Proving Trigonometry Identities $tan(frac{1}{2}x+45)+cot(frac{1}{2}x+45)=2sec x$












0












$begingroup$


How do i prove that




$tan(frac{1}{2}x+45)+cot(frac{1}{2}x+45)=2sec x$?




I managed to arrive at



$$frac{2 sec^2 (frac{1}{2}x)}{1- tan^2 (frac{1}{2}x)}$$



but I got stuck here. Any help is appreciated.






trigonometry







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  • 1




    $begingroup$
    Be careful! 45 (radians) is not $45^circ$. Multiply top and bottom by $cos^2(x/2)$ and use double angle formula on the denominator.
    $endgroup$
    – user10354138
    Jan 23 at 15:38


















0












$begingroup$


How do i prove that




$tan(frac{1}{2}x+45)+cot(frac{1}{2}x+45)=2sec x$?




I managed to arrive at



$$frac{2 sec^2 (frac{1}{2}x)}{1- tan^2 (frac{1}{2}x)}$$



but I got stuck here. Any help is appreciated.






trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Be careful! 45 (radians) is not $45^circ$. Multiply top and bottom by $cos^2(x/2)$ and use double angle formula on the denominator.
    $endgroup$
    – user10354138
    Jan 23 at 15:38
















0












0








0





$begingroup$


How do i prove that




$tan(frac{1}{2}x+45)+cot(frac{1}{2}x+45)=2sec x$?




I managed to arrive at



$$frac{2 sec^2 (frac{1}{2}x)}{1- tan^2 (frac{1}{2}x)}$$



but I got stuck here. Any help is appreciated.






trigonometry







share|cite|improve this question











$endgroup$




How do i prove that




$tan(frac{1}{2}x+45)+cot(frac{1}{2}x+45)=2sec x$?




I managed to arrive at



$$frac{2 sec^2 (frac{1}{2}x)}{1- tan^2 (frac{1}{2}x)}$$



but I got stuck here. Any help is appreciated.






trigonometry




trigonometry






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share|cite|improve this question













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edited Jan 23 at 15:47









mrtaurho

5,98551641




5,98551641










asked Jan 23 at 15:34









XingXiroXingXiro

12




12








  • 1




    $begingroup$
    Be careful! 45 (radians) is not $45^circ$. Multiply top and bottom by $cos^2(x/2)$ and use double angle formula on the denominator.
    $endgroup$
    – user10354138
    Jan 23 at 15:38
















  • 1




    $begingroup$
    Be careful! 45 (radians) is not $45^circ$. Multiply top and bottom by $cos^2(x/2)$ and use double angle formula on the denominator.
    $endgroup$
    – user10354138
    Jan 23 at 15:38










1




1




$begingroup$
Be careful! 45 (radians) is not $45^circ$. Multiply top and bottom by $cos^2(x/2)$ and use double angle formula on the denominator.
$endgroup$
– user10354138
Jan 23 at 15:38






$begingroup$
Be careful! 45 (radians) is not $45^circ$. Multiply top and bottom by $cos^2(x/2)$ and use double angle formula on the denominator.
$endgroup$
– user10354138
Jan 23 at 15:38












2 Answers
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$begingroup$

Because
$$tanleft(frac{x}{2}+45^{circ}right)+cotleft(frac{x}{2}+45^{circ}right)=frac{1}{sinleft(frac{x}{2}+45^{circ}right)cosleft(frac{x}{2}+45^{circ}right)}=frac{2}{sin(x+90^{circ})}=frac{2}{cos{x}}.$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    begin{align*}
    tanleft(frac{1}{2}x+45^{circ}right)+cotleft(frac{1}{2}x+45^{circ}right)&=tanleft(frac{1}{2}x+45^{circ}right)+frac1{tanleft(frac{1}{2}x+45^{circ}right)}\
    &=frac{1+tan^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
    &=frac{sec^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
    &=frac{sec(frac{1}{2}x+45^{circ})}{sin(frac{1}{2}x+45^{circ})}\
    &=frac{2}{2sin(frac{1}{2}x+45^{circ})cos(frac{1}{2}x+45^{circ})}\
    &=frac{2}{sin(x+90^{circ})}\
    &=frac{2}{cos(x)}\
    &=2sec x
    end{align*}






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      active

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      1












      $begingroup$

      Because
      $$tanleft(frac{x}{2}+45^{circ}right)+cotleft(frac{x}{2}+45^{circ}right)=frac{1}{sinleft(frac{x}{2}+45^{circ}right)cosleft(frac{x}{2}+45^{circ}right)}=frac{2}{sin(x+90^{circ})}=frac{2}{cos{x}}.$$






      share|cite|improve this answer









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        1












        $begingroup$

        Because
        $$tanleft(frac{x}{2}+45^{circ}right)+cotleft(frac{x}{2}+45^{circ}right)=frac{1}{sinleft(frac{x}{2}+45^{circ}right)cosleft(frac{x}{2}+45^{circ}right)}=frac{2}{sin(x+90^{circ})}=frac{2}{cos{x}}.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Because
          $$tanleft(frac{x}{2}+45^{circ}right)+cotleft(frac{x}{2}+45^{circ}right)=frac{1}{sinleft(frac{x}{2}+45^{circ}right)cosleft(frac{x}{2}+45^{circ}right)}=frac{2}{sin(x+90^{circ})}=frac{2}{cos{x}}.$$






          share|cite|improve this answer









          $endgroup$



          Because
          $$tanleft(frac{x}{2}+45^{circ}right)+cotleft(frac{x}{2}+45^{circ}right)=frac{1}{sinleft(frac{x}{2}+45^{circ}right)cosleft(frac{x}{2}+45^{circ}right)}=frac{2}{sin(x+90^{circ})}=frac{2}{cos{x}}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 15:45









          Michael RozenbergMichael Rozenberg

          108k1895200




          108k1895200























              1












              $begingroup$

              begin{align*}
              tanleft(frac{1}{2}x+45^{circ}right)+cotleft(frac{1}{2}x+45^{circ}right)&=tanleft(frac{1}{2}x+45^{circ}right)+frac1{tanleft(frac{1}{2}x+45^{circ}right)}\
              &=frac{1+tan^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
              &=frac{sec^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
              &=frac{sec(frac{1}{2}x+45^{circ})}{sin(frac{1}{2}x+45^{circ})}\
              &=frac{2}{2sin(frac{1}{2}x+45^{circ})cos(frac{1}{2}x+45^{circ})}\
              &=frac{2}{sin(x+90^{circ})}\
              &=frac{2}{cos(x)}\
              &=2sec x
              end{align*}






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                begin{align*}
                tanleft(frac{1}{2}x+45^{circ}right)+cotleft(frac{1}{2}x+45^{circ}right)&=tanleft(frac{1}{2}x+45^{circ}right)+frac1{tanleft(frac{1}{2}x+45^{circ}right)}\
                &=frac{1+tan^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
                &=frac{sec^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
                &=frac{sec(frac{1}{2}x+45^{circ})}{sin(frac{1}{2}x+45^{circ})}\
                &=frac{2}{2sin(frac{1}{2}x+45^{circ})cos(frac{1}{2}x+45^{circ})}\
                &=frac{2}{sin(x+90^{circ})}\
                &=frac{2}{cos(x)}\
                &=2sec x
                end{align*}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  begin{align*}
                  tanleft(frac{1}{2}x+45^{circ}right)+cotleft(frac{1}{2}x+45^{circ}right)&=tanleft(frac{1}{2}x+45^{circ}right)+frac1{tanleft(frac{1}{2}x+45^{circ}right)}\
                  &=frac{1+tan^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
                  &=frac{sec^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
                  &=frac{sec(frac{1}{2}x+45^{circ})}{sin(frac{1}{2}x+45^{circ})}\
                  &=frac{2}{2sin(frac{1}{2}x+45^{circ})cos(frac{1}{2}x+45^{circ})}\
                  &=frac{2}{sin(x+90^{circ})}\
                  &=frac{2}{cos(x)}\
                  &=2sec x
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$



                  begin{align*}
                  tanleft(frac{1}{2}x+45^{circ}right)+cotleft(frac{1}{2}x+45^{circ}right)&=tanleft(frac{1}{2}x+45^{circ}right)+frac1{tanleft(frac{1}{2}x+45^{circ}right)}\
                  &=frac{1+tan^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
                  &=frac{sec^2(frac{1}{2}x+45^{circ})}{tan(frac{1}{2}x+45^{circ})}\
                  &=frac{sec(frac{1}{2}x+45^{circ})}{sin(frac{1}{2}x+45^{circ})}\
                  &=frac{2}{2sin(frac{1}{2}x+45^{circ})cos(frac{1}{2}x+45^{circ})}\
                  &=frac{2}{sin(x+90^{circ})}\
                  &=frac{2}{cos(x)}\
                  &=2sec x
                  end{align*}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 16:05









                  Thomas ShelbyThomas Shelby

                  4,0442625




                  4,0442625






























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