if sina + 2cosa = 1
then prove
LHS 2SINA-COSA = RHS 2
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i got it-
(sinθ + 2cosθ)2 = 12
(sinθ + 2cosθ)2 + (2sinθ – cosθ)2 = 1 + (2sinθ – cosθ)2
5sin2θ + 5cos2θ = 1 + (2sinθ – cosθ)2
5 - 1 = (2sinθ – cosθ)2
root4 = 2sinθ – cosθ
or, 2sinθ – cosθ = 2.
(sinθ + 2cosθ)2 = 12
(sinθ + 2cosθ)2 + (2sinθ – cosθ)2 = 1 + (2sinθ – cosθ)2
5sin2θ + 5cos2θ = 1 + (2sinθ – cosθ)2
5 - 1 = (2sinθ – cosθ)2
root4 = 2sinθ – cosθ
or, 2sinθ – cosθ = 2.
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