If θ\thetaθ is an acute angle, find the denominator A, when (cosecθ−cotθ)2=1−cotθA(cosec\theta-cot\theta)^2=\frac{1-cot\theta}{A}(cosecθ−cotθ)2=A1−cotθ Acosecθ−1cosec\theta-1cosecθ−1B1+sinθ1+sin\theta1+sinθCcotθcot\thetacotθD1+cosθ1+cos\theta1+cosθAnswer: 1+cosθ1+cos\theta1+cosθ Read Explanation: (cosecθ−cotθ)2=1−cosθA(cosec\theta-cot\theta)^2=\frac{1-cos\theta}{A}(cosecθ−cotθ)2=A1−cosθ(cosecθ−cotθ)2=(1sinθ−cosθsinθ)2(cosec\theta-cot\theta)^2=(\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta})^2(cosecθ−cotθ)2=(sinθ1−sinθcosθ)2=(1−cosθsinθ)2=(\frac{1-cos\theta}{sin\theta})^2=(sinθ1−cosθ)2=(1−cosθ)2sin2θ=\frac{(1-cos\theta)^2}{sin^2\theta}=sin2θ(1−cosθ)2=(1−cosθ)2(1−cos2θ)=\frac{(1-cos\theta)^2}{(1-cos^2\theta)}=(1−cos2θ)(1−cosθ)2=(1−cosθ)(1−cosθ)((1−cosθ)(1+cosθ)=\frac{(1-cos\theta)(1-cos\theta)}{((1-cos\theta)(1+cos\theta)}=((1−cosθ)(1+cosθ)(1−cosθ)(1−cosθ)=(1−cosθ)(1+cosθ)=\frac{(1-cos\theta)}{(1+cos\theta)}=(1+cosθ)(1−cosθ)which is equal to (1−cosθ)A\frac{(1-cos\theta)}{A}A(1−cosθ) Read more in App
