{sin x} + dfrac{cos x}{sin x} = dfrac{1 + cos x}{sin x}$
    $endgroup$
    – Chaitanya Tappu
    Nov 30 '18 at 2:29














4












4








4


0



$begingroup$


I am currently working on understanding trig identities.
A question has me stumped, and no matter how I look at it, it never leads to the proof. I believe I am making a mistake when dividing multiple fractions.



$$ frac{csc x + cot x}{tan x + sin x} = cot xcsc x $$



For my first step I break up the $csc x$ and $cot x$ in the numerator and add them together to make:



$$frac{frac{1+cos x}{sin xcos x}}{tan x+sin x}$$



I then simplify further and end up at:



$$ frac{cos x+cos^2 x}{sin^2 xcos^2 x} $$



From here on I don't see any identities, or possible ways to decompose this further.






trigonometry problem-solving







share|cite|improve this question











$endgroup$




I am currently working on understanding trig identities.
A question has me stumped, and no matter how I look at it, it never leads to the proof. I believe I am making a mistake when dividing multiple fractions.



$$ frac{csc x + cot x}{tan x + sin x} = cot xcsc x $$



For my first step I break up the $csc x$ and $cot x$ in the numerator and add them together to make:



$$frac{frac{1+cos x}{sin xcos x}}{tan x+sin x}$$



I then simplify further and end up at:



$$ frac{cos x+cos^2 x}{sin^2 xcos^2 x} $$



From here on I don't see any identities, or possible ways to decompose this further.






trigonometry problem-solving




trigonometry problem-solving






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 5:06









Blue

48.4k870154




48.4k870154










asked Nov 30 '18 at 2:20









Rawley FowlerRawley Fowler

47116




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  • 2




    $begingroup$
    Note that $csc x + cot x = dfrac{1}{sin x} + dfrac{cos x}{sin x} = dfrac{1 + cos x}{sin x}$
    $endgroup$
    – Chaitanya Tappu
    Nov 30 '18 at 2:29














  • 2




    $begingroup$
    Note that $csc x + cot x = dfrac{1}{sin x} + dfrac{cos x}{sin x} = dfrac{1 + cos x}{sin x}$
    $endgroup$
    – Chaitanya Tappu
    Nov 30 '18 at 2:29








2




2




$begingroup$
Note that $csc x + cot x = dfrac{1}{sin x} + dfrac{cos x}{sin x} = dfrac{1 + cos x}{sin x}$
$endgroup$
– Chaitanya Tappu
Nov 30 '18 at 2:29




$begingroup$
Note that $csc x + cot x = dfrac{1}{sin x} + dfrac{cos x}{sin x} = dfrac{1 + cos x}{sin x}$
$endgroup$
– Chaitanya Tappu
Nov 30 '18 at 2:29










2 Answers
2






active

oldest

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3












$begingroup$

$require{cancel}$
As Chaitanya Tappu noted, you made a mistake when adding $csc x$ and $cot x$.



$$frac{csc x+cot x}{tan x+sin x}=frac{frac{1}{sin x}+frac{cos }{sin x}}{frac{sin x}{cos x}+frac{sin xcos x}{cos x}}=frac{frac{1+cos x}{sin x}}{frac{sin x(1+cos x)}{cos x}}=frac{cancel{1+cos x}}{sin x}cdotfrac{cos x}{sin xcancel{(1+cos x)}}$$
$$=frac{cos x}{sin x}cdotfrac{1}{sin x}=cot xcsc x$$






share|cite|improve this answer











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  • 1




    $begingroup$
    you beat me to it. (+1)
    $endgroup$
    – clathratus
    Nov 30 '18 at 2:37



















3












$begingroup$

$$dfrac{a+b}{dfrac1a+dfrac1b}=cdots=ab$$ for $a+bne0$



$tan x=dfrac1?,sin x=dfrac1?$






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    $require{cancel}$
    As Chaitanya Tappu noted, you made a mistake when adding $csc x$ and $cot x$.



    $$frac{csc x+cot x}{tan x+sin x}=frac{frac{1}{sin x}+frac{cos }{sin x}}{frac{sin x}{cos x}+frac{sin xcos x}{cos x}}=frac{frac{1+cos x}{sin x}}{frac{sin x(1+cos x)}{cos x}}=frac{cancel{1+cos x}}{sin x}cdotfrac{cos x}{sin xcancel{(1+cos x)}}$$
    $$=frac{cos x}{sin x}cdotfrac{1}{sin x}=cot xcsc x$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      you beat me to it. (+1)
      $endgroup$
      – clathratus
      Nov 30 '18 at 2:37
















    3












    $begingroup$

    $require{cancel}$
    As Chaitanya Tappu noted, you made a mistake when adding $csc x$ and $cot x$.



    $$frac{csc x+cot x}{tan x+sin x}=frac{frac{1}{sin x}+frac{cos }{sin x}}{frac{sin x}{cos x}+frac{sin xcos x}{cos x}}=frac{frac{1+cos x}{sin x}}{frac{sin x(1+cos x)}{cos x}}=frac{cancel{1+cos x}}{sin x}cdotfrac{cos x}{sin xcancel{(1+cos x)}}$$
    $$=frac{cos x}{sin x}cdotfrac{1}{sin x}=cot xcsc x$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      you beat me to it. (+1)
      $endgroup$
      – clathratus
      Nov 30 '18 at 2:37














    3












    3








    3





    $begingroup$

    $require{cancel}$
    As Chaitanya Tappu noted, you made a mistake when adding $csc x$ and $cot x$.



    $$frac{csc x+cot x}{tan x+sin x}=frac{frac{1}{sin x}+frac{cos }{sin x}}{frac{sin x}{cos x}+frac{sin xcos x}{cos x}}=frac{frac{1+cos x}{sin x}}{frac{sin x(1+cos x)}{cos x}}=frac{cancel{1+cos x}}{sin x}cdotfrac{cos x}{sin xcancel{(1+cos x)}}$$
    $$=frac{cos x}{sin x}cdotfrac{1}{sin x}=cot xcsc x$$






    share|cite|improve this answer











    $endgroup$



    $require{cancel}$
    As Chaitanya Tappu noted, you made a mistake when adding $csc x$ and $cot x$.



    $$frac{csc x+cot x}{tan x+sin x}=frac{frac{1}{sin x}+frac{cos }{sin x}}{frac{sin x}{cos x}+frac{sin xcos x}{cos x}}=frac{frac{1+cos x}{sin x}}{frac{sin x(1+cos x)}{cos x}}=frac{cancel{1+cos x}}{sin x}cdotfrac{cos x}{sin xcancel{(1+cos x)}}$$
    $$=frac{cos x}{sin x}cdotfrac{1}{sin x}=cot xcsc x$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 12:42

























    answered Nov 30 '18 at 2:34









    Robert HowardRobert Howard

    1,9261822




    1,9261822








    • 1




      $begingroup$
      you beat me to it. (+1)
      $endgroup$
      – clathratus
      Nov 30 '18 at 2:37














    • 1




      $begingroup$
      you beat me to it. (+1)
      $endgroup$
      – clathratus
      Nov 30 '18 at 2:37








    1




    1




    $begingroup$
    you beat me to it. (+1)
    $endgroup$
    – clathratus
    Nov 30 '18 at 2:37




    $begingroup$
    you beat me to it. (+1)
    $endgroup$
    – clathratus
    Nov 30 '18 at 2:37











    3












    $begingroup$

    $$dfrac{a+b}{dfrac1a+dfrac1b}=cdots=ab$$ for $a+bne0$



    $tan x=dfrac1?,sin x=dfrac1?$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      $$dfrac{a+b}{dfrac1a+dfrac1b}=cdots=ab$$ for $a+bne0$



      $tan x=dfrac1?,sin x=dfrac1?$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        $$dfrac{a+b}{dfrac1a+dfrac1b}=cdots=ab$$ for $a+bne0$



        $tan x=dfrac1?,sin x=dfrac1?$






        share|cite|improve this answer









        $endgroup$



        $$dfrac{a+b}{dfrac1a+dfrac1b}=cdots=ab$$ for $a+bne0$



        $tan x=dfrac1?,sin x=dfrac1?$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 3:52









        lab bhattacharjeelab bhattacharjee

        225k15157275




        225k15157275






























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