How can you solve and equation with inverse functions?












2












$begingroup$


$$arctan(x) - arctan(2/x) = arctan(7/9)$$ where $x$ is positive .



The answer should be 3. Thanks






trigonometry







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$endgroup$












  • $begingroup$
    Welcome to MSE! Please show your working, including where you are stuck. In addition, it is always better to format your questions with MathJax.
    $endgroup$
    – KM101
    Jan 28 at 17:28
















2












$begingroup$


$$arctan(x) - arctan(2/x) = arctan(7/9)$$ where $x$ is positive .



The answer should be 3. Thanks






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE! Please show your working, including where you are stuck. In addition, it is always better to format your questions with MathJax.
    $endgroup$
    – KM101
    Jan 28 at 17:28














2












2








2





$begingroup$


$$arctan(x) - arctan(2/x) = arctan(7/9)$$ where $x$ is positive .



The answer should be 3. Thanks






trigonometry







share|cite|improve this question











$endgroup$




$$arctan(x) - arctan(2/x) = arctan(7/9)$$ where $x$ is positive .



The answer should be 3. Thanks






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 17:27









Adrian Keister

5,28171933




5,28171933










asked Jan 28 at 17:25









Doris VellaDoris Vella

133




133












  • $begingroup$
    Welcome to MSE! Please show your working, including where you are stuck. In addition, it is always better to format your questions with MathJax.
    $endgroup$
    – KM101
    Jan 28 at 17:28


















  • $begingroup$
    Welcome to MSE! Please show your working, including where you are stuck. In addition, it is always better to format your questions with MathJax.
    $endgroup$
    – KM101
    Jan 28 at 17:28
















$begingroup$
Welcome to MSE! Please show your working, including where you are stuck. In addition, it is always better to format your questions with MathJax.
$endgroup$
– KM101
Jan 28 at 17:28




$begingroup$
Welcome to MSE! Please show your working, including where you are stuck. In addition, it is always better to format your questions with MathJax.
$endgroup$
– KM101
Jan 28 at 17:28










2 Answers
2






active

oldest

votes


















2












$begingroup$

Using
$$tan(A-B)=frac{tan A-tan B}{1+tan Atan B}$$
we get the equation
$$frac{x-2/x}{1+2}=frac79$$
which looks like a nice quadratic equation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 28 at 17:33










  • $begingroup$
    Take the tangent of both sides.
    $endgroup$
    – KM101
    Jan 28 at 17:35










  • $begingroup$
    @KM101 thank you for the reply.
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 28 at 17:38










  • $begingroup$
    Thank you sooo much!
    $endgroup$
    – Doris Vella
    Jan 28 at 17:41



















0












$begingroup$

Using Inverse trigonometric function identity doubt: $tan^{-1}x+tan^{-1}y =-pi+tan^{-1}left(frac{x+y}{1-xy}right)$, when $x<0$, $y<0$, and $xy>1$,



$$arctandfrac2x+arctandfrac79=arctandfrac{18+7x}{9x-14}$$ for $dfrac{2cdot7}{xcdot9}<1$ which happens if $x<0$ or if $x>dfrac{14}9$



What if $dfrac{14}{9x}ge1?$






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Using
    $$tan(A-B)=frac{tan A-tan B}{1+tan Atan B}$$
    we get the equation
    $$frac{x-2/x}{1+2}=frac79$$
    which looks like a nice quadratic equation.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:33










    • $begingroup$
      Take the tangent of both sides.
      $endgroup$
      – KM101
      Jan 28 at 17:35










    • $begingroup$
      @KM101 thank you for the reply.
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:38










    • $begingroup$
      Thank you sooo much!
      $endgroup$
      – Doris Vella
      Jan 28 at 17:41
















    2












    $begingroup$

    Using
    $$tan(A-B)=frac{tan A-tan B}{1+tan Atan B}$$
    we get the equation
    $$frac{x-2/x}{1+2}=frac79$$
    which looks like a nice quadratic equation.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:33










    • $begingroup$
      Take the tangent of both sides.
      $endgroup$
      – KM101
      Jan 28 at 17:35










    • $begingroup$
      @KM101 thank you for the reply.
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:38










    • $begingroup$
      Thank you sooo much!
      $endgroup$
      – Doris Vella
      Jan 28 at 17:41














    2












    2








    2





    $begingroup$

    Using
    $$tan(A-B)=frac{tan A-tan B}{1+tan Atan B}$$
    we get the equation
    $$frac{x-2/x}{1+2}=frac79$$
    which looks like a nice quadratic equation.






    share|cite|improve this answer









    $endgroup$



    Using
    $$tan(A-B)=frac{tan A-tan B}{1+tan Atan B}$$
    we get the equation
    $$frac{x-2/x}{1+2}=frac79$$
    which looks like a nice quadratic equation.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 28 at 17:29









    Lord Shark the UnknownLord Shark the Unknown

    107k1162135




    107k1162135












    • $begingroup$
      I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:33










    • $begingroup$
      Take the tangent of both sides.
      $endgroup$
      – KM101
      Jan 28 at 17:35










    • $begingroup$
      @KM101 thank you for the reply.
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:38










    • $begingroup$
      Thank you sooo much!
      $endgroup$
      – Doris Vella
      Jan 28 at 17:41


















    • $begingroup$
      I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:33










    • $begingroup$
      Take the tangent of both sides.
      $endgroup$
      – KM101
      Jan 28 at 17:35










    • $begingroup$
      @KM101 thank you for the reply.
      $endgroup$
      – Mohammad Zuhair Khan
      Jan 28 at 17:38










    • $begingroup$
      Thank you sooo much!
      $endgroup$
      – Doris Vella
      Jan 28 at 17:41
















    $begingroup$
    I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 28 at 17:33




    $begingroup$
    I apologise for my lack of knowledge but could you elaborate on how you got that $tan (A-B)=frac 79, tan A=x$ and $tan B=frac 2x$?
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 28 at 17:33












    $begingroup$
    Take the tangent of both sides.
    $endgroup$
    – KM101
    Jan 28 at 17:35




    $begingroup$
    Take the tangent of both sides.
    $endgroup$
    – KM101
    Jan 28 at 17:35












    $begingroup$
    @KM101 thank you for the reply.
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 28 at 17:38




    $begingroup$
    @KM101 thank you for the reply.
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 28 at 17:38












    $begingroup$
    Thank you sooo much!
    $endgroup$
    – Doris Vella
    Jan 28 at 17:41




    $begingroup$
    Thank you sooo much!
    $endgroup$
    – Doris Vella
    Jan 28 at 17:41











    0












    $begingroup$

    Using Inverse trigonometric function identity doubt: $tan^{-1}x+tan^{-1}y =-pi+tan^{-1}left(frac{x+y}{1-xy}right)$, when $x<0$, $y<0$, and $xy>1$,



    $$arctandfrac2x+arctandfrac79=arctandfrac{18+7x}{9x-14}$$ for $dfrac{2cdot7}{xcdot9}<1$ which happens if $x<0$ or if $x>dfrac{14}9$



    What if $dfrac{14}{9x}ge1?$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Using Inverse trigonometric function identity doubt: $tan^{-1}x+tan^{-1}y =-pi+tan^{-1}left(frac{x+y}{1-xy}right)$, when $x<0$, $y<0$, and $xy>1$,



      $$arctandfrac2x+arctandfrac79=arctandfrac{18+7x}{9x-14}$$ for $dfrac{2cdot7}{xcdot9}<1$ which happens if $x<0$ or if $x>dfrac{14}9$



      What if $dfrac{14}{9x}ge1?$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Using Inverse trigonometric function identity doubt: $tan^{-1}x+tan^{-1}y =-pi+tan^{-1}left(frac{x+y}{1-xy}right)$, when $x<0$, $y<0$, and $xy>1$,



        $$arctandfrac2x+arctandfrac79=arctandfrac{18+7x}{9x-14}$$ for $dfrac{2cdot7}{xcdot9}<1$ which happens if $x<0$ or if $x>dfrac{14}9$



        What if $dfrac{14}{9x}ge1?$






        share|cite|improve this answer









        $endgroup$



        Using Inverse trigonometric function identity doubt: $tan^{-1}x+tan^{-1}y =-pi+tan^{-1}left(frac{x+y}{1-xy}right)$, when $x<0$, $y<0$, and $xy>1$,



        $$arctandfrac2x+arctandfrac79=arctandfrac{18+7x}{9x-14}$$ for $dfrac{2cdot7}{xcdot9}<1$ which happens if $x<0$ or if $x>dfrac{14}9$



        What if $dfrac{14}{9x}ge1?$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 17:59









        lab bhattacharjeelab bhattacharjee

        228k15158278




        228k15158278






























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