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












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











$endgroup$



closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 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
    Mar 12 at 16:57






  • 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
    Mar 12 at 16:58












  • $begingroup$
    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
    Mar 12 at 17:00












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    Mar 12 at 17:03










  • $begingroup$
    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
    Mar 12 at 17:07
















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











$endgroup$



closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 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
    Mar 12 at 16:57






  • 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
    Mar 12 at 16:58












  • $begingroup$
    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
    Mar 12 at 17:00












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    Mar 12 at 17:03










  • $begingroup$
    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
    Mar 12 at 17:07














1












1








1


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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 12 at 22:33









Blue

49.1k870156




49.1k870156










asked Mar 12 at 16:32









Nour GNour G

164




164




closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 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
    Mar 12 at 16:57






  • 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
    Mar 12 at 16:58












  • $begingroup$
    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
    Mar 12 at 17:00












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    Mar 12 at 17:03










  • $begingroup$
    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
    Mar 12 at 17:07














  • 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
    Mar 12 at 16:57






  • 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
    Mar 12 at 16:58












  • $begingroup$
    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
    Mar 12 at 17:00












  • $begingroup$
    Does your question insist on the identity being used ?
    $endgroup$
    – Martin Hansen
    Mar 12 at 17:03










  • $begingroup$
    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
    Mar 12 at 17:07








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
Mar 12 at 16:57




$begingroup$
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
Mar 12 at 16:57




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
Mar 12 at 16:58






$begingroup$
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
Mar 12 at 16:58














$begingroup$
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
Mar 12 at 17:00






$begingroup$
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
Mar 12 at 17:00














$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03




$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03












$begingroup$
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
Mar 12 at 17:07




$begingroup$
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
Mar 12 at 17:07










5 Answers
5






active

oldest

votes


















2












$begingroup$

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
    Mar 12 at 17:52










  • $begingroup$
    @Teepeemm Thanks for the suggested edit
    $endgroup$
    – Martin Hansen
    Mar 12 at 22:06



















6












$begingroup$

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
    Mar 12 at 17:37








  • 1




    $begingroup$
    Thank you so much! Makes so much sense now.
    $endgroup$
    – Nour G
    Mar 12 at 17:53



















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











$endgroup$





















    1












    $begingroup$

    $alpha+beta=90^circ$
    $implies2alpha+2beta=180^circ$
    $implies2beta=180 ^circ-2alpha$



    Now
    $$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

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






      share|cite|improve this answer









      $endgroup$




















        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        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
          Mar 12 at 17:52










        • $begingroup$
          @Teepeemm Thanks for the suggested edit
          $endgroup$
          – Martin Hansen
          Mar 12 at 22:06
















        2












        $begingroup$

        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
          Mar 12 at 17:52










        • $begingroup$
          @Teepeemm Thanks for the suggested edit
          $endgroup$
          – Martin Hansen
          Mar 12 at 22:06














        2












        2








        2





        $begingroup$

        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 Mar 12 at 22:05









        Teepeemm

        69459




        69459










        answered Mar 12 at 17:28









        Martin HansenMartin Hansen

        670114




        670114








        • 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
          Mar 12 at 17:52










        • $begingroup$
          @Teepeemm Thanks for the suggested edit
          $endgroup$
          – Martin Hansen
          Mar 12 at 22:06














        • 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
          Mar 12 at 17:52










        • $begingroup$
          @Teepeemm Thanks for the suggested edit
          $endgroup$
          – Martin Hansen
          Mar 12 at 22:06








        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
        Mar 12 at 17:52




        $begingroup$
        Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
        $endgroup$
        – Nour G
        Mar 12 at 17:52












        $begingroup$
        @Teepeemm Thanks for the suggested edit
        $endgroup$
        – Martin Hansen
        Mar 12 at 22:06




        $begingroup$
        @Teepeemm Thanks for the suggested edit
        $endgroup$
        – Martin Hansen
        Mar 12 at 22:06











        6












        $begingroup$

        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
          Mar 12 at 17:37








        • 1




          $begingroup$
          Thank you so much! Makes so much sense now.
          $endgroup$
          – Nour G
          Mar 12 at 17:53
















        6












        $begingroup$

        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
          Mar 12 at 17:37








        • 1




          $begingroup$
          Thank you so much! Makes so much sense now.
          $endgroup$
          – Nour G
          Mar 12 at 17:53














        6












        6








        6





        $begingroup$

        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 Mar 12 at 17:25









        Yves DaoustYves Daoust

        131k676229




        131k676229








        • 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
          Mar 12 at 17:37








        • 1




          $begingroup$
          Thank you so much! Makes so much sense now.
          $endgroup$
          – Nour G
          Mar 12 at 17:53














        • 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
          Mar 12 at 17:37








        • 1




          $begingroup$
          Thank you so much! Makes so much sense now.
          $endgroup$
          – Nour G
          Mar 12 at 17:53








        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
        Mar 12 at 17:37






        $begingroup$
        Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
        $endgroup$
        – Martin Hansen
        Mar 12 at 17:37






        1




        1




        $begingroup$
        Thank you so much! Makes so much sense now.
        $endgroup$
        – Nour G
        Mar 12 at 17:53




        $begingroup$
        Thank you so much! Makes so much sense now.
        $endgroup$
        – Nour G
        Mar 12 at 17:53











        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 Mar 12 at 17:17

























            answered Mar 12 at 17:01









            VasyaVasya

            4,1081618




            4,1081618























                1












                $begingroup$

                $alpha+beta=90^circ$
                $implies2alpha+2beta=180^circ$
                $implies2beta=180 ^circ-2alpha$



                Now
                $$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  $alpha+beta=90^circ$
                  $implies2alpha+2beta=180^circ$
                  $implies2beta=180 ^circ-2alpha$



                  Now
                  $$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    $alpha+beta=90^circ$
                    $implies2alpha+2beta=180^circ$
                    $implies2beta=180 ^circ-2alpha$



                    Now
                    $$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$






                    share|cite|improve this answer











                    $endgroup$



                    $alpha+beta=90^circ$
                    $implies2alpha+2beta=180^circ$
                    $implies2beta=180 ^circ-2alpha$



                    Now
                    $$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 13 at 3:46

























                    answered Mar 13 at 3:34









                    saket kumarsaket kumar

                    143113




                    143113























                        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 Mar 12 at 17:31









                            Maria MazurMaria Mazur

                            47.9k1260120




                            47.9k1260120















                                -

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