Question

Show that $\frac{{\sin }^3({180}^o+A).\tan ({360}^o-A).{\sec }^2({180}^o-A)}{{\cos }^2({90}^o+A){\csc }^{2}A.\sin ({180}^o-A)}={\tan }^{3}A$.

Answer 1

$\begin{array}{c}LHS=\frac{{\sin }^3\left(180+A\right)\tan \left(360-A\right)\cdot {\sec }^2\left(180-A\right)}{{\cos }^2\left(90+A\right)\cos e{c}^{2}A\cdot \sin \left(180-A\right)}\\ \frac{-{\sin }^{3}A\times -\tan A\times {\sec }^{2}A}{{\sin }^{2}A\times \cos e{c}^{2}A\times \sin A}\\ =\tan A\times \frac{{\sin }^{2}A}{{\cos }^{2}A}\\ =\tan A\times {\tan }^{2}A\\ ={\tan }^{3}A=RHS\end{array}$