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Home / Doubts and Solutions/JEE/Mathematics

If A+B+C=180°, then prove that tan^2theta/2=tanB/2tanC/2, when costheta(sinB+sinC)=sinA

Asked by azizrajil03 1st March 2019, 6:26 AM
Answered by Expert
Answer:
begin mathsize 12px style G i v e n space colon space space cos theta space equals space fraction numerator sin space A over denominator sin space B space plus space sin space C end fraction space equals space fraction numerator sin left square bracket space 180 space minus left parenthesis B plus C right parenthesis space right square bracket over denominator sin space B space plus space sin space C end fraction space equals space fraction numerator sin space left parenthesis B plus C right parenthesis over denominator sin space B space plus space sin space C end fraction h e n c e space s e c space theta space equals space fraction numerator sin space B space plus space sin space C over denominator sin space left parenthesis B plus C right parenthesis end fraction space equals space fraction numerator sin B over denominator sin open parentheses B plus C close parentheses end fraction space plus space fraction numerator sin C over denominator sin open parentheses B plus C close parentheses end fraction space space................... left parenthesis 1 right parenthesis fraction numerator sin B over denominator sin left parenthesis B plus c right parenthesis end fraction space equals space fraction numerator sin B over denominator sin B space cos C plus cos B sin C end fraction space equals space fraction numerator begin display style bevelled fraction numerator 2 tan space begin display style B over 2 end style over denominator 1 plus tan squared begin display style B over 2 end style end fraction end style over denominator begin display style fraction numerator 2 tan begin display style B over 2 end style open parentheses 1 minus tan squared begin display style C over 2 end style close parentheses over denominator open parentheses 1 plus tan squared begin display style B over 2 end style close parentheses space open parentheses 1 plus tan squared C over 2 close parentheses end fraction end style space plus space fraction numerator 2 tan begin display style C over 2 end style open parentheses 1 minus tan squared begin display style B over 2 end style close parentheses over denominator open parentheses 1 plus tan squared begin display style B over 2 end style close parentheses space open parentheses 1 plus tan squared C over 2 close parentheses end fraction end fraction fraction numerator sin space B over denominator sin left parenthesis B plus c right parenthesis end fraction space equals space fraction numerator 2 tan space B over 2 over denominator fraction numerator 2 tan begin display style B over 2 end style open parentheses 1 minus tan squared begin display style C over 2 end style close parentheses over denominator space open parentheses 1 plus tan squared C over 2 close parentheses end fraction space plus space fraction numerator 2 tan begin display style C over 2 end style open parentheses 1 minus tan squared begin display style B over 2 end style close parentheses over denominator space open parentheses 1 plus tan squared C over 2 close parentheses end fraction end fraction space equals space fraction numerator t a n space B over 2 space open parentheses 1 plus tan squared C over 2 close parentheses over denominator tan B over 2 open parentheses 1 minus tan squared C over 2 close parentheses plus tan C over 2 open parentheses 1 minus tan squared B over 2 close parentheses end fraction space........... left parenthesis 2 right parenthesis S i m i l a r l y space w e space c a n space p r o v e space fraction numerator sin space C over denominator sin left parenthesis B plus c right parenthesis end fraction space equals space space fraction numerator t a n space C over 2 space open parentheses 1 plus tan squared B over 2 close parentheses over denominator tan B over 2 open parentheses 1 minus tan squared C over 2 close parentheses plus tan C over 2 open parentheses 1 minus tan squared B over 2 close parentheses end fraction space space............... left parenthesis 3 right parenthesis B y space u sin g space e q n. left parenthesis 2 right parenthesis space a n d space e q n. left parenthesis 3 right parenthesis comma space space space e q n. left parenthesis 1 right parenthesis space c a n space b e space w r i t t e n space a s s e c space theta space equals space space fraction numerator t a n space B over 2 space open parentheses 1 plus tan squared C over 2 close parentheses space plus space t a n space C over 2 space open parentheses 1 plus tan squared B over 2 close parentheses over denominator tan B over 2 open parentheses 1 minus tan squared C over 2 close parentheses plus tan C over 2 open parentheses 1 minus tan squared B over 2 close parentheses end fraction space equals space fraction numerator 1 plus tan B over 2 space tan C over 2 over denominator 1 minus tan begin display style B over 2 end style tan begin display style C over 2 end style end fraction c os space theta space equals space fraction numerator 1 minus tan squared begin display style theta over 2 end style over denominator 1 plus tan squared theta over 2 end fraction space equals space fraction numerator 1 minus tan B over 2 tan C over 2 over denominator 1 plus tan B over 2 tan C over 2 end fraction space space space space space o r space space tan squared theta over 2 space equals space tan B over 2 tan C over 2 end style
Answered by Expert 1st March 2019, 12:32 PM
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