Answer:
[tex]1+\cos x[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{4cm}\underline{Trigonometric Identities}\\\\$\csc \theta=\dfrac{1}{\sin \theta}\\\\\\\cot \theta=\dfrac{\cos \theta}{\sin\theta}\\\\\\\cos^2 \theta + \sin^2 \theta = 1$\\\end{minipage}}[/tex]
[tex]\begin{aligned}\implies \dfrac{ \sin x}{ \csc x - \cot x} & = \dfrac{ \sin x}{\dfrac{1}{\sin x} - \dfrac{\cos x}{ \sin x}}\\\\& = \dfrac{ \sin x}{\dfrac{1 - \cos x}{\sin x}}\\\\& = \sin x \times \dfrac{\sin x}{1 - \cos x}\\\\& = \dfrac{\sin^2 x}{1-\cos x}\\\\& = \dfrac{1-\cos^2x}{1-\cos x}\\\\& = \dfrac{(1-\cos x)(1+ \cos x)}{1-\cos x}\\\\& = 1 + \cos x\end{aligned}[/tex]
