The value of tan 75° – cot75° is

$\tan 75^{\circ}-\cot 75^{\circ}$

$=\tan 75^{\circ}-\frac{1}{\tan 75^{\circ}}$

$=\frac{\tan ^{2} 75^{\circ}-1}{\tan 75^{\circ}}$

$=-\frac{\left(1-\tan ^{2} 75^{\circ}\right)}{\tan 75^{\circ}} \times \frac{2}{2} \quad$ (multiply and divide by 2 )

$=-2\left[\frac{1-\tan ^{2} 75^{\circ}}{2 \tan 75^{\circ}}\right]$

$=-2\left[\frac{1}{\tan 150^{\circ}}\right] \quad$ (Using identity $\left.\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)$

$=-2 \cot 150^{\circ}$

$=-2 \cot \left(90^{\circ}+60^{\circ}\right)$

$=-2\left(-\tan 60^{\circ}\right)$   $\left(\because \cot \left(90^{\circ}+\theta\right)=-\tan \theta\right)$

$=2 \times \sqrt{3}$

$\therefore \tan 75^{\circ}-\cot 75^{\circ}=2 \sqrt{3}$

Hence, the correct answer is A.