(a) Prove that: (sec^3 θ)/(sec^2 θ-1)+(cosec^3 θ)/(cosec^2 θ-1)= sec θ · cosec θ( sec θ+ cosec θ)

\[\begin{aligned} \mathrm{LHS} & =\frac{\sec ^{3} \theta}{\left(\sec ^{2} \theta-1\right)}+\frac{\operatorname{cosec}^{3} \theta}{\left(\operatorname{cosec}^{2} \theta-1\right)} \\ & =\frac{\sec ^{3} \theta}{\tan ^{2} \theta}+\frac{\operatorname{cosec}^{3} \theta}{\cot ^{2} \theta} \\ & =\frac{1}{\cos ^{3} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta}+\frac{1}{\sin ^{3} \theta} \times \frac{\sin ^{2} \theta}{\cos ^{2} \theta} \\ & =\frac{1}{\cos \theta \sin ^{2} \theta}+\frac{1}{\sin \theta \cos ^{2} \theta} \end{aligned} \]\[\begin{array}{l} =\frac{1}{\sin \theta \cos \theta}\left[\frac{1}{\sin \theta}+\frac{1}{\cos \theta}\right] \\ =\sec \theta \cdot \operatorname{cosec} \theta(\sec \theta+\operatorname{cosec} \theta)=\mathrm{RHS} \\ \text { OR } \end{array} \]If $\displaystyle \frac{\sec \alpha}{\operatorname{cosec} \beta}=p$ and $\displaystyle \frac{\tan \alpha}{\operatorname{cosec} \beta}=q$, then prove that $\displaystyle \left(p^{2}-q^{2}\right) \sec ^{2} \alpha=p^{2}$.LHS $\displaystyle =\left(p^{2}-q^{2}\right) \sec ^{2} \alpha$ \[\begin{array}{l} =\left(\frac{\sec ^{2} \alpha}{\operatorname{cosec}^{2} \beta}-\frac{\tan ^{2} \alpha}{\operatorname{cosec}^{2} \beta}\right) \times \sec ^{2} \alpha \\ =\left(\frac{\sec ^{2} \alpha-\tan ^{2} \alpha}{\operatorname{cosec}^{2} \beta}\right) \times \sec ^{2} \alpha \\ =\left(\frac{1}{\operatorname{cosec}^{2} \beta}\right) \times \sec ^{2} \alpha \\ =p^{2}=\text { RHS } \end{array} \]