$\begin{align} & \text{Show that }\frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{2}{{{\cos }^{2}}x}-\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}, \\ & \text{Hence or otherwise evaluate}\int{\frac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}} \\\end{align}$
$\begin{align} & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{d}{dx}\left( \operatorname{cosec}x\sec x \right) \\ & \text{Using the product rule} \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\operatorname{cosec}x\frac{d}{dx}(\sec x)+\sec x\frac{d}{dx}(\operatorname{cosec}x) \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\operatorname{cosec}x(\sec x\tan x)+\sec x(-\operatorname{cosec}x\cot x) \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{1}{\sin x}\left( \frac{1}{\cos x}\times \frac{\sin x}{\cos x} \right)-\frac{1}{\cos x}\left( \frac{1}{\sin x}\times \frac{\cos x}{\sin x} \right) \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{1}{{{\cos }^{2}}x}-\frac{1}{{{\sin }^{2}}x}=\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\cos }^{2}}x{{\sin }^{2}}x} \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{-\cos 2x}{{{\cos }^{2}}x{{\sin }^{2}}x}=\frac{-(1-2{{\sin }^{2}}x)}{{{\cos }^{2}}x{{\sin }^{2}}x} \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{2{{\sin }^{2}}x}{{{\cos }^{2}}x{{\sin }^{2}}x}-\frac{1}{{{\cos }^{2}}x{{\sin }^{2}}x} \\ & \frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{2}{{{\cos }^{2}}x}-\frac{1}{{{\cos }^{2}}x{{\sin }^{2}}x} \\ & \\ & \int{\frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)dx}=\int{\frac{2}{{{\cos }^{2}}x}dx-\int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}} \\ & \int{\frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)dx}=\int{2{{\sec }^{2}}x}-\int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx} \\ & \frac{1}{\sin x\cos x}=2\tan x-\int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx} \\ & \int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=2\tan x-\frac{1}{\sin x\cos x} \\ & \int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=2\tan x-\operatorname{cosec}x\sec x+C \\\end{align}$