Explanation:
The given equation is False, so cannot be proven to be true.
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Perhaps you want to prove ...
[tex]2\tan{x}=\dfrac{\cos{x}}{\csc{(x)}-1}+\dfrac{\cos{x}}{\csc{(x)}+1}[/tex]
This is one way to show it:
[tex]2\tan{x}=\cos{(x)}\dfrac{(\csc{(x)}+1)+(\csc{(x)}-1)}{(\csc{(x)}-1)(\csc{(x)}+1)}\\\\=\cos{(x)}\dfrac{2\csc{(x)}}{\csc{(x)}^2-1}=2\cos{(x)}\dfrac{\csc{x}}{\cot{(x)}^2}=2\dfrac{\cos{(x)}\sin{(x)}^2}{\cos{(x)}^2\sin{(x)}}\\\\=2\dfrac{\sin{x}}{\cos{x}}\\\\2\tan{x}=2\tan{x}\qquad\text{QED}[/tex]
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We have used the identities ...
csc = 1/sin
cot = cos/sin
csc^2 -1 = cot^2
tan = sin/cos
