[tex]\tan x=\dfrac{\sin x}{\cos x}\qquad\qquad\cot x=\dfrac{\cos x}{\sin x}\\\\\\
\sin^2x+\cos^2x=1\implies\sin^2x=1-\cos^2x\\\\\\\\ R\,=\,\tan x-\cot x=\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}=\dfrac{\sin x}{\cos x}\cdot\dfrac{\sin x}{\sin x}-\dfrac{\cos x}{\sin x}\cdot\dfrac{\cos x}{\cos x}=\\\\\\=
\dfrac{\sin^2x}{\sin x\cdot\cos x}-\dfrac{\cos^2x}{\sin x\cdot\cos x}=\dfrac{\sin^2x-\cos^2x}{\sin x\cdot\cos x}=\\\\\\=
\dfrac{(1-\cos^2x)-\cos^2x}{\sin x\cdot\cos x}=\dfrac{1-2\cos^2x}{\sin x\cdot\cos x}=L[/tex]
