Question

if X =a cos theta + b sin theta and y =a sin theta-B cos theta prove that x^2+ y^2= a^2 + b^2
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Answer 1

Answer:

Given that ;

• x = a cosθ + b sinθ
• y = a sinθ - b cosθ

At first, find the value of x².

⇒ x² = (a cosθ + b sinθ )²

⇒ x² = a² cos²θ + b² sin²θ + 2 * a cosθ * b sinθ .... (i)

Now, find the value of y².

⇒ y² = (a sinθ - b cosθ )²

⇒ y² = a² sin²θ + b² cos²θ - 2 * a sinθ * b cosθ .... (ii)

On adding equation (i) and (ii) -

⇒ x² + y² = a² cos²θ + b² sin²θ + 2 * a cosθ * b sinθ + a² sin²θ + b² cos²θ - 2 * a sinθ * b cosθ

⇒ x² + y² = a² ( cos²θ + sin²θ ) + b² ( sin²θ + cos²θ )

We know that -

• sin²θ + cos²θ = 1

⇒ x² + y² = a² + b²

Hence, it is proved.

Answer 2

Answer:

Step-by-step explanation:

Given :-

x = a cosθ + b sinθ

y = a sinθ - b cosθ

To Prove :-

x² + y² = a² + b²

Formula to be used :-

Some Identities :-

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b²-2ab

sin²θ + cos²θ = 1

Solution :-

L.H.S = x² + y²

x² + y²

= (a sinθ + b cosθ)² + (a cosθ + b sinθ)²

= (a²sin²θ + b²cos²θ a + ab sinθ cosθ ) + (a²cos²θ a + b²sin²θ - ab sinθ cosθ)

[Using (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab]

= a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ)

= a²+b² [sin²θ + cos²θ = 1]

= x² + y² = a²+b²

= L.H.S = R.H.S

Hence, Proved.