Question

Prove:

$\frac{1+\cot \theta +\tan \theta \left)\right(\sin \theta -\cos \theta )}{{\sec }^3\theta -{\text{cosec}}^3\theta }={\sin }^2\theta {\cos }^2\theta$.

Answer 1

On LHS, write $\cot \theta =\frac{\cos \theta }{\sin \theta }$ and $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Also, $\sec \theta =\frac{1}{\cos \theta }$ , $\text{cosec}\theta =\frac{1}{\sin \theta }$

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

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

= $\frac{{\sin }^3\theta -{\cos }^3\theta }{\left(\sin \theta \cos \theta \right)\frac{({\sin }^3\theta -{\cos }^3\theta )}{{\sin }^3\theta {\cos }^3\theta }}$

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

Hence proved.