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    Maths-
    General
    Easy

    Question

    In triangle A B C comma left parenthesis a left parenthesis cos invisible function application B plus cos invisible function application C right parenthesis right parenthesis divided by left parenthesis 2 left parenthesis b plus c right parenthesis right parenthesis equals

    1. c o s squared invisible function application A divided by 2
    2. s i m squared invisible function application A divided by 2
    3. t a n squared invisible function application A divided by 2
    4. c o t squared invisible function application A divided by 2

    hintHint:

    In this question first we have to use Sin Rule a space equals space k sin open parentheses A close parentheses comma space b equals k sin open parentheses B close parentheses comma space c equals k sin open parentheses C close parentheses. Then we will use the C-D formula and Half- Angle formula
    sin open parentheses C close parentheses plus sin open parentheses D close parentheses equals 2 sin open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses cos open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses cos open parentheses C close parentheses plus cos open parentheses D close parentheses equals 2 cos open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses cos open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses
    sin open parentheses 2 x close parentheses equals 2 sin open parentheses x close parentheses cos open parentheses x close parentheses

    The correct answer is: s i m squared invisible function application A divided by 2

      In this question we have to find the value of fraction numerator a left parenthesis cos open parentheses B close parentheses plus cos open parentheses C close parentheses right parenthesis over denominator 2 left parenthesis b plus c right parenthesis end fraction
      By using Sine Rule
      fraction numerator a over denominator sin open parentheses A close parentheses end fraction equals fraction numerator b over denominator sin open parentheses B close parentheses end fraction equals fraction numerator c over denominator sin open parentheses C close parentheses end fraction equals k
      =>a space equals space k sin open parentheses A close parentheses comma space b equals k sin open parentheses B close parentheses comma space c equals k sin open parentheses C close parentheses
      => fraction numerator k sin open parentheses A close parentheses left parenthesis cos open parentheses B close parentheses plus cos open parentheses C close parentheses right parenthesis over denominator 2 left parenthesis k sin open parentheses B close parentheses plus k sin open parentheses C close parentheses right parenthesis end fraction
      =>fraction numerator sin open parentheses A close parentheses left parenthesis cos open parentheses B close parentheses plus cos open parentheses C close parentheses right parenthesis over denominator 2 left parenthesis sin open parentheses B close parentheses plus sin open parentheses C close parentheses right parenthesis end fraction
      By Using C-D formula and half-angle formula
      sin open parentheses C close parentheses plus sin open parentheses D close parentheses equals 2 sin open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses cos open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses cos open parentheses C close parentheses plus cos open parentheses D close parentheses equals 2 cos open parentheses fraction numerator C plus D over denominator 2 end fraction close parentheses cos open parentheses fraction numerator C minus D over denominator 2 end fraction close parentheses
      sin open parentheses 2 x close parentheses equals 2 sin open parentheses x close parentheses cos open parentheses x close parentheses
      => fraction numerator 2 sin open parentheses begin display style A over 2 end style close parentheses cos open parentheses begin display style A over 2 end style close parentheses left parenthesis 2 cos open parentheses begin display style fraction numerator B plus C over denominator 2 end fraction end style close parentheses cos open parentheses begin display style fraction numerator B minus C over denominator 2 end fraction end style close parentheses right parenthesis over denominator 2 left parenthesis 2 sin open parentheses begin display style fraction numerator B plus C over denominator 2 end fraction end style close parentheses cos open parentheses begin display style fraction numerator B minus C over denominator 2 end fraction end style close parentheses right parenthesis end fraction
      We know that
      A plus B plus C space equals space pi
      B plus C equals space pi space minus space A
      fraction numerator B plus C over denominator 2 end fraction equals pi over 2 minus A over 2
      =>sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses equals sin open parentheses pi over 2 minus A over 2 close parentheses
      =>sin open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses equals cos open parentheses A over 2 close parentheses
      Similarly cos open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses equals sin open parentheses A over 2 close parentheses
      By substituting these values in the given expression, we get
      => fraction numerator sin open parentheses begin display style A over 2 end style close parentheses cos open parentheses begin display style A over 2 end style close parentheses sin open parentheses begin display style A over 2 end style close parentheses over denominator cos open parentheses begin display style A over 2 end style close parentheses end fraction
      =>sin squared open parentheses A over 2 close parentheses

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