Prove the following identity: `((1+cottheta+t a ntheta)(sin theta-costheta))/(sec^3theta-cos e c^3theta)=sin^2thetacos^2theta`

Explanation

`
\text { LHS }=\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-\operatorname{cosec}^{3} \theta}
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=\frac{(1+\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta})(\sin \theta-\cos \theta)}{\frac{1}{\cos ^{3} \theta}-\frac{1}{\sin ^{3} \theta}} \quad(\sin ^{2} \theta+\cos ^{2} \theta=1)
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=\frac{(\cos \theta \cdot \sin \theta+1)(\sin \theta-\cos \theta)}{(\sin ^{3} \theta-\cos ^{3} \theta) \cdot \sin \theta \cdot \cos \theta} \cdot \sin ^{3} \theta \cdot \cos ^{3} \theta
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