tan 75^circ - cot 75^circ = | vedclass

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Question

(A) We know that $	an 75^\circ = 	an(45^\circ + 30^\circ) = \frac{	an 45^\circ + 	an 30^\circ}{1 - 	an 45^\circ 	an 30^\circ} = \frac{1 + 1/\sqr

Vedclass · Accepted Answer

$2\sqrt{3}$

Vedclass · Answer

(A) We know that $	an 75^\circ = 	an(45^\circ + 30^\circ) = \frac{	an 45^\circ + 	an 30^\circ}{1 - 	an 45^\circ 	an 30^\circ} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.Since $\cot 75^\circ = \frac{1}{	an 75^\circ} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.Therefore,$	an 75^\circ - \cot 75^\circ = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}$.

$\tan 75^\circ - \cot 75^\circ = $

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