[tex]\dfrac{\tan x}{\csc x}+\dfrac{\sin x}{\tan x}=\dfrac{\sin x}{\cos x}:\dfrac{1}{\sin x}+\sin x:\dfrac{\sin x}{\cos x}\\\\=\dfrac{\sin x}{\cos x}\cdot\dfrac{\sin x}{1}+\sin x\cdot\dfrac{\cos x}{\sin x}=\dfrac{\sin^2x}{\cos x}+\cos x\\\\=\dfrac{\sin^2x}{\cos x}+\dfrac{\cos x\cdot\cos x}{\cos x}=\dfrac{\sin^2x}{\cos x}+\dfrac{\cos^2x}{\cos x}\\\\=\dfrac{\sin^2x+\cos^2x}{\cos x}=\dfrac{1}{\cos x}=\sec x\\\\Answer:\boxed{\dfrac{\tan x}{\csc x}+\dfrac{\sin x}{\tan x}=\sec x}[/tex]
[tex]Used:\\\csc x=\dfrac{1}{\sin x};\ \sec x=\dfrac{1}{\cos x}\\\\\tan x=\dfrac{\sin x}{\cos x}\\\\\sin^2x+\cos^2x=1[/tex]
