Question

x²+y²=1 than prove that (3x-4x³)²+(3y-4y³)²=1​

Answer 1

Answer:

Given : x² + y² = 1

=> 4x² = 4 - 4y² and 4y² = 4 - 4x²

. . .(4x³ - 3x)² + (3y - 4y³)² = [ x * (4x² - 3) ]² + [ y * (3 - 4y²) ]²

=> x² * (4 - 4y² - 3)² + y² * (3 - 4 + 4x²)²

=> x² * (1 - 4y²)² + y² * (4x² - 1)²

=> x² * (1 + 16y^4 - 8y² ) + y² * (16x^4 + 1 - 8x²)

=> x² + 16x²y^4 - 8x²y² + 16x^4 y² + y² - 8x²y²

=> (x² + y²) + 16x²y² ( x² + y²) - 16x²y²

=> 1 + 16x²y² - 16x²y² . . . {As x² + y² = 1 }

=> 1 ===> ANS

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ALTERNATIVELY

=> x² * (4x² - 3)² + y² * (3 - 4y²)²

=> x² * (4x² - 3)² + (1 - x²) * [3 - (4 - 4x²)]²

=> x² * (4x² - 3)² - (x² - 1) * ( 4x² - 1)²

=> x² * (4x² - 3)² - x² * ( 4x² - 1)² + ( 4x² - 1)²

=> x² [ (4x² - 3)² - ( 4x² - 1)² ]+ ( 4x² - 1)²

=> x² [ (4x² - 3 + 4x² - 1)(4x² - 3 - 4x² + 1) ] + ( 4x² - 1)²

=> x² * (8x² - 4)(-2) + ( 4x² - 1)²

=> -16x^4 + 8x² + 16x^4 - 8x² + 1

=> 1 (ANS)

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ALTERNATIVELY

Given : x² + y² = 1

then x can be considered as sin θ and y as cos θ

{Since sin² θ + cos² θ = 1 }

Now, 4x³ - 3x = 4 sin³ θ - 3 sin θ = sin 3θ

and 3y - 4y³ = 3 cos θ - 4 cos³ θ = cos 3θ

=> (4x³ - 3x)² + (3y - 4y³)²

= sin² 3θ + cos² 3θ

= 1 ===> (Ans)