Question

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

Answer 1

HELLO DEAR,


(1 + sinθ - cosθ)²/(1 + sinθ + cosθ)²

=> {1 + (sinθ - cosθ)² + 2(sinθ - cosθ)}/{1 + (sinθ + cosθ)² + 2(sinθ + cosθ)}

=> {1 + (sin²θ + cos²θ - 2sinθcosθ) + 2(sinθ - cosθ}/{1 + (sin²θ + cos²θ + 2sinθcosθ) + 2(sinθ + cosθ)}

=> {2 - 2sinθcosθ + 2(sinθ - cosθ)}/{2 + 2sinθcosθ + 2(sinθ + cosθ)}

=> {1 - sinθcosθ + sinθ - cosθ}/{1 + sinθcosθ + sinθ + cosθ}

=> {(1 - cosθ) + sinθ(1 - cosθ)}/{(1 + cosθ) + sinθ(1 + cosθ)}

=> {(1 - cosθ)(sinθ)}/{(1 + cosθ)(sinθ)}

=> (1 - cosθ)/(1 + cosθ)


I HOPE IT'S HELP YOU DEAR,
THANKS