Question

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Answer 1

Given :

xsin³(theta) + y cos³ (theta) = sin(theta).cos(theta),

xsin(theta) = ycos(theta)

To prove : x² + y² = 1

Proof : xsin(theta) = ycos(theta)

-> xsin³(theta) + y cos³ (theta) =sin(theta).cos(theta)

-> Divide by y on both the sides

->( x/y) sin³(theta) + cos³(theta) = sin(theta).cos(theta) / y

->cos(theta).sin²(theta) + cos³(theta) = sin(theta).cos(theta) / y

-> cos(theta) = sin(theta).cos(theta) / y

-> sin(theta)/y = 1 = cos(theta) / x

therefore , xsin(theta) = ycos(theta) = 1

.............(1)

Now, xsin³(theta) + y cos³ (theta) = sin(theta).cos(theta)

-> sin²(theta) + cos²(theta) = 1 = sin(theta).cos(theta) - ----------- (2)

-> tan(theta) + cot(theta) = 1

Now ,xsin³(theta) + y cos³ (theta) = sin(theta) .cos(theta)

-> x (tan(theta)).sin²(theta) + y cot(theta).cos²(theta)

= 1

-> xtan(theta) - xsin(theta).cos(theta) + y cot(theta) - ysin(theta).cos(theta) = 1

-> xtan(theta) - cos(theta) + y cot(theta) - sin(theta) = 1

-> x tan(theta) + y cot (theta) = sin(theta) + cos (theta)

-> x sin²(theta) + y cos²(theta) =(sin(theta) + cos (theta) ) (sin(theta).cos(theta))

-> x sin²(theta) + y cos²(theta) = (sin(theta)+cos(theta)

-> x² sin⁴(theta) + y² cos⁴(theta) + 2xy(sin²theta)(cos²theta) = 2sin(theta) .cos(theta)

-> x² sin⁴(theta) + y² cos⁴(theta) = 1

-> x² (ycos(theta)/x)⁴ + y²(x.sin(theta)/y)⁴ =1

-> 1/x²+ 1/ y²= 1 = 1/1

-> y² + x² = x⁴ y⁴

-> y² + x² = (xsin(theta)/cos(theta))⁴ (y.cos(theta)/sin(theta))⁴

-> y² + x² = (1 / cos⁴(theta)) (1/sin⁴(theta))

........... from 1

-> y² + x² = 1 / (sin²(theta).cos²(theta))²

-> x² + y² = 1 ... (from 2)

-> x² + y² = 1

Hence proved !

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