Which of the following identities, wherever defined, hold(s) good?

$\left(1\right) \frac{{\cos }^2\mathrm{\alpha }-{\sin }^2\mathrm{\alpha }}{\sin \mathrm{\alpha cos}\mathrm{\alpha }}=2\cot 2\mathrm{\alpha }$

$\left(2\right) \frac{1+t}{1-t}-\frac{1-t}{1+t},\text{ where }$

       $\begin{array}{l}\mathrm{t}=\tan \mathrm{\alpha }=\frac{(1+\mathrm{t}{)}^2-(1-\mathrm{t}{)}^2}{1-{\mathrm{t}}^2}\\ =\frac{4\mathrm{t}}{1-{\mathrm{t}}^2}=\frac{2\times 2\tan \mathrm{\alpha }}{1-{\tan }^2\mathrm{\alpha }}=2\tan 2\mathrm{\alpha }\end{array}$

         Hence, (2) is incorrect. 

$\begin{array}{c}\left(3\right) \frac{1+\mathrm{t}}{1-\mathrm{t}}+\frac{1-\mathrm{t}}{1+\mathrm{t}}=\frac{(1+\mathrm{t}{)}^2+(1-\mathrm{t}{)}^2}{1-{\mathrm{t}}^2}\\ =\frac{2\left(1+{\mathrm{t}}^2\right)}{1-{\mathrm{t}}^2}\\ =\frac{2}{\cos 2\mathrm{\alpha }}=2\sec 2\mathrm{\alpha }\end{array}$

           Hence, (3) is correct. 

$\left(4\right) \tan \mathrm{\alpha }+\cot \mathrm{\alpha }=\frac{1}{\cos \mathrm{\alpha sin}\mathrm{\alpha }}=\frac{2}{\sin 2\mathrm{\alpha }}=2\mathrm{cosec}2\mathrm{\alpha }$

         Hence, (4) is incorrect.