Question

Prove that
$(\frac{1+sin\theta-cos\theta}{1+sin\theta+cos\theta})^2=\frac{1-cos\theta}{1+cos\theta}$

Answer 1

Step-by-step explanation:

Given: [(1 + sinθ - cosθ)/(1 + sinθ + cosθ)]²

∴ (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ca

∴ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

= [(1 + sin²θ + cos²θ + 2sinθ - 2cosθ - sinθcosθ)/(1 + sin²θ + cos²θ + 2sinθ + 2cosθ + sinθcosθ]

= [(1 + 1 + 2sinθ - 2cosθ - sinθcosθ)/(1 + 1 + 2sinθ + 2cosθ + sinθcosθ)

= [(2 + 2sinθ - 2cosθ - 2sinθcosθ)/(2 + 2sinθ + 2cosθ + sinθcosθ)]

= [2(1 + sinθ) - 2cosθ(1 + sinθ)]/[2(1 + sinθ) + 2cosθ(1 + sinθ)]

= [1 + sinθ(2 - 2cosθ)]/[1 + sinθ(2 + 2cosθ)]

= [(2 - 2cosθ)/(2 + 2cosθ)]

= [2(1 - cosθ)/2(1 + cosθ)]

= (1 - cosθ)/(1 + cosθ).

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Answer 2

Answer:

mark as a brainlist answer