Question

Prove the following identities (1-17)

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

Answer 1

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


$$\begin{gathered} \mathrm{LHS}=\frac{\left(1+\mathrm{co}t\theta +\tan \theta \right)\left(\sin \theta -\cos \theta \right)}{se{c}^3\theta -\cos e{c}^3\theta } \\ =\frac{\left(1+{\displaystyle \frac{\cos \theta }{\sin \theta }}+{\displaystyle \frac{\sin \theta }{\cos \theta }}\right)\left(\sin \theta -\cos \theta \right)}{{\displaystyle \frac{1}{{\cos }^3\theta }}-{\displaystyle \frac{1}{si{n}^3\theta }}} \\ =\frac{\left(\sin \theta \cos \theta +{\cos }^2\theta +{\sin }^2\theta \right)\left(\sin \theta -\cos \theta \right)\left({\sin }^2\theta {\cos }^2\theta \right)}{\left({\sin }^3\theta -{\cos }^3\theta \right)} \\ =\frac{\left(1+\sin \theta \cos \theta \right)\left(\sin \theta -\cos \theta \right)\left({\sin }^2\theta {\cos }^2\theta \right)}{\left(\sin \theta -\cos \theta \right)\left({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta \right)} \\ ={\sin }^2\theta {\cos }^2\theta \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$