Question

If 2cos theta - sin theta = x and cos theta - 3sin theta = y .

( Prove that 2 x sq + y sq. - 2xy = 5 )

Plz answer it !!!!

urgent !!!!

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Answer 1

Hey friend, Harish here.

Here is your answer:

Given that :

1) 2 cos θ - sin θ = x  

2) cos θ - 3 sin θ = y

To Prove :

2x² + y² - 2xy = 5

Proof :

2 x² = 2 ( 2cos θ - sin θ )²  =  2 + 6 cos² θ - 8 sin θ. cos θ    - (i)

y² = ( cos θ - 3 sin θ )² = 1 + sin² θ - 6 sin θ cos θ    - (ii)

2 xy = 2 × (2cos θ - sin θ) (cos θ - 3 sin θ) = 4 + 2 sin² θ - 14 sin θ . cos θ   - (iii)

Now adding (i) , (ii) and (iii) we get :

2x² + y² - 2xy = [2 + 6 cos² θ - 8 sin θ. cos θ] + [1 + sin² θ - 6 sin θ cos θ] - [4 + 2 sin² θ - 14 sin θ . cos θ]

⇒ 2x² + y² - 2xy   = 6 cos² θ + 6 sin² θ - 1 = 6 (cos² θ + sin² θ) - 1 = 6 (1) - 1 = 5

IDENTITY USED :

$\boxed{\mathrm{\sin^2\theta + \cos ^2 \theta = 1}}$

HENCE PROVED : )

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Hope my answer is helpful to you. All the best.

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