Question

Prove that: ${\cot }^{2}A-{\tan }^{2}A=4\cot 2A\mathrm{cosec}2A$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\cot }^{2}A-{\tan }^{2}A \\ =\frac{{\cos }^{2}A}{{\sin }^{2}A}-\frac{{\sin }^{2}A}{{\cos }^{2}A} \\ =\frac{{\left({\cos }^{2}A\right)}^2-{\left({\sin }^{2}A\right)}^2}{{\sin }^{2}A{\cos }^{2}A} \end{gathered}$$

$$\begin{gathered} =\frac{\left({\cos }^{2}A+{\sin }^{2}A\right)\left({\cos }^{2}A-{\sin }^{2}A\right)}{{\sin }^{2}A{\cos }^{2}A} \\ =\frac{1\times \left(\cos 2A\right)}{{\sin }^{2}A{\cos }^{2}A}\left(\because {\cos }^{2}A-{\sin }^{2}A=\cos 2A\right) \\ =\frac{4\cos 2A}{{\displaystyle 4{\sin }^{2}A{\cos }^{2}A}} \end{gathered}$$

$$\begin{gathered} =\frac{4\left(\cos 2\mathrm{A}\right)}{{\left(\sin 2A\right)}^2} \\ =\frac{4\left(\cos 2A\right)}{\left(\sin 2A\right)}\times \frac{1}{\left(\sin 2A\right)} \\ =4\cot 2A\mathrm{cosec}2A=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$