If 2cos theta - sin theta = x and cos theta - 3sin theta =y . Prove that 2x^2+y^2 - 2xy=5
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HELLO DEAR,
GIVEN:-
2cos theta - sin theta = x
and
(cosθ - 3 sinθ) = y
PUT THE VALUES OF X AND Y IN
2x2 + y2 − 2xy (LHS)
= 2(2 cosθ − sinθ)2 + (cosθ − 3 sinθ)2
− 2(2 cosθ − sinθ)(cosθ − 3 sinθ)
= 2(4cos2θ − 4cosθ sinθ + sin2θ)
+ (cos2θ − 6cosθ sinθ + 9sin2θ)
− 2(2cos2θ − 7cosθ sinθ + 3sin2θ)
= 8cos2θ − 8cosθ sinθ + 2sin2θ + cos2θ
− 6cosθ sinθ + 9sin2θ − 4cos2θ
+ 14cosθ sinθ − 6sin2θ
= 5cos2θ + 5sin2θ
= 5(cos2θ + sin2θ)
= 5(1) = 5 (Since cos2θ + sin2θ = 1)
= RHS
. I HOPE ITS HELP YOU DEAR,
THANKS
GIVEN:-
2cos theta - sin theta = x
and
(cosθ - 3 sinθ) = y
PUT THE VALUES OF X AND Y IN
2x2 + y2 − 2xy (LHS)
= 2(2 cosθ − sinθ)2 + (cosθ − 3 sinθ)2
− 2(2 cosθ − sinθ)(cosθ − 3 sinθ)
= 2(4cos2θ − 4cosθ sinθ + sin2θ)
+ (cos2θ − 6cosθ sinθ + 9sin2θ)
− 2(2cos2θ − 7cosθ sinθ + 3sin2θ)
= 8cos2θ − 8cosθ sinθ + 2sin2θ + cos2θ
− 6cosθ sinθ + 9sin2θ − 4cos2θ
+ 14cosθ sinθ − 6sin2θ
= 5cos2θ + 5sin2θ
= 5(cos2θ + sin2θ)
= 5(1) = 5 (Since cos2θ + sin2θ = 1)
= RHS
. I HOPE ITS HELP YOU DEAR,
THANKS
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