Question

please ans me fast....its urgent​

Answer 1

$Question:$

[[[[[Should be]]]]]

If a cos∅ - b sin∅ = x

 a sin∅ + b cos∅ = y

then prove that a²+ b² = x² + y².

$Solution:$

Given :

a cos∅ - b sin∅ = x

Squaring both side we get ,

⇒ a² cos²∅ + b² sin²∅ - 2 · a cos∅ · b sin∅ = x²

∴ a² cos²∅ + b² sin²∅ - 2 · a cos∅ · b sin∅ = x²...........(i)

Again

a sin∅ + b cos∅ = y

Squaring both side we get ,

⇒ a² sin²∅ + b² cos²∅ + 2 · a cos∅ · b sin∅ = y²

∴  a² sin²∅ + b² cos²∅ + 2 · a cos∅ · b sin∅ = y²................(ii)

Now

By adding eq.(i) and eq.(ii) we get ,

⇒ a²(cos²∅ + sin²∅) + b² (cos²∅ + sin²∅) =  x² + y²

We know that (cos²∅ + sin²∅) = 1

∴ a² + b² = x² + y².

Hence proved.

Thanks..

Answer 2

SOLUTION ☺️

Given:-

$= > x = a \: cos \theta - b \: sin \theta \\ = > therefore \\ = > x {}^{2} = (a \: cos \theta - b \: sin \theta) {}^{2} \\ = > x {}^{2} = (a {}^{2} \: cos {}^{2} \theta) + (b {}^{2} \: sin {}^{2} \theta) - (2ab \: cos \theta \: sin \theta) - - - - (1) \\ \\ = >also \: given \\ = > y = a \: sin \theta + b \: cos \theta \\ therefore \\ = > y {}^{2} = (a \: sin \theta + b \: cos \theta) {}^{2} \\ = > y {}^{2} = (a {}^{2} sin {}^{2} \theta) + (b {}^{2} \: cos {}^{2} \theta) - (2ab \: cos \theta \: sin \theta) - - - - - (2) \\ \\ = > adding \: (1) \: and \: (2) \\ = > x {}^{2} + y {}^{2} = (a {}^{2} \: cos {}^{2} \theta) + (b {}^{2} sin {}^{2} \theta) - (2ab \: cos \theta \: sin \theta) + (a {}^{2} sin {}^{2} \theta) + (b {}^{2} cos {}^{2} \theta) + (2ab \: cos \theta \: sin \theta) \\ = > cancelling(2ab \: cos \theta \: sin \theta) \: and \: - (2ab \: cos \theta \: sin \theta) \\ = > x {}^{2} + y {}^{2} = (a {}^{2} cos {}^{2} \theta) + (b{}^{2} sin {}^{2} \theta) + (a {}^{2} sin {}^{2} \theta) + (b {}^{2} cos {}^{2} \theta) \\ \\ = > bringing \: a {}^{2} \: terms \: and \: b {}^{2} \: terms \: together \\ = > x {}^{2} + y {}^{2} = (a {}^{2} cos {}^{2} \theta) + (a {}^{2} sin {}^{2} \theta) + (a {}^{2} sin {}^{2} \theta) + (b {}^{2}cos {}^{2} \theta) \\ = > x {}^{2} + y {}^{2} = a {}^{2} (cos {}^{2} \theta + sin {}^{2} \theta ) +b {}^{2} (sin {}^{2} \theta + cos {}^{2} \: theta) \\ = > by \: the \: identify \: sin {}^{2} \theta + cos {}^{2} \theta = 1 \\ = > x {}^{2} + y {}^{2} = a {}^{2} (1) + b {}^{2} (1) \\ \\ = > {x}^{2} + {y}^{2} = a {}^{2} + b {}^{2} \: \: proved$

hope it helps ✔️