Question

If xsinθ = ycosθ and ysin³θ + xcos³θ = sinθcosθ. Then find the value of x² + y².

Answer 1

Correct Question

If xsinθ = ycosθ and xsin³θ + ycos³θ = sinθcosθ. Then find the value of x² + y².

Answer

1

Explanation

Given

xsin∅ = ycos∅

xsin³∅ + ycos³∅ = sin∅cos∅

The second equation can be written as

xsin∅(sin²∅) + ycos³∅ = sin∅cos∅

Put xsin∅ as ycos∅

→ ycos∅(sin²∅) + ycos³∅ = sin∅cos∅

→ ycos∅(sin²∅ + cos²∅) = sin∅cos∅

→ y cos∅ = sin∅cos∅

(using, sin²∅ + cos²∅ = 1)

→ y = sin∅

And, xsin∅ = ycos∅

→ xsin∅ = sin∅cos∅ (since y = sin∅)

→ x = cos∅

So, x² + y² = cos²∅ + sin²∅ = 1

Answer 2

Correct Question :--- If xsinθ = ycosθ and xsin³θ + ycos³θ = sinθcosθ. Then find the value of x² + y². ?

Solution :---

it is Given That,

→ xsinθ = ycosθ ------------------- Equation (1)

→ xsin³θ + ycos³θ = sinθcosθ

→ (xsinθ)*sin²θ + ycos³θ = sinθcosθ

Putting value of (xsinθ = ycosθ) From Equation (1) we get,

→ ycosθ*sin²θ + ycos³θ = sinθcosθ

Taking ycosθ common From LHS now,

→ ycosθ( sin²θ + cos²θ ) = sinθcosθ

Now, we know that , sin²θ + cos²θ = 1..

So,

→ ycosθ * 1 = sinθcosθ

Dividing both sides by cosθ now,

→ y = sinθ ----------------- Equation (2)

_____________________________

Putting value of Equation (2) in Equation (1) now, we get,

→ xsinθ = sinθ * cosθ

Diving sinθ both sides now, we get,

→ x = cosθ ------------------ Equation(3).

_____________________

So,

→ x² + y²

putting value of Equation (2) and Equation (3) we get,

→ (cosθ)² + (sinθ)²

→ cos²θ + sin²θ

→ 1

Hence, value of x² + y² is 1.