Question

Prove that sin A - cos A +1 / sin A + cos A -1 = cos A / 1-sinA​

Answer 1

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Question :-

→ If cos θ + sin θ = √2cos θ , then prove that cos θ - sin θ = √2sin θ .

Answer :-

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 .

Answer 2

Step-by-step explanation:

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