How to prove $frac{tan (A)}{tan (A)}+frac{cot (A)}{cot (A)}=frac{1}{1-2cos(A)^2}$












1












$begingroup$


I am unable to prove this trigonometric identity



$$frac{tan (A)}{tan (A)}+frac{cot (A)}{cot (A)}=frac{1}{1-2cos^2(A)}$$



I have tried to transform the left-hand side and stuck with this



$$frac{2sin(A)cos(A)}{sin(A)cos(A)}$$



And I have tried to transform the right-hand side by changing the $$2cos^2(A)$$ to $$frac{2}{sec^2(A)}$$, and used the trigonometric identity $$1+tan^2(A)=sec^2(A)$$ and got this instead



$$frac{1+tan^2(A)}{tan^2(A)-1}$$ which I can transform to $$frac{cot(A)+tan(A)}{tan(A)-cot(A)}$$.



I cannot get both sides equal, help please?






trigonometry







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 1:09












  • $begingroup$
    @JohnOmielan but that's exactly how the question on textbook is written?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:11






  • 3




    $begingroup$
    Then your textbook must have a typo, because this identity clearly can not be proven true.
    $endgroup$
    – Noble Mushtak
    Dec 30 '18 at 1:12










  • $begingroup$
    @JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:14










  • $begingroup$
    @NobleMushtak I see, thank you for the answer. That must be the case, the book is so old
    $endgroup$
    – user569622
    Dec 30 '18 at 1:15
















1












$begingroup$


I am unable to prove this trigonometric identity



$$frac{tan (A)}{tan (A)}+frac{cot (A)}{cot (A)}=frac{1}{1-2cos^2(A)}$$



I have tried to transform the left-hand side and stuck with this



$$frac{2sin(A)cos(A)}{sin(A)cos(A)}$$



And I have tried to transform the right-hand side by changing the $$2cos^2(A)$$ to $$frac{2}{sec^2(A)}$$, and used the trigonometric identity $$1+tan^2(A)=sec^2(A)$$ and got this instead



$$frac{1+tan^2(A)}{tan^2(A)-1}$$ which I can transform to $$frac{cot(A)+tan(A)}{tan(A)-cot(A)}$$.



I cannot get both sides equal, help please?






trigonometry







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 1:09












  • $begingroup$
    @JohnOmielan but that's exactly how the question on textbook is written?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:11






  • 3




    $begingroup$
    Then your textbook must have a typo, because this identity clearly can not be proven true.
    $endgroup$
    – Noble Mushtak
    Dec 30 '18 at 1:12










  • $begingroup$
    @JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:14










  • $begingroup$
    @NobleMushtak I see, thank you for the answer. That must be the case, the book is so old
    $endgroup$
    – user569622
    Dec 30 '18 at 1:15














1












1








1





$begingroup$


I am unable to prove this trigonometric identity



$$frac{tan (A)}{tan (A)}+frac{cot (A)}{cot (A)}=frac{1}{1-2cos^2(A)}$$



I have tried to transform the left-hand side and stuck with this



$$frac{2sin(A)cos(A)}{sin(A)cos(A)}$$



And I have tried to transform the right-hand side by changing the $$2cos^2(A)$$ to $$frac{2}{sec^2(A)}$$, and used the trigonometric identity $$1+tan^2(A)=sec^2(A)$$ and got this instead



$$frac{1+tan^2(A)}{tan^2(A)-1}$$ which I can transform to $$frac{cot(A)+tan(A)}{tan(A)-cot(A)}$$.



I cannot get both sides equal, help please?






trigonometry







share|cite|improve this question











$endgroup$




I am unable to prove this trigonometric identity



$$frac{tan (A)}{tan (A)}+frac{cot (A)}{cot (A)}=frac{1}{1-2cos^2(A)}$$



I have tried to transform the left-hand side and stuck with this



$$frac{2sin(A)cos(A)}{sin(A)cos(A)}$$



And I have tried to transform the right-hand side by changing the $$2cos^2(A)$$ to $$frac{2}{sec^2(A)}$$, and used the trigonometric identity $$1+tan^2(A)=sec^2(A)$$ and got this instead



$$frac{1+tan^2(A)}{tan^2(A)-1}$$ which I can transform to $$frac{cot(A)+tan(A)}{tan(A)-cot(A)}$$.



I cannot get both sides equal, help please?






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 2:05









bjcolby15

1,52111016




1,52111016










asked Dec 30 '18 at 1:07









user569622user569622

345




345








  • 7




    $begingroup$
    I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 1:09












  • $begingroup$
    @JohnOmielan but that's exactly how the question on textbook is written?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:11






  • 3




    $begingroup$
    Then your textbook must have a typo, because this identity clearly can not be proven true.
    $endgroup$
    – Noble Mushtak
    Dec 30 '18 at 1:12










  • $begingroup$
    @JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:14










  • $begingroup$
    @NobleMushtak I see, thank you for the answer. That must be the case, the book is so old
    $endgroup$
    – user569622
    Dec 30 '18 at 1:15














  • 7




    $begingroup$
    I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 1:09












  • $begingroup$
    @JohnOmielan but that's exactly how the question on textbook is written?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:11






  • 3




    $begingroup$
    Then your textbook must have a typo, because this identity clearly can not be proven true.
    $endgroup$
    – Noble Mushtak
    Dec 30 '18 at 1:12










  • $begingroup$
    @JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven?
    $endgroup$
    – user569622
    Dec 30 '18 at 1:14










  • $begingroup$
    @NobleMushtak I see, thank you for the answer. That must be the case, the book is so old
    $endgroup$
    – user569622
    Dec 30 '18 at 1:15








7




7




$begingroup$
I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant.
$endgroup$
– John Omielan
Dec 30 '18 at 1:09






$begingroup$
I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant.
$endgroup$
– John Omielan
Dec 30 '18 at 1:09














$begingroup$
@JohnOmielan but that's exactly how the question on textbook is written?
$endgroup$
– user569622
Dec 30 '18 at 1:11




$begingroup$
@JohnOmielan but that's exactly how the question on textbook is written?
$endgroup$
– user569622
Dec 30 '18 at 1:11




3




3




$begingroup$
Then your textbook must have a typo, because this identity clearly can not be proven true.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:12




$begingroup$
Then your textbook must have a typo, because this identity clearly can not be proven true.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:12












$begingroup$
@JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven?
$endgroup$
– user569622
Dec 30 '18 at 1:14




$begingroup$
@JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven?
$endgroup$
– user569622
Dec 30 '18 at 1:14












$begingroup$
@NobleMushtak I see, thank you for the answer. That must be the case, the book is so old
$endgroup$
– user569622
Dec 30 '18 at 1:15




$begingroup$
@NobleMushtak I see, thank you for the answer. That must be the case, the book is so old
$endgroup$
– user569622
Dec 30 '18 at 1:15










1 Answer
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1












$begingroup$

One way we can prove the identity false is as follows:



$$begin {align}
dfrac {tan A} {tan A} + dfrac{cot A}{cot A} = dfrac {1}{1-2cos^2 2A} \
2 = dfrac {1}{1-2cos^2 2A} \
2 (1-2cos^2 2A) = 1 \
2 - 4cos^2 2A = 1 \
- 4cos^2 2A = dfrac {1}{2} \
cos^2 2A = -dfrac {1}{8}
end {align}$$



Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    One way we can prove the identity false is as follows:



    $$begin {align}
    dfrac {tan A} {tan A} + dfrac{cot A}{cot A} = dfrac {1}{1-2cos^2 2A} \
    2 = dfrac {1}{1-2cos^2 2A} \
    2 (1-2cos^2 2A) = 1 \
    2 - 4cos^2 2A = 1 \
    - 4cos^2 2A = dfrac {1}{2} \
    cos^2 2A = -dfrac {1}{8}
    end {align}$$



    Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      One way we can prove the identity false is as follows:



      $$begin {align}
      dfrac {tan A} {tan A} + dfrac{cot A}{cot A} = dfrac {1}{1-2cos^2 2A} \
      2 = dfrac {1}{1-2cos^2 2A} \
      2 (1-2cos^2 2A) = 1 \
      2 - 4cos^2 2A = 1 \
      - 4cos^2 2A = dfrac {1}{2} \
      cos^2 2A = -dfrac {1}{8}
      end {align}$$



      Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        One way we can prove the identity false is as follows:



        $$begin {align}
        dfrac {tan A} {tan A} + dfrac{cot A}{cot A} = dfrac {1}{1-2cos^2 2A} \
        2 = dfrac {1}{1-2cos^2 2A} \
        2 (1-2cos^2 2A) = 1 \
        2 - 4cos^2 2A = 1 \
        - 4cos^2 2A = dfrac {1}{2} \
        cos^2 2A = -dfrac {1}{8}
        end {align}$$



        Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.






        share|cite|improve this answer











        $endgroup$



        One way we can prove the identity false is as follows:



        $$begin {align}
        dfrac {tan A} {tan A} + dfrac{cot A}{cot A} = dfrac {1}{1-2cos^2 2A} \
        2 = dfrac {1}{1-2cos^2 2A} \
        2 (1-2cos^2 2A) = 1 \
        2 - 4cos^2 2A = 1 \
        - 4cos^2 2A = dfrac {1}{2} \
        cos^2 2A = -dfrac {1}{8}
        end {align}$$



        Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 21:22

























        answered Dec 30 '18 at 18:15









        bjcolby15bjcolby15

        1,52111016




        1,52111016






























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