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Home / Doubts and Solutions/JEE/Mathematics

Plz solve this?

Asked by dasratna_72 1st July 2017, 7:54 PM
Answered by Expert
Answer:
begin mathsize 16px style Correct space option colon space left parenthesis straight b right parenthesis sinθ plus sinϕ equals straight a Squaring space on space both space sides comma sin squared straight theta plus 2 sinθsinϕ plus sin squared straight ϕ equals straight a squared............. left parenthesis straight i right parenthesis cosθ plus cosϕ equals straight b open parentheses cosθ plus cosϕ close parentheses squared equals straight b squared cos squared straight theta plus 2 cosθcosϕ plus cos squared straight ϕ equals straight b squared......... left parenthesis ii right parenthesis Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis 1 plus 2 sinθsinϕ plus 2 cosθcosϕ plus 1 equals straight a squared plus straight b squared 2 plus 2 cos open parentheses straight theta minus straight ϕ close parentheses equals straight a squared plus straight b squared 2 open square brackets 2 cos squared open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses close square brackets equals straight a squared plus straight b squared cos squared open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals fraction numerator straight a squared plus straight b squared over denominator 4 end fraction sec squared open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals fraction numerator 4 over denominator straight a squared plus straight b squared end fraction sec open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals square root of fraction numerator 4 over denominator straight a squared plus straight b squared end fraction end root sec squared open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals fraction numerator 4 over denominator straight a squared plus straight b squared end fraction Using space identity space tan squared straight theta equals sec squared straight theta minus 1 tan squared open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals fraction numerator 4 over denominator straight a squared plus straight b squared end fraction minus 1 tan squared open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals fraction numerator 4 minus straight a squared minus straight b squared over denominator straight a squared plus straight b squared end fraction tan open parentheses fraction numerator straight theta minus straight ϕ over denominator 2 end fraction close parentheses equals square root of fraction numerator 4 minus straight a squared minus straight b squared over denominator straight a squared plus straight b squared end fraction end root end style
Answered by Expert 3rd July 2017, 10:26 AM
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