Question

$\frac{(1+\cot \theta +\tan \theta )(\sin \theta -\cos \theta )}{{\sec }^3\theta -{\csc }^3\theta }={\sin }^2\theta {\cos }^2\theta$

Answer 1

$LHS=\frac{(1+\cot \theta +\tan \theta )(\sin \theta -\cos \theta )}{{\sec }^3\theta -{\csc }^3\theta }$

$=\frac{(1+\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta })(\sin \theta -\cos \theta )}{\frac{1}{{\cos }^3\theta }-\frac{1}{{\sin }^3\theta }}$

$=\frac{(1+\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta })(\sin \theta -\cos \theta )}{\frac{{\sin }^3\theta -{\cos }^3\theta }{{\sin }^3\theta {\cos }^3\theta }}$

$=\frac{(1+\frac{1}{\sin \theta \cos \theta })(\sin \theta -\cos \theta )}{\frac{{\sin }^3\theta -{\cos }^3\theta }{{\sin }^3\theta {\cos }^3\theta }}$

$=\frac{(1+\frac{1}{\sin \theta \cos \theta })(\sin \theta -\cos \theta )\times {\sin }^3\theta {\cos }^3\theta }{(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )}$

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

$={\sin }^2\theta {\cos }^2\theta =RHS$

Hence proved.