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Q. The value of $\tan 75^{\circ}-\cot 75^{\circ}$ is

MHT CETMHT CET 2021

Solution:

$\tan 75^{\circ}-\cot 75^{\circ} $
$ =\frac{\sin 75^{\circ}}{\cos 75^{\circ}}-\frac{\cos 75^{\circ}}{\sin 75^{\circ}} $
$ =\frac{\sin ^2 75^{\circ}-\cos ^2 75^{\circ}}{\sin 75^{\circ} \cdot \cos 75^{\circ}}=\frac{-2 \cos 150^{\circ}}{\sin 150^{\circ}} $
$ =\frac{-2 \cos \left(90^{\circ}+60^{\circ}\right)}{\sin \left(90+60^{\circ}\right)}$
$ =\frac{2 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}}=2 \sqrt{3}$