Question

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Answer 1

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GIVEN,

cos$\theta$ + sin$\theta$ = √2 cos$\theta$

Squaring on both sides , we get

cos²$\theta$ + sin²$\theta$ + 2sin$\theta$ = 2cos² $\theta$

cos² $\theta$ - sin² $\theta$ = 2cos $\theta$ 2sin$\theta$

(cos$\theta$ + sin $\theta$) (cos $\theta$ - sin $\theta$) = 2cos $\theta$ sin $\theta$

√2cos$\theta$cos$\theta$ - sin$\theta$= 2cos $\theta$ - sin$\theta$= 2cos $\theta$ sin $\theta$ [given cos $\theta$+sin$\theta$=√2cos$\theta$ ]

$\therefore$ cos$\theta$ - sin$\theta$ = √2sin $\theta$

HENCE PROVED :)

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Answer 2

Answer:

GIVEN,

cos\thetaθ + sin\thetaθ = √2 cos\thetaθ

Squaring on both sides , we get

cos²\thetaθ + sin²\thetaθ + 2sin\thetaθ = 2cos² \thetaθ

cos² \thetaθ - sin² \thetaθ = 2cos \thetaθ 2sin\thetaθ

(cos\thetaθ + sin \thetaθ ) (cos \thetaθ - sin \thetaθ ) = 2cos \thetaθ sin \thetaθ

√2cos\thetaθ cos\thetaθ - sin\thetaθ = 2cos \thetaθ - sin\thetaθ = 2cos \thetaθ sin \thetaθ [given cos \thetaθ +sin\thetaθ =√2cos\thetaθ ]

therefore∴ costhetaθ - sin\thetaθ = √2sin thetaθ

HENCE PROVED

Step-by-step explanation:

Hope this answer will help you.