How to prove a trigonometric identity $tan(A)=frac{sin2A}{1+cos 2A}$












5












$begingroup$


Show that
$$
tan(A)=frac{sin2A}{1+cos 2A}
$$



I've tried a few methods, and it stumped my teacher.






trigonometry







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  • 1




    $begingroup$
    Failure to prove is not surprising. It is quite false. For example, take $A=pi/4$ ($45$ degrees).
    $endgroup$
    – André Nicolas
    Mar 15 '12 at 23:48












  • $begingroup$
    I think you want $cos(2A)$ there...
    $endgroup$
    – David Mitra
    Mar 15 '12 at 23:53










  • $begingroup$
    This is not a valid identity: try $A=pi/4$ for example.
    $endgroup$
    – Shane O Rourke
    Mar 15 '12 at 23:54
















5












$begingroup$


Show that
$$
tan(A)=frac{sin2A}{1+cos 2A}
$$



I've tried a few methods, and it stumped my teacher.






trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Failure to prove is not surprising. It is quite false. For example, take $A=pi/4$ ($45$ degrees).
    $endgroup$
    – André Nicolas
    Mar 15 '12 at 23:48












  • $begingroup$
    I think you want $cos(2A)$ there...
    $endgroup$
    – David Mitra
    Mar 15 '12 at 23:53










  • $begingroup$
    This is not a valid identity: try $A=pi/4$ for example.
    $endgroup$
    – Shane O Rourke
    Mar 15 '12 at 23:54














5












5








5


4



$begingroup$


Show that
$$
tan(A)=frac{sin2A}{1+cos 2A}
$$



I've tried a few methods, and it stumped my teacher.






trigonometry







share|cite|improve this question











$endgroup$




Show that
$$
tan(A)=frac{sin2A}{1+cos 2A}
$$



I've tried a few methods, and it stumped my teacher.






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 14:26









Martin Sleziak

44.7k10119272




44.7k10119272










asked Mar 15 '12 at 23:44









Rollo Montgomery Konig-BrockRollo Montgomery Konig-Brock

646




646








  • 1




    $begingroup$
    Failure to prove is not surprising. It is quite false. For example, take $A=pi/4$ ($45$ degrees).
    $endgroup$
    – André Nicolas
    Mar 15 '12 at 23:48












  • $begingroup$
    I think you want $cos(2A)$ there...
    $endgroup$
    – David Mitra
    Mar 15 '12 at 23:53










  • $begingroup$
    This is not a valid identity: try $A=pi/4$ for example.
    $endgroup$
    – Shane O Rourke
    Mar 15 '12 at 23:54














  • 1




    $begingroup$
    Failure to prove is not surprising. It is quite false. For example, take $A=pi/4$ ($45$ degrees).
    $endgroup$
    – André Nicolas
    Mar 15 '12 at 23:48












  • $begingroup$
    I think you want $cos(2A)$ there...
    $endgroup$
    – David Mitra
    Mar 15 '12 at 23:53










  • $begingroup$
    This is not a valid identity: try $A=pi/4$ for example.
    $endgroup$
    – Shane O Rourke
    Mar 15 '12 at 23:54








1




1




$begingroup$
Failure to prove is not surprising. It is quite false. For example, take $A=pi/4$ ($45$ degrees).
$endgroup$
– André Nicolas
Mar 15 '12 at 23:48






$begingroup$
Failure to prove is not surprising. It is quite false. For example, take $A=pi/4$ ($45$ degrees).
$endgroup$
– André Nicolas
Mar 15 '12 at 23:48














$begingroup$
I think you want $cos(2A)$ there...
$endgroup$
– David Mitra
Mar 15 '12 at 23:53




$begingroup$
I think you want $cos(2A)$ there...
$endgroup$
– David Mitra
Mar 15 '12 at 23:53












$begingroup$
This is not a valid identity: try $A=pi/4$ for example.
$endgroup$
– Shane O Rourke
Mar 15 '12 at 23:54




$begingroup$
This is not a valid identity: try $A=pi/4$ for example.
$endgroup$
– Shane O Rourke
Mar 15 '12 at 23:54










5 Answers
5






active

oldest

votes


















51












$begingroup$

Proof without words: $tan(A)=dfrac{color{red}{sin(2A)}}{color{blue}{1}+color{green}{cos(2A)}}$



$hspace{4cm}$Proof






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    I've never seen anyone prove a trig identity like that, but they should!
    $endgroup$
    – The Substitute
    Mar 29 '12 at 23:54



















11












$begingroup$

The given equality is false. Set $A = pi/2$. (Note: this applied to an earlier version of the problem).



Perhaps what you meant was



$$ tan frac{A}{2} = frac{sin A}{1 + cos A}$$



or



$$ tan A = frac{sin 2A}{1 + cos 2A}$$



which is true, by using the half/double angle formulas.



$$frac{sin A}{1 + cos A} = frac{ 2 sin A/2 cos A/2}{2 cos^2 A/2} = tan A/2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
    $endgroup$
    – robjohn♦
    Mar 29 '12 at 22:40












  • $begingroup$
    Nice proof, too :-)
    $endgroup$
    – robjohn♦
    Mar 29 '12 at 22:48



















3












$begingroup$

$$sin 2A = 2 sin A cos A$$



$$cos 2A = 2 cos^2A - 1$$



Substitute these identities and you will get $tan A$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    duplicate?
    $endgroup$
    – draks ...
    Mar 29 '12 at 7:55






  • 1




    $begingroup$
    I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
    $endgroup$
    – user21436
    Mar 29 '12 at 22:28



















3












$begingroup$

We need to prove that:
$$frac{sin(2A)}{1+cos(2A)}=tan(A)$$
Let's do LHS-RHS to prove it. I will try to make the left side equal the right side.
$$frac{sin(2A)}{1+cos(2A)}$$
Using double angle identites for both sine and cosine:
$$frac{2sin(A)cos(A)}{1+2cos^2(A)-1}$$
How nice. The $1$ and $-1$ in the denominator cancel out.
$$frac{2sin(A)cos(A)}{2cos^2(A)}$$
Cancelling out the $cos(A)$ in the numerator and the denominator yields:
$$frac{2sin(A)}{2cos(A)}$$
We can also cancel out the $2$ in the numerator and the denominator.
$$frac{sin(A)}{cos(A)}$$
$$=tan(A)$$
$$text{LHS=RHS}$$
$$displaystyle boxed{therefore dfrac{sin(2A)}{1+cos(2A)}=tan(A)}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    First, lets develop a couple of identities.
    Given that $sin 2A = 2sin Acos A$, and $cos 2A = cos^2A - sin^2 A$ we have



    $$begin{array}{lll}
    tan 2A &=& frac{sin 2A}{cos 2A}\
    &=&frac{2sin Acos A}{cos^2 A-sin^2A}\
    &=&frac{2sin Acos A}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
    &=&frac{2tan A}{1-tan^2A}
    end{array}$$
    Similarly, we have
    $$begin{array}{lll}
    sec 2A &=& frac{1}{cos 2A}\
    &=&frac{1}{cos^2 A-sin^2A}\
    &=&frac{1}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
    &=&frac{sec^2 A}{1-tan^2A}
    end{array}$$



    But sometimes it is just as easy to represent these identities as




    $$begin{array}{lll}
    (1-tan^2 A)sec 2A &=& sec^2 A\
    (1-tan^2 A)tan 2A &=& 2tan A
    end{array}$$


    Applying these identities to the problem at hand we have
    $$begin{array}{lll}
    frac{sin 2A}{1+cos 2A}&=& frac{sin 2A}{1+cos 2A}cdotfrac{frac{1}{cos 2A}}{frac{1}{cos 2A}}\
    &=& frac{tan 2A}{sec 2A +1}\
    &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)(sec 2A +1)}\
    &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)sec 2A +(1-tan^2 A)}\
    &=& frac{2tan A}{sec^2 A +(1-tan^2 A)}\
    &=& frac{2tan A}{(tan^2 A+1) +(1-tan^2 A)}\
    &=& frac{2tan A}{2}\
    &=& tan A\
    end{array}$$
    discalimer: I like JChau's answer better :)






    share|cite|improve this answer









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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      51












      $begingroup$

      Proof without words: $tan(A)=dfrac{color{red}{sin(2A)}}{color{blue}{1}+color{green}{cos(2A)}}$



      $hspace{4cm}$Proof






      share|cite|improve this answer









      $endgroup$









      • 3




        $begingroup$
        I've never seen anyone prove a trig identity like that, but they should!
        $endgroup$
        – The Substitute
        Mar 29 '12 at 23:54
















      51












      $begingroup$

      Proof without words: $tan(A)=dfrac{color{red}{sin(2A)}}{color{blue}{1}+color{green}{cos(2A)}}$



      $hspace{4cm}$Proof






      share|cite|improve this answer









      $endgroup$









      • 3




        $begingroup$
        I've never seen anyone prove a trig identity like that, but they should!
        $endgroup$
        – The Substitute
        Mar 29 '12 at 23:54














      51












      51








      51





      $begingroup$

      Proof without words: $tan(A)=dfrac{color{red}{sin(2A)}}{color{blue}{1}+color{green}{cos(2A)}}$



      $hspace{4cm}$Proof






      share|cite|improve this answer









      $endgroup$



      Proof without words: $tan(A)=dfrac{color{red}{sin(2A)}}{color{blue}{1}+color{green}{cos(2A)}}$



      $hspace{4cm}$Proof







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Mar 29 '12 at 22:23









      robjohn♦robjohn

      268k27308634




      268k27308634








      • 3




        $begingroup$
        I've never seen anyone prove a trig identity like that, but they should!
        $endgroup$
        – The Substitute
        Mar 29 '12 at 23:54














      • 3




        $begingroup$
        I've never seen anyone prove a trig identity like that, but they should!
        $endgroup$
        – The Substitute
        Mar 29 '12 at 23:54








      3




      3




      $begingroup$
      I've never seen anyone prove a trig identity like that, but they should!
      $endgroup$
      – The Substitute
      Mar 29 '12 at 23:54




      $begingroup$
      I've never seen anyone prove a trig identity like that, but they should!
      $endgroup$
      – The Substitute
      Mar 29 '12 at 23:54











      11












      $begingroup$

      The given equality is false. Set $A = pi/2$. (Note: this applied to an earlier version of the problem).



      Perhaps what you meant was



      $$ tan frac{A}{2} = frac{sin A}{1 + cos A}$$



      or



      $$ tan A = frac{sin 2A}{1 + cos 2A}$$



      which is true, by using the half/double angle formulas.



      $$frac{sin A}{1 + cos A} = frac{ 2 sin A/2 cos A/2}{2 cos^2 A/2} = tan A/2$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:40












      • $begingroup$
        Nice proof, too :-)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:48
















      11












      $begingroup$

      The given equality is false. Set $A = pi/2$. (Note: this applied to an earlier version of the problem).



      Perhaps what you meant was



      $$ tan frac{A}{2} = frac{sin A}{1 + cos A}$$



      or



      $$ tan A = frac{sin 2A}{1 + cos 2A}$$



      which is true, by using the half/double angle formulas.



      $$frac{sin A}{1 + cos A} = frac{ 2 sin A/2 cos A/2}{2 cos^2 A/2} = tan A/2$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:40












      • $begingroup$
        Nice proof, too :-)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:48














      11












      11








      11





      $begingroup$

      The given equality is false. Set $A = pi/2$. (Note: this applied to an earlier version of the problem).



      Perhaps what you meant was



      $$ tan frac{A}{2} = frac{sin A}{1 + cos A}$$



      or



      $$ tan A = frac{sin 2A}{1 + cos 2A}$$



      which is true, by using the half/double angle formulas.



      $$frac{sin A}{1 + cos A} = frac{ 2 sin A/2 cos A/2}{2 cos^2 A/2} = tan A/2$$






      share|cite|improve this answer











      $endgroup$



      The given equality is false. Set $A = pi/2$. (Note: this applied to an earlier version of the problem).



      Perhaps what you meant was



      $$ tan frac{A}{2} = frac{sin A}{1 + cos A}$$



      or



      $$ tan A = frac{sin 2A}{1 + cos 2A}$$



      which is true, by using the half/double angle formulas.



      $$frac{sin A}{1 + cos A} = frac{ 2 sin A/2 cos A/2}{2 cos^2 A/2} = tan A/2$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 29 '12 at 20:52

























      answered Mar 15 '12 at 23:51









      AryabhataAryabhata

      70k6156246




      70k6156246












      • $begingroup$
        This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:40












      • $begingroup$
        Nice proof, too :-)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:48


















      • $begingroup$
        This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:40












      • $begingroup$
        Nice proof, too :-)
        $endgroup$
        – robjohn♦
        Mar 29 '12 at 22:48
















      $begingroup$
      This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
      $endgroup$
      – robjohn♦
      Mar 29 '12 at 22:40






      $begingroup$
      This is one of my favorite identities. It is the basis for one form of $operatorname{atan2}(x,y)=2operatorname{atan}left(dfrac{y}{r+x}right)$ which is useful if you have to compute $r=sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1)
      $endgroup$
      – robjohn♦
      Mar 29 '12 at 22:40














      $begingroup$
      Nice proof, too :-)
      $endgroup$
      – robjohn♦
      Mar 29 '12 at 22:48




      $begingroup$
      Nice proof, too :-)
      $endgroup$
      – robjohn♦
      Mar 29 '12 at 22:48











      3












      $begingroup$

      $$sin 2A = 2 sin A cos A$$



      $$cos 2A = 2 cos^2A - 1$$



      Substitute these identities and you will get $tan A$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        duplicate?
        $endgroup$
        – draks ...
        Mar 29 '12 at 7:55






      • 1




        $begingroup$
        I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
        $endgroup$
        – user21436
        Mar 29 '12 at 22:28
















      3












      $begingroup$

      $$sin 2A = 2 sin A cos A$$



      $$cos 2A = 2 cos^2A - 1$$



      Substitute these identities and you will get $tan A$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        duplicate?
        $endgroup$
        – draks ...
        Mar 29 '12 at 7:55






      • 1




        $begingroup$
        I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
        $endgroup$
        – user21436
        Mar 29 '12 at 22:28














      3












      3








      3





      $begingroup$

      $$sin 2A = 2 sin A cos A$$



      $$cos 2A = 2 cos^2A - 1$$



      Substitute these identities and you will get $tan A$.






      share|cite|improve this answer











      $endgroup$



      $$sin 2A = 2 sin A cos A$$



      $$cos 2A = 2 cos^2A - 1$$



      Substitute these identities and you will get $tan A$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 29 '12 at 19:21







      user21436

















      answered Mar 29 '12 at 7:48









      RohanRohan

      945




      945








      • 1




        $begingroup$
        duplicate?
        $endgroup$
        – draks ...
        Mar 29 '12 at 7:55






      • 1




        $begingroup$
        I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
        $endgroup$
        – user21436
        Mar 29 '12 at 22:28














      • 1




        $begingroup$
        duplicate?
        $endgroup$
        – draks ...
        Mar 29 '12 at 7:55






      • 1




        $begingroup$
        I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
        $endgroup$
        – user21436
        Mar 29 '12 at 22:28








      1




      1




      $begingroup$
      duplicate?
      $endgroup$
      – draks ...
      Mar 29 '12 at 7:55




      $begingroup$
      duplicate?
      $endgroup$
      – draks ...
      Mar 29 '12 at 7:55




      1




      1




      $begingroup$
      I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
      $endgroup$
      – user21436
      Mar 29 '12 at 22:28




      $begingroup$
      I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name.
      $endgroup$
      – user21436
      Mar 29 '12 at 22:28











      3












      $begingroup$

      We need to prove that:
      $$frac{sin(2A)}{1+cos(2A)}=tan(A)$$
      Let's do LHS-RHS to prove it. I will try to make the left side equal the right side.
      $$frac{sin(2A)}{1+cos(2A)}$$
      Using double angle identites for both sine and cosine:
      $$frac{2sin(A)cos(A)}{1+2cos^2(A)-1}$$
      How nice. The $1$ and $-1$ in the denominator cancel out.
      $$frac{2sin(A)cos(A)}{2cos^2(A)}$$
      Cancelling out the $cos(A)$ in the numerator and the denominator yields:
      $$frac{2sin(A)}{2cos(A)}$$
      We can also cancel out the $2$ in the numerator and the denominator.
      $$frac{sin(A)}{cos(A)}$$
      $$=tan(A)$$
      $$text{LHS=RHS}$$
      $$displaystyle boxed{therefore dfrac{sin(2A)}{1+cos(2A)}=tan(A)}$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        We need to prove that:
        $$frac{sin(2A)}{1+cos(2A)}=tan(A)$$
        Let's do LHS-RHS to prove it. I will try to make the left side equal the right side.
        $$frac{sin(2A)}{1+cos(2A)}$$
        Using double angle identites for both sine and cosine:
        $$frac{2sin(A)cos(A)}{1+2cos^2(A)-1}$$
        How nice. The $1$ and $-1$ in the denominator cancel out.
        $$frac{2sin(A)cos(A)}{2cos^2(A)}$$
        Cancelling out the $cos(A)$ in the numerator and the denominator yields:
        $$frac{2sin(A)}{2cos(A)}$$
        We can also cancel out the $2$ in the numerator and the denominator.
        $$frac{sin(A)}{cos(A)}$$
        $$=tan(A)$$
        $$text{LHS=RHS}$$
        $$displaystyle boxed{therefore dfrac{sin(2A)}{1+cos(2A)}=tan(A)}$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          We need to prove that:
          $$frac{sin(2A)}{1+cos(2A)}=tan(A)$$
          Let's do LHS-RHS to prove it. I will try to make the left side equal the right side.
          $$frac{sin(2A)}{1+cos(2A)}$$
          Using double angle identites for both sine and cosine:
          $$frac{2sin(A)cos(A)}{1+2cos^2(A)-1}$$
          How nice. The $1$ and $-1$ in the denominator cancel out.
          $$frac{2sin(A)cos(A)}{2cos^2(A)}$$
          Cancelling out the $cos(A)$ in the numerator and the denominator yields:
          $$frac{2sin(A)}{2cos(A)}$$
          We can also cancel out the $2$ in the numerator and the denominator.
          $$frac{sin(A)}{cos(A)}$$
          $$=tan(A)$$
          $$text{LHS=RHS}$$
          $$displaystyle boxed{therefore dfrac{sin(2A)}{1+cos(2A)}=tan(A)}$$






          share|cite|improve this answer









          $endgroup$



          We need to prove that:
          $$frac{sin(2A)}{1+cos(2A)}=tan(A)$$
          Let's do LHS-RHS to prove it. I will try to make the left side equal the right side.
          $$frac{sin(2A)}{1+cos(2A)}$$
          Using double angle identites for both sine and cosine:
          $$frac{2sin(A)cos(A)}{1+2cos^2(A)-1}$$
          How nice. The $1$ and $-1$ in the denominator cancel out.
          $$frac{2sin(A)cos(A)}{2cos^2(A)}$$
          Cancelling out the $cos(A)$ in the numerator and the denominator yields:
          $$frac{2sin(A)}{2cos(A)}$$
          We can also cancel out the $2$ in the numerator and the denominator.
          $$frac{sin(A)}{cos(A)}$$
          $$=tan(A)$$
          $$text{LHS=RHS}$$
          $$displaystyle boxed{therefore dfrac{sin(2A)}{1+cos(2A)}=tan(A)}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 15 '14 at 3:31









          TrueDefaultTrueDefault

          3,81172950




          3,81172950























              1












              $begingroup$

              First, lets develop a couple of identities.
              Given that $sin 2A = 2sin Acos A$, and $cos 2A = cos^2A - sin^2 A$ we have



              $$begin{array}{lll}
              tan 2A &=& frac{sin 2A}{cos 2A}\
              &=&frac{2sin Acos A}{cos^2 A-sin^2A}\
              &=&frac{2sin Acos A}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
              &=&frac{2tan A}{1-tan^2A}
              end{array}$$
              Similarly, we have
              $$begin{array}{lll}
              sec 2A &=& frac{1}{cos 2A}\
              &=&frac{1}{cos^2 A-sin^2A}\
              &=&frac{1}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
              &=&frac{sec^2 A}{1-tan^2A}
              end{array}$$



              But sometimes it is just as easy to represent these identities as




              $$begin{array}{lll}
              (1-tan^2 A)sec 2A &=& sec^2 A\
              (1-tan^2 A)tan 2A &=& 2tan A
              end{array}$$


              Applying these identities to the problem at hand we have
              $$begin{array}{lll}
              frac{sin 2A}{1+cos 2A}&=& frac{sin 2A}{1+cos 2A}cdotfrac{frac{1}{cos 2A}}{frac{1}{cos 2A}}\
              &=& frac{tan 2A}{sec 2A +1}\
              &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)(sec 2A +1)}\
              &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)sec 2A +(1-tan^2 A)}\
              &=& frac{2tan A}{sec^2 A +(1-tan^2 A)}\
              &=& frac{2tan A}{(tan^2 A+1) +(1-tan^2 A)}\
              &=& frac{2tan A}{2}\
              &=& tan A\
              end{array}$$
              discalimer: I like JChau's answer better :)






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                First, lets develop a couple of identities.
                Given that $sin 2A = 2sin Acos A$, and $cos 2A = cos^2A - sin^2 A$ we have



                $$begin{array}{lll}
                tan 2A &=& frac{sin 2A}{cos 2A}\
                &=&frac{2sin Acos A}{cos^2 A-sin^2A}\
                &=&frac{2sin Acos A}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
                &=&frac{2tan A}{1-tan^2A}
                end{array}$$
                Similarly, we have
                $$begin{array}{lll}
                sec 2A &=& frac{1}{cos 2A}\
                &=&frac{1}{cos^2 A-sin^2A}\
                &=&frac{1}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
                &=&frac{sec^2 A}{1-tan^2A}
                end{array}$$



                But sometimes it is just as easy to represent these identities as




                $$begin{array}{lll}
                (1-tan^2 A)sec 2A &=& sec^2 A\
                (1-tan^2 A)tan 2A &=& 2tan A
                end{array}$$


                Applying these identities to the problem at hand we have
                $$begin{array}{lll}
                frac{sin 2A}{1+cos 2A}&=& frac{sin 2A}{1+cos 2A}cdotfrac{frac{1}{cos 2A}}{frac{1}{cos 2A}}\
                &=& frac{tan 2A}{sec 2A +1}\
                &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)(sec 2A +1)}\
                &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)sec 2A +(1-tan^2 A)}\
                &=& frac{2tan A}{sec^2 A +(1-tan^2 A)}\
                &=& frac{2tan A}{(tan^2 A+1) +(1-tan^2 A)}\
                &=& frac{2tan A}{2}\
                &=& tan A\
                end{array}$$
                discalimer: I like JChau's answer better :)






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  First, lets develop a couple of identities.
                  Given that $sin 2A = 2sin Acos A$, and $cos 2A = cos^2A - sin^2 A$ we have



                  $$begin{array}{lll}
                  tan 2A &=& frac{sin 2A}{cos 2A}\
                  &=&frac{2sin Acos A}{cos^2 A-sin^2A}\
                  &=&frac{2sin Acos A}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
                  &=&frac{2tan A}{1-tan^2A}
                  end{array}$$
                  Similarly, we have
                  $$begin{array}{lll}
                  sec 2A &=& frac{1}{cos 2A}\
                  &=&frac{1}{cos^2 A-sin^2A}\
                  &=&frac{1}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
                  &=&frac{sec^2 A}{1-tan^2A}
                  end{array}$$



                  But sometimes it is just as easy to represent these identities as




                  $$begin{array}{lll}
                  (1-tan^2 A)sec 2A &=& sec^2 A\
                  (1-tan^2 A)tan 2A &=& 2tan A
                  end{array}$$


                  Applying these identities to the problem at hand we have
                  $$begin{array}{lll}
                  frac{sin 2A}{1+cos 2A}&=& frac{sin 2A}{1+cos 2A}cdotfrac{frac{1}{cos 2A}}{frac{1}{cos 2A}}\
                  &=& frac{tan 2A}{sec 2A +1}\
                  &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)(sec 2A +1)}\
                  &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)sec 2A +(1-tan^2 A)}\
                  &=& frac{2tan A}{sec^2 A +(1-tan^2 A)}\
                  &=& frac{2tan A}{(tan^2 A+1) +(1-tan^2 A)}\
                  &=& frac{2tan A}{2}\
                  &=& tan A\
                  end{array}$$
                  discalimer: I like JChau's answer better :)






                  share|cite|improve this answer









                  $endgroup$



                  First, lets develop a couple of identities.
                  Given that $sin 2A = 2sin Acos A$, and $cos 2A = cos^2A - sin^2 A$ we have



                  $$begin{array}{lll}
                  tan 2A &=& frac{sin 2A}{cos 2A}\
                  &=&frac{2sin Acos A}{cos^2 A-sin^2A}\
                  &=&frac{2sin Acos A}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
                  &=&frac{2tan A}{1-tan^2A}
                  end{array}$$
                  Similarly, we have
                  $$begin{array}{lll}
                  sec 2A &=& frac{1}{cos 2A}\
                  &=&frac{1}{cos^2 A-sin^2A}\
                  &=&frac{1}{cos^2 A-sin^2A}cdotfrac{frac{1}{cos^2 A}}{frac{1}{cos^2 A}}\
                  &=&frac{sec^2 A}{1-tan^2A}
                  end{array}$$



                  But sometimes it is just as easy to represent these identities as




                  $$begin{array}{lll}
                  (1-tan^2 A)sec 2A &=& sec^2 A\
                  (1-tan^2 A)tan 2A &=& 2tan A
                  end{array}$$


                  Applying these identities to the problem at hand we have
                  $$begin{array}{lll}
                  frac{sin 2A}{1+cos 2A}&=& frac{sin 2A}{1+cos 2A}cdotfrac{frac{1}{cos 2A}}{frac{1}{cos 2A}}\
                  &=& frac{tan 2A}{sec 2A +1}\
                  &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)(sec 2A +1)}\
                  &=& frac{(1-tan^2 A)tan 2A}{(1-tan^2 A)sec 2A +(1-tan^2 A)}\
                  &=& frac{2tan A}{sec^2 A +(1-tan^2 A)}\
                  &=& frac{2tan A}{(tan^2 A+1) +(1-tan^2 A)}\
                  &=& frac{2tan A}{2}\
                  &=& tan A\
                  end{array}$$
                  discalimer: I like JChau's answer better :)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 27 '14 at 14:27









                  John JoyJohn Joy

                  6,31811727




                  6,31811727






























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