Question

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

Answer 1

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

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

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