Question

if 2cos theta - sin theta=x and cos theta - 3sin theta=y , prove that 2x2 +y2- 2xy=5

Answer 1

Hey here u go

2cosθ-sinθ=x
or, x²=4cos²θ-4sinθcosθ+sin²θ
or, x²=3cos²θ- and
cosθ-3sinθ=y
or, y²=cos²θ-6sinθcosθ+9sin²θ
xy=(2cosθ-sinθ)(cosθ-3sinθ)
or, xy=2cos²θ-sinθcosθ-6sinθcosθ+3sin²θ
or, xy=2cos²θ-7sinθcosθ+3sin²θ
∴, 2x²+y²-2xy
=2(4cos²θ-4sinθcosθ+sin²θ)+cos²θ-6sinθcosθ+9sin²θ-2(2cos²θ-7sinθcosθ+3sin²θ)
=8cos²θ-8sinθcosθ+2sin²θ+cos²θ-6sinθcosθ+9sin²θ-4cos²θ+14sinθcosθ-6sin²θ
=5cos²θ+5sin²θ
=5(sin²θ+cos²θ)
=5  (Proved) [∵, sin²θ+cos²θ=1]

Answer 2

According to the Question

(2 cosθ - sinθ) = x   (cosθ - 3 sinθ) = y

Substitute the values of x and y

2x^2 + y^2 − 2xy (Left Hand Side)

= 2(2 cosθ − sinθ)2 + (cosθ − 3 sinθ)2 − 2(2 cosθ − sinθ)(cosθ − 3 sinθ)

= 2(4cos^2θ − 4cosθ sinθ + sin^2θ) + (cos^2θ − 6cosθ sinθ + 9sin^2θ) - 2(2cos^2θ − 7cosθ sinθ + 3sin^2θ)

= 8cos^2θ − 8cosθ sinθ + 2sin^2θ + cos^2θ − 6cosθ sinθ + 9sin^2θ − 4cos^2θ + 14cosθ sinθ − 6sin^2θ

= 5cos^2θ + 5sin^2θ

= 5(cos^2θ + sin^2θ)

= 5(1) = 5    ⇒   ( cos^2θ + sin^2θ = 1)

LHS = RHS