Find the value of $frac (a^2+b^2+c^2+d^2)(sin 20°)(bd-ac).$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Given that $a,b,c,d in R$, if $$ a sec(200°) - c tan(200°) =d$$ and $$b sec(200°) + d tan(200°) = c$$ then find the value of $$dfrac (a^2+b^2+c^2+d^2)(sin 20°)(bd-ac).$$



My attempt:Let $theta=200^o$ then our equations become $a sectheta - c tantheta =d$ and $b sectheta+ d tantheta= c$

Squaring and adding them,we get

$a^2sec^2theta+c^2tan^2theta-2acsecthetatantheta+b^2sec^2theta+d^2tan^2theta+2bdsecthetatantheta=d^2+c^2$


$a^2sec^2theta+c^2tan^2theta+b^2sec^2theta+d^2tan^2theta-d^2-c^2=2acsecthetatantheta-2bdsecthetatantheta$


I am stuck here.




trigonometry




share|cite|improve this question

























    up vote
    0
    down vote

    favorite












    Given that $a,b,c,d in R$, if $$ a sec(200°) - c tan(200°) =d$$ and $$b sec(200°) + d tan(200°) = c$$ then find the value of $$dfrac (a^2+b^2+c^2+d^2)(sin 20°)(bd-ac).$$



    My attempt:Let $theta=200^o$ then our equations become $a sectheta - c tantheta =d$ and $b sectheta+ d tantheta= c$

    Squaring and adding them,we get

    $a^2sec^2theta+c^2tan^2theta-2acsecthetatantheta+b^2sec^2theta+d^2tan^2theta+2bdsecthetatantheta=d^2+c^2$


    $a^2sec^2theta+c^2tan^2theta+b^2sec^2theta+d^2tan^2theta-d^2-c^2=2acsecthetatantheta-2bdsecthetatantheta$


    I am stuck here.




    trigonometry




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Given that $a,b,c,d in R$, if $$ a sec(200°) - c tan(200°) =d$$ and $$b sec(200°) + d tan(200°) = c$$ then find the value of $$dfrac (a^2+b^2+c^2+d^2)(sin 20°)(bd-ac).$$



      My attempt:Let $theta=200^o$ then our equations become $a sectheta - c tantheta =d$ and $b sectheta+ d tantheta= c$

      Squaring and adding them,we get

      $a^2sec^2theta+c^2tan^2theta-2acsecthetatantheta+b^2sec^2theta+d^2tan^2theta+2bdsecthetatantheta=d^2+c^2$


      $a^2sec^2theta+c^2tan^2theta+b^2sec^2theta+d^2tan^2theta-d^2-c^2=2acsecthetatantheta-2bdsecthetatantheta$


      I am stuck here.




      trigonometry




      share|cite|improve this question













      Given that $a,b,c,d in R$, if $$ a sec(200°) - c tan(200°) =d$$ and $$b sec(200°) + d tan(200°) = c$$ then find the value of $$dfrac (a^2+b^2+c^2+d^2)(sin 20°)(bd-ac).$$



      My attempt:Let $theta=200^o$ then our equations become $a sectheta - c tantheta =d$ and $b sectheta+ d tantheta= c$

      Squaring and adding them,we get

      $a^2sec^2theta+c^2tan^2theta-2acsecthetatantheta+b^2sec^2theta+d^2tan^2theta+2bdsecthetatantheta=d^2+c^2$


      $a^2sec^2theta+c^2tan^2theta+b^2sec^2theta+d^2tan^2theta-d^2-c^2=2acsecthetatantheta-2bdsecthetatantheta$


      I am stuck here.





      trigonometry





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Martin Sleziak

      43.4k6111259




      43.4k6111259









      asked 2 days ago









      learner_avid

      654212




      654212




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Hint:



          Method$#1:$



          Use $sec(180^circ+y)=-sec y,tan(180^circ+y)=+tan y$



          Solve for $sec20^circ,tan20^circ$



          Finally $sin t=dfractan tsec t (1)$



          Here $t=20^circ$



          Method$#2:$



          Solve for $sec200^circ,tan200^circ$



          Use $(1),$ here $t=200^circ$



          Finally $sin(180^circ+y)=-sin y $






          share|cite|improve this answer





















          • I got it,thank you
            – learner_avid
            2 days ago










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );








           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871691%2ffind-the-value-of-frac-a2b2c2d2-sin-20bd-ac%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Hint:



          Method$#1:$



          Use $sec(180^circ+y)=-sec y,tan(180^circ+y)=+tan y$



          Solve for $sec20^circ,tan20^circ$



          Finally $sin t=dfractan tsec t (1)$



          Here $t=20^circ$



          Method$#2:$



          Solve for $sec200^circ,tan200^circ$



          Use $(1),$ here $t=200^circ$



          Finally $sin(180^circ+y)=-sin y $






          share|cite|improve this answer





















          • I got it,thank you
            – learner_avid
            2 days ago














          up vote
          1
          down vote



          accepted










          Hint:



          Method$#1:$



          Use $sec(180^circ+y)=-sec y,tan(180^circ+y)=+tan y$



          Solve for $sec20^circ,tan20^circ$



          Finally $sin t=dfractan tsec t (1)$



          Here $t=20^circ$



          Method$#2:$



          Solve for $sec200^circ,tan200^circ$



          Use $(1),$ here $t=200^circ$



          Finally $sin(180^circ+y)=-sin y $






          share|cite|improve this answer





















          • I got it,thank you
            – learner_avid
            2 days ago












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hint:



          Method$#1:$



          Use $sec(180^circ+y)=-sec y,tan(180^circ+y)=+tan y$



          Solve for $sec20^circ,tan20^circ$



          Finally $sin t=dfractan tsec t (1)$



          Here $t=20^circ$



          Method$#2:$



          Solve for $sec200^circ,tan200^circ$



          Use $(1),$ here $t=200^circ$



          Finally $sin(180^circ+y)=-sin y $






          share|cite|improve this answer













          Hint:



          Method$#1:$



          Use $sec(180^circ+y)=-sec y,tan(180^circ+y)=+tan y$



          Solve for $sec20^circ,tan20^circ$



          Finally $sin t=dfractan tsec t (1)$



          Here $t=20^circ$



          Method$#2:$



          Solve for $sec200^circ,tan200^circ$



          Use $(1),$ here $t=200^circ$



          Finally $sin(180^circ+y)=-sin y $







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 2 days ago









          lab bhattacharjee

          214k14152263




          214k14152263











          • I got it,thank you
            – learner_avid
            2 days ago
















          • I got it,thank you
            – learner_avid
            2 days ago















          I got it,thank you
          – learner_avid
          2 days ago




          I got it,thank you
          – learner_avid
          2 days ago












           

          draft saved


          draft discarded


























           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871691%2ffind-the-value-of-frac-a2b2c2d2-sin-20bd-ac%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What is the equation of a 3D cone with generalised tilt?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
          Read more

          Color the edges and diagonals of a regular polygon

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
          Read more

          Relationship between determinant of matrix and determinant of adjoint?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
          Read more