Question

only ans if u know q30

Answer 1

Solution:(Instead of θ I use A)

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Given:

2cos A - sin A = x,

cos A - 3 sin A = y

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To Prove:

 value of 2x² + y² - 2xy = 5
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Proof:

x² = (2cos A - sin A)²

=> 4cos²A - 4sin A cos A + sin²A

=> 3cos²A - 4sin A cos A  + 1

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y² = ( cos A - 3sin A )²

=> cos²A - 6sin A cos A + 9 sin² A

=> 8sin²A - 6sinA cos A + 1

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xy = (2cos A - sin A) (cos A - 3 sin A)

=> 2cos²A - 6sin A cos A - sin A cos A + 3sin²A

=> sin²A - 7 sin A cos A + 2

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2x² + y² - 2xy = 5,..

=> 2 ( 3cos²A - 4sin A cos A  + 1) + ( 8sin²A - 6sinA cos A + 1) - 2 (sin²A - 7 sin A cos A + 2) = 5,.

=> 6cos²A - 8sinAcosA +2 +

 8sin²A - 6sinAcosA + 1

 - 2sin²A + 14sinAcosA - 4 = 5


=> 6cos²A + 8sin²A -2sin²A + 2 + 1 - 4 -14sinAcosA + 14sinAcosA = 5
 
=>  6 sin²A + 6 cos²A +3 -4 = 5

=> 6(sin²A + cos²A) - 1 = 5

=> 6(1) - 1 = 5

=> 6-1 =5

=>  5 = 5,.

=> ∴ LHS = RHS,.

                                         Hence proved,.

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                        Hope it Helps!!