1. If 3sinθ+ 4cosθ = 5 (0< θ





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  • By: anil on 05 May 2019 01.49 am
    Expression : $$3sin heta + 4cos heta=5$$ => $$4cos heta=5-3sin heta$$
    Squaring both sides,
    => $$(4cos heta)^2=(5-3sin heta)^2$$ => $$16cos^2 heta = 25+9sin^2 heta-30sin heta$$ => $$16(1-sin^2 heta)=25+9sin^2 heta-30sin heta$$ => $$16-16sin^2 heta=25+9sin^2 heta-30sin heta$$ => $$25sin^2 heta-30sin heta+9=0$$ => $$(5sin heta-3)^2=0$$ => $$5sin heta=3$$ => $$sin heta=frac{3}{5}$$ => Ans - (C)
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