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

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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




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    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
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      up vote
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      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





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      edited 2 days ago









      Martin Sleziak

      43.4k6111259




      43.4k6111259









      asked 2 days ago









      learner_avid

      654212




      654212




















          1 Answer
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          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










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          1 Answer
          1






          active

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          active

          oldest

          votes






          active

          oldest

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          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












           

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