solve it in form of cos and sin pls

[tex]\text{L.H.S}\\\\=\dfrac{\cot A}{1- \tan A} + \dfrac{\tan A}{1- \cot A}\\\\\\=\dfrac{\tfrac{\cos A}{\sin A}}{1- \tfrac{\sin A}{\cos A}} + \dfrac{\tfrac{\sin A}{\cos A}}{1-\tfrac{\cos A}{\sin A}}\\\\\\=\dfrac{\tfrac{\cos A}{\sin A}}{\tfrac{\cos A- \sin A}{\cos A}} + \dfrac{\tfrac{\sin A}{\cos A}}{\tfrac{\sin A - \cos A}{\sin A}}\\\\\\=\dfrac{\cos^2 A}{\sin A(\cos A-\sin A)} + \dfrac{\sin^2 A}{\cos A(\sin A - \cos A)}\\\\\\[/tex]
[tex]=\dfrac{\cos^2 A}{\sin A(\cos A-\sin A)} - \dfrac{\sin^2 A}{\cos A(\cos A - \cos A)}\\\\\\=\dfrac 1{\cos A - \sin A} \left( \dfrac{\cos^2 A}{\sin A} - \dfrac{\sin^2 A}{\cos A}\right)\\\\\\=\dfrac 1{\cos A - \sin A} \left(\dfrac{\cos^3 A - \sin^3 A}{\sin A \cos A} \right)\\\\\\=\dfrac 1{\cos A - \sin A} \left[\dfrac{(\cos A - \sin A)(\cos^2 A + \sin^2 A + \sin A \cos A)}{\sin A \cos A} \right]\\\\\\=\dfrac{\cos^2 A + \sin^2 A+\sin A \cos A}{\sin A \cos A}\\\\\\[/tex]
[tex]=\dfrac{\cos^2 A}{\sin A \cos A}+\dfrac{\sin^2 A}{\sin A \cos A}+\dfrac{\sin A \cos A}{\sin A \cos A}\\\\\\=\dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A} + 1\\\\\\=\cot A + \tan A + 1\\\\\\=1 + \tan A + \cot A\\\\=\text{R.H.S}\\\\\text{Proved.}[/tex]