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If θ\theta is an acute angle, find the denominator A, when (cosecθcotθ)2=1cotθA(cosec\theta-cot\theta)^2=\frac{1-cot\theta}{A}

Acosecθ1cosec\theta-1

B1+sinθ1+sin\theta

Ccotθcot\theta

D1+cosθ1+cos\theta

Answer:

1+cosθ1+cos\theta

Read Explanation:

(cosecθcotθ)2=1cosθA(cosec\theta-cot\theta)^2=\frac{1-cos\theta}{A}

(cosecθcotθ)2=(1sinθcosθsinθ)2(cosec\theta-cot\theta)^2=(\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta})^2

=(1cosθsinθ)2=(\frac{1-cos\theta}{sin\theta})^2

=(1cosθ)2sin2θ=\frac{(1-cos\theta)^2}{sin^2\theta}

=(1cosθ)2(1cos2θ)=\frac{(1-cos\theta)^2}{(1-cos^2\theta)}

=(1cosθ)(1cosθ)((1cosθ)(1+cosθ)=\frac{(1-cos\theta)(1-cos\theta)}{((1-cos\theta)(1+cos\theta)}

=(1cosθ)(1+cosθ)=\frac{(1-cos\theta)}{(1+cos\theta)}

which is equal to (1cosθ)A\frac{(1-cos\theta)}{A}


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