Question

How may I solve this​

Answer 1

Answer:

Here is your answer mate! Pls mark as brainliest!

Step-by-step explanation:

Given :-

x = a cos ∅ - b sin ∅

Therefore,

x² = (a cos∅ - b sin∅)²  

x²= (a² Cos²∅) + (b² Sin2∅) - (2ab Cos∅ Sin∅)   ---------------( 1)

Also given ,

y = a sin ∅ + b cos ∅

Therefore,

y² = (a sin ∅  + b cos∅)y² = (a² Sin²∅) + (b² Cos²∅) + (2ab Cos∅  Sin∅)   ------(2)

Adding  (1 )and( 2),

x² + y² = (a² Cos²∅) + (b² Sin²∅)-(2ab Cos∅  Sin∅)+(a²Sin²∅) + (b² Cos²∅) +(2ab Cos∅  Sin∅)

Cancelling (2ab Cos∅  Sin∅) and - (2ab Cos∅  Sin∅)

 x² + y² = (a² Cos²∅) + (b²Sin²∅) + (a² Sin²∅) + (b² Cos²∅)

 

Bringing a² terms and b² terms together,

x² + y² = (a² Cos²∅) + (a² Sin²∅) + (b² Sin²∅) + (b² Cos²∅)

x² + y² = a² (Cos2∅  + Sin2∅) + b2 (Sin2∅  + Cos2∅)

By the identity Sin²∅  + Cos²∅  = 1

x² + y² = a2 (1) + b2 (1)

x² + y² = a² + b²

Hence, proved.

Answer 2

Step-by-step explanation:

$\bf \: x = ☠☠acos \theta + bsin\theta \\ \bf y = asin\theta - bcos\theta \\ \\ {x}^{2} = {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2} \theta + 2acos\theta \: bsin\theta \\ \\ {y}^{2} = {a}^{2} {sin}^{2} \theta + {b}^{2} {cos}^{2} \theta - 2asin\theta \: bcos\theta \\ \\ {x}^{2} +{y} ^{2}= {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2} \theta \cancel{+ 2acos\theta \: bsin\theta }\\ + {a}^{2} {sin}^{2} \theta + {b}^{2} {cos}^{2} \theta \cancel{ - 2asin\theta \: bcos \theta} \\ \\ \tt \implies {a}^{2} \{ {sin}^{2} \theta + {cos}^{2} \theta \} + {b}^{2} \{ {sin}^{2}\theta + {cos}^{2} \theta\} \\ \\ \boxed{ \tt {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}}$