Question

solve 55th questions ​

Answer 1

Here is your answer

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Answer 2

Sinθ + Cosθ = $\sqrt{2}$Cosθ

Squaring on both sides

Sin²θ + 2SinθCosθ + Cos²θ = 2Cos²θ

Since, Sin²θ + Cos²θ = 1,

2SinθCosθ = 2Cos²θ - 1

Now, we have

( Cosθ - Sinθ )² = Sin²θ - 2SinθCosθ + Cos²θ

( Cosθ - Sinθ )² = 1 - 2SinθCosθ

But,  2SinθCosθ = 2Cos²θ - 1

Hence,

( Cosθ - Sinθ )² = 1 - ( 2Cos²θ - 1 )

( Cosθ - Sinθ )² = 1 -  2Cos²θ + 1

( Cosθ - Sinθ )² = 2 - 2Cos²θ

( Cosθ - Sinθ )² = 2( 1  - Cos²θ )

( Cosθ - Sinθ )² = 2Sin²θ

Cosθ - Sinθ  =  $\sqrt{2}$Sinθ

Hence proved

HOPE IT HELPS !!