If $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$












2












$begingroup$


Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$



Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.






trigonometry







share|cite|improve this question









New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 3




    $begingroup$
    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
    $endgroup$
    – saulspatz
    6 hours ago






  • 1




    $begingroup$
    You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
    $endgroup$
    – Martin Hansen
    6 hours ago












  • $begingroup$
    You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
    $endgroup$
    – Barry Cipra
    6 hours ago












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    6 hours ago










  • $begingroup$
    Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
    $endgroup$
    – Nour G
    6 hours ago
















2












$begingroup$


Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$



Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.






trigonometry







share|cite|improve this question









New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
    $endgroup$
    – saulspatz
    6 hours ago






  • 1




    $begingroup$
    You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
    $endgroup$
    – Martin Hansen
    6 hours ago












  • $begingroup$
    You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
    $endgroup$
    – Barry Cipra
    6 hours ago












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    6 hours ago










  • $begingroup$
    Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
    $endgroup$
    – Nour G
    6 hours ago














2












2








2


1



$begingroup$


Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$



Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.






trigonometry







share|cite|improve this question









New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$



Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.






trigonometry




trigonometry






share|cite|improve this question









New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Blue

49k870156




49k870156






New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









Nour GNour G

163




163




New contributor




Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Nour G is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    $begingroup$
    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
    $endgroup$
    – saulspatz
    6 hours ago






  • 1




    $begingroup$
    You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
    $endgroup$
    – Martin Hansen
    6 hours ago












  • $begingroup$
    You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
    $endgroup$
    – Barry Cipra
    6 hours ago












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    6 hours ago










  • $begingroup$
    Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
    $endgroup$
    – Nour G
    6 hours ago














  • 3




    $begingroup$
    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
    $endgroup$
    – saulspatz
    6 hours ago






  • 1




    $begingroup$
    You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
    $endgroup$
    – Martin Hansen
    6 hours ago












  • $begingroup$
    You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
    $endgroup$
    – Barry Cipra
    6 hours ago












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    6 hours ago










  • $begingroup$
    Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
    $endgroup$
    – Nour G
    6 hours ago








3




3




$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
6 hours ago




$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
6 hours ago




1




1




$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
6 hours ago






$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
6 hours ago














$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
6 hours ago






$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
6 hours ago














$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
6 hours ago




$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
6 hours ago












$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
6 hours ago




$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
6 hours ago










4 Answers
4






active

oldest

votes


















1












$begingroup$

Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
and
$$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
Thus
$$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$



Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$



Observe that
$$tan alpha=frac{x}{y}$$
and
$$tan beta=frac{y}{x}$$
Thus
$$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
$$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
$$=-1$$
$$=RHS$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
    $endgroup$
    – Nour G
    6 hours ago










  • $begingroup$
    @Teepeemm Thanks for the suggested edit
    $endgroup$
    – Martin Hansen
    1 hour ago



















4












$begingroup$

If



$$2alpha+2beta=180°$$



then



$$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
    $endgroup$
    – Martin Hansen
    6 hours ago








  • 1




    $begingroup$
    Thank you so much! Makes so much sense now.
    $endgroup$
    – Nour G
    6 hours ago



















2












$begingroup$


Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.


Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$



Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)







share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Assuming that the question is insisting on the $tan(A+B)$ identity being used;
      $$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
      and
      $$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
      Thus
      $$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$



      Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$



      Observe that
      $$tan alpha=frac{x}{y}$$
      and
      $$tan beta=frac{y}{x}$$
      Thus
      $$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
      Multiply both the numerator and the denominator by $x^2y^2$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
      $$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
      $$=-1$$
      $$=RHS$$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
        $endgroup$
        – Nour G
        6 hours ago










      • $begingroup$
        @Teepeemm Thanks for the suggested edit
        $endgroup$
        – Martin Hansen
        1 hour ago
















      1












      $begingroup$

      Assuming that the question is insisting on the $tan(A+B)$ identity being used;
      $$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
      and
      $$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
      Thus
      $$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$



      Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$



      Observe that
      $$tan alpha=frac{x}{y}$$
      and
      $$tan beta=frac{y}{x}$$
      Thus
      $$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
      Multiply both the numerator and the denominator by $x^2y^2$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
      $$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
      $$=-1$$
      $$=RHS$$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
        $endgroup$
        – Nour G
        6 hours ago










      • $begingroup$
        @Teepeemm Thanks for the suggested edit
        $endgroup$
        – Martin Hansen
        1 hour ago














      1












      1








      1





      $begingroup$

      Assuming that the question is insisting on the $tan(A+B)$ identity being used;
      $$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
      and
      $$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
      Thus
      $$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$



      Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$



      Observe that
      $$tan alpha=frac{x}{y}$$
      and
      $$tan beta=frac{y}{x}$$
      Thus
      $$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
      Multiply both the numerator and the denominator by $x^2y^2$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
      $$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
      $$=-1$$
      $$=RHS$$






      share|cite|improve this answer











      $endgroup$



      Assuming that the question is insisting on the $tan(A+B)$ identity being used;
      $$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
      and
      $$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
      Thus
      $$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$



      Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$



      Observe that
      $$tan alpha=frac{x}{y}$$
      and
      $$tan beta=frac{y}{x}$$
      Thus
      $$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
      Multiply both the numerator and the denominator by $x^2y^2$
      $$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
      $$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
      $$=-1$$
      $$=RHS$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 1 hour ago









      Teepeemm

      69459




      69459










      answered 6 hours ago









      Martin HansenMartin Hansen

      27713




      27713








      • 1




        $begingroup$
        Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
        $endgroup$
        – Nour G
        6 hours ago










      • $begingroup$
        @Teepeemm Thanks for the suggested edit
        $endgroup$
        – Martin Hansen
        1 hour ago














      • 1




        $begingroup$
        Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
        $endgroup$
        – Nour G
        6 hours ago










      • $begingroup$
        @Teepeemm Thanks for the suggested edit
        $endgroup$
        – Martin Hansen
        1 hour ago








      1




      1




      $begingroup$
      Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
      $endgroup$
      – Nour G
      6 hours ago




      $begingroup$
      Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
      $endgroup$
      – Nour G
      6 hours ago












      $begingroup$
      @Teepeemm Thanks for the suggested edit
      $endgroup$
      – Martin Hansen
      1 hour ago




      $begingroup$
      @Teepeemm Thanks for the suggested edit
      $endgroup$
      – Martin Hansen
      1 hour ago











      4












      $begingroup$

      If



      $$2alpha+2beta=180°$$



      then



      $$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
        $endgroup$
        – Martin Hansen
        6 hours ago








      • 1




        $begingroup$
        Thank you so much! Makes so much sense now.
        $endgroup$
        – Nour G
        6 hours ago
















      4












      $begingroup$

      If



      $$2alpha+2beta=180°$$



      then



      $$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
        $endgroup$
        – Martin Hansen
        6 hours ago








      • 1




        $begingroup$
        Thank you so much! Makes so much sense now.
        $endgroup$
        – Nour G
        6 hours ago














      4












      4








      4





      $begingroup$

      If



      $$2alpha+2beta=180°$$



      then



      $$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.






      share|cite|improve this answer









      $endgroup$



      If



      $$2alpha+2beta=180°$$



      then



      $$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 6 hours ago









      Yves DaoustYves Daoust

      130k676227




      130k676227








      • 1




        $begingroup$
        Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
        $endgroup$
        – Martin Hansen
        6 hours ago








      • 1




        $begingroup$
        Thank you so much! Makes so much sense now.
        $endgroup$
        – Nour G
        6 hours ago














      • 1




        $begingroup$
        Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
        $endgroup$
        – Martin Hansen
        6 hours ago








      • 1




        $begingroup$
        Thank you so much! Makes so much sense now.
        $endgroup$
        – Nour G
        6 hours ago








      1




      1




      $begingroup$
      Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
      $endgroup$
      – Martin Hansen
      6 hours ago






      $begingroup$
      Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
      $endgroup$
      – Martin Hansen
      6 hours ago






      1




      1




      $begingroup$
      Thank you so much! Makes so much sense now.
      $endgroup$
      – Nour G
      6 hours ago




      $begingroup$
      Thank you so much! Makes so much sense now.
      $endgroup$
      – Nour G
      6 hours ago











      2












      $begingroup$


      Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.


      Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$



      Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)







      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$


        Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.


        Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$



        Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)







        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$


          Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.


          Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$



          Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)







          share|cite|improve this answer











          $endgroup$




          Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.


          Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$



          Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 hours ago

























          answered 6 hours ago









          VasyaVasya

          4,0881618




          4,0881618























              0












              $begingroup$

              $$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$






                  share|cite|improve this answer









                  $endgroup$



                  $$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  Maria MazurMaria Mazur

                  46.6k1260119




                  46.6k1260119






















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