if cos theta + sin theta is equal to root 2 sin theta then prove that cos theta minus sin theta is equal 1
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Answer:
--> cos θ + sin θ = √2 cos θ
--> sin θ = √2 cos θ - cos θ
=> sin θ = ( √2 - 1 ) cos θ
=> [ sin θ / ( √2 - 1 ) ] = cos θ
=> [ sin θ ( √2 + 1 ) / ( 2 - 1 ) ] = cos θ
0_0 --> We rationalized the denominator in the 2nd step ^_^
=> [ √2 sin θ + sin θ ] = cos θ
=> cos θ - sin θ = √2 sin θ
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Step-by-step explanation:
We have,
→ cos θ + sin θ = √2cos θ .
[ Squaring both side, we get ] .
⇒ ( cos θ + sin θ )² = 2cos²θ .
⇒ cos²θ + sin²θ + 2cosθsinθ = 2cos² .
⇒ sin²θ + 2cosθsinθ = 2cos²θ - cos²θ .
⇒ sin²θ + 2cosθsinθ = cos²θ .
⇒ cos²θ - 2cosθsinθ = sin²θ .
[ Adding sin²θ both side, we get ] .
⇒ cos²θ - 2cosθsinθ + sin²θ = sin²θ + sin²θ .
⇒ ( cos θ - sin θ )² = 2sin²θ .
⇒ cos θ - sin θ = √( 2sin²θ ) .
∴ cos θ - sin θ = √2sin θ .
Hence, it is proved .
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