Question

if X=aSin∅+bCos∅ and y=aCos∅-bSin∅ then prove that x²+y²=a²+b²​

Answer 1

Answer:-

Step-by-step explanation:Given :-

→ a cos∅ + b sin∅ = x .

→ a sin∅ - b cos∅ = y .

Now,

We have,

x² + y² .

= ( a cos ∅ + b sin ∅ )² + ( a sin∅ - b cos∅ )² .

= a²cos²∅ + b²cos²∅ + 2 ab cos∅ sin∅ + a²sin²∅ + b²sin²∅ - 2 ab sin∅ cos∅ .

= a²cos²∅ + b²cos²∅ + a²sin²∅ + b²sin²∅ .

= a²cos²∅ + a²sin²∅ + b²cos²∅ + b²cos²∅ .

= a²( cos²∅ + sin²∅ ) + b²( cos²∅ + sin²∅ ) .

= a² + b² .

LHS = RHS .

Hence, it is proved .

Answer 2

Answer:

Given: x = a sin theta + b cos theta

y = a cos theta - b sin theta

To prove: x² +y² = a² + b²

Solution:

x = a sin theta + b cos theta..............(Given)

y = a cos theta - b sin theta ...............(Given)

L.H.S = x² + y²

= (a sin theta + b cos theta)² + ( a cos theta - b sin theta)²

= a² sin²theta + 2asin theta b cos theta + b² cos²theta

+ a² cos²theta - 2a cos theta b sin theta + b²sin²theta

..................[ since, ( a+b)² = a² +2ab +b²

(a-b)² = a² -2ab + b²]

= a² ( sin ² theta + cos ²theta) + b² ( cos ² theta + sin ² theta)

= a² (1) + b²(1) .........[ since, sin²theta + cos²theta =1]

= a²+b²

= R.H.S

Hence,

L.H.S = R.H.S

THEREFORE,

x²+y² = a² + b²