Finding $dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}$












1












$begingroup$



find the :
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}$$




My Try :
$$dfrac{1}{1+dfrac{sin 70^{circ}}{cos 70^{circ}}}+dfrac{1}{1+dfrac{sin 20^{circ}}{cos 20^{circ}}}$$



$$dfrac{cos70^{circ}}{cos 70^{circ}+sin 70^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}$$



now what do i do ?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $cos20=sin70$
    $endgroup$
    – Nosrati
    Jan 24 '18 at 20:24










  • $begingroup$
    $$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
    $endgroup$
    – Jack D'Aurizio
    Jan 24 '18 at 20:34
















1












$begingroup$



find the :
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}$$




My Try :
$$dfrac{1}{1+dfrac{sin 70^{circ}}{cos 70^{circ}}}+dfrac{1}{1+dfrac{sin 20^{circ}}{cos 20^{circ}}}$$



$$dfrac{cos70^{circ}}{cos 70^{circ}+sin 70^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}$$



now what do i do ?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $cos20=sin70$
    $endgroup$
    – Nosrati
    Jan 24 '18 at 20:24










  • $begingroup$
    $$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
    $endgroup$
    – Jack D'Aurizio
    Jan 24 '18 at 20:34














1












1








1


1



$begingroup$



find the :
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}$$




My Try :
$$dfrac{1}{1+dfrac{sin 70^{circ}}{cos 70^{circ}}}+dfrac{1}{1+dfrac{sin 20^{circ}}{cos 20^{circ}}}$$



$$dfrac{cos70^{circ}}{cos 70^{circ}+sin 70^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}$$



now what do i do ?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$





find the :
$$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}$$




My Try :
$$dfrac{1}{1+dfrac{sin 70^{circ}}{cos 70^{circ}}}+dfrac{1}{1+dfrac{sin 20^{circ}}{cos 20^{circ}}}$$



$$dfrac{cos70^{circ}}{cos 70^{circ}+sin 70^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}$$



now what do i do ?






algebra-precalculus trigonometry




algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 22:02









Maria Mazur

50.4k1361126




50.4k1361126










asked Jan 24 '18 at 20:22









Fricul38Fricul38

36918




36918








  • 1




    $begingroup$
    $cos20=sin70$
    $endgroup$
    – Nosrati
    Jan 24 '18 at 20:24










  • $begingroup$
    $$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
    $endgroup$
    – Jack D'Aurizio
    Jan 24 '18 at 20:34














  • 1




    $begingroup$
    $cos20=sin70$
    $endgroup$
    – Nosrati
    Jan 24 '18 at 20:24










  • $begingroup$
    $$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
    $endgroup$
    – Jack D'Aurizio
    Jan 24 '18 at 20:34








1




1




$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24




$begingroup$
$cos20=sin70$
$endgroup$
– Nosrati
Jan 24 '18 at 20:24












$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34




$begingroup$
$$frac{1}{1+z}+frac{1}{1+frac{1}{z}}=1.$$
$endgroup$
– Jack D'Aurizio
Jan 24 '18 at 20:34










4 Answers
4






active

oldest

votes


















3












$begingroup$

Let $x=20^{circ}$



begin{eqnarray}dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}&=&dfrac{1}{1+cot 20^{circ}}+dfrac{1}{1+tan 20^{circ}}\&=& dfrac{sin x}{sin x+cos x}+dfrac{cos x}{cos x+sin x}\ &=&1end{eqnarray}






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      $$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$


        See that $$tan (x) = frac{sin x}{cos x}=frac{cos (90^{circ}-x)}{sin (90^{circ}-x)}=frac{1}{tan (90^{circ}-x)}$$
        what happen if you take $x=70$




        $$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$






        share|cite|improve this answer











        $endgroup$














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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Let $x=20^{circ}$



          begin{eqnarray}dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}&=&dfrac{1}{1+cot 20^{circ}}+dfrac{1}{1+tan 20^{circ}}\&=& dfrac{sin x}{sin x+cos x}+dfrac{cos x}{cos x+sin x}\ &=&1end{eqnarray}






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            Let $x=20^{circ}$



            begin{eqnarray}dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}&=&dfrac{1}{1+cot 20^{circ}}+dfrac{1}{1+tan 20^{circ}}\&=& dfrac{sin x}{sin x+cos x}+dfrac{cos x}{cos x+sin x}\ &=&1end{eqnarray}






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              Let $x=20^{circ}$



              begin{eqnarray}dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}&=&dfrac{1}{1+cot 20^{circ}}+dfrac{1}{1+tan 20^{circ}}\&=& dfrac{sin x}{sin x+cos x}+dfrac{cos x}{cos x+sin x}\ &=&1end{eqnarray}






              share|cite|improve this answer









              $endgroup$



              Let $x=20^{circ}$



              begin{eqnarray}dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}}&=&dfrac{1}{1+cot 20^{circ}}+dfrac{1}{1+tan 20^{circ}}\&=& dfrac{sin x}{sin x+cos x}+dfrac{cos x}{cos x+sin x}\ &=&1end{eqnarray}







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 24 '18 at 20:26









              Maria MazurMaria Mazur

              50.4k1361126




              50.4k1361126























                  2












                  $begingroup$

                  It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$






                      share|cite|improve this answer









                      $endgroup$



                      It's $$dfrac{sin20^{circ}}{sin 20^{circ}+cos 20^{circ}}+dfrac{cos20^{circ}}{cos 20^{circ}+sin 20^{circ}}=1$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 24 '18 at 20:24









                      Michael RozenbergMichael Rozenberg

                      111k1897201




                      111k1897201























                          2












                          $begingroup$

                          $$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            $$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              $$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$






                              share|cite|improve this answer









                              $endgroup$



                              $$frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+tan { { 20 }^{ circ } } } =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+cot { { 70 }^{ circ } } } =\ =frac { 1 }{ 1+tan { { 70 }^{ circ } } } +frac { 1 }{ 1+frac { 1 }{ tan { { 70 }^{ circ } } } } =frac { 1+tan { { 70 }^{ circ } } }{ 1+tan { { 70 }^{ circ } } } =1$$







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                              share|cite|improve this answer










                              answered Jan 24 '18 at 20:29









                              haqnaturalhaqnatural

                              20.8k72457




                              20.8k72457























                                  2












                                  $begingroup$


                                  See that $$tan (x) = frac{sin x}{cos x}=frac{cos (90^{circ}-x)}{sin (90^{circ}-x)}=frac{1}{tan (90^{circ}-x)}$$
                                  what happen if you take $x=70$




                                  $$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    2












                                    $begingroup$


                                    See that $$tan (x) = frac{sin x}{cos x}=frac{cos (90^{circ}-x)}{sin (90^{circ}-x)}=frac{1}{tan (90^{circ}-x)}$$
                                    what happen if you take $x=70$




                                    $$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$


                                      See that $$tan (x) = frac{sin x}{cos x}=frac{cos (90^{circ}-x)}{sin (90^{circ}-x)}=frac{1}{tan (90^{circ}-x)}$$
                                      what happen if you take $x=70$




                                      $$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$






                                      share|cite|improve this answer











                                      $endgroup$




                                      See that $$tan (x) = frac{sin x}{cos x}=frac{cos (90^{circ}-x)}{sin (90^{circ}-x)}=frac{1}{tan (90^{circ}-x)}$$
                                      what happen if you take $x=70$




                                      $$dfrac{1}{1+tan 70^{circ}}+dfrac{1}{1+tan 20^{circ}} = dfrac{1}{1+frac{1}{tan 20^{circ}}}+dfrac{1}{1+tan 20^{circ}}= dfrac{tan 20^{circ}}{1+tan 20^{circ}}+dfrac{1}{1+tan 20^{circ}} = 1$$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Jan 24 '18 at 21:11

























                                      answered Jan 24 '18 at 20:40









                                      Guy FsoneGuy Fsone

                                      17.3k43074




                                      17.3k43074






























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