tan 15^circ = | vedclass

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Question

(C) We know that $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.$	an 15^\circ = 	an(45^\circ - 30^\circ)$$= \frac{	an 45^\circ - 	an 30

Vedclass · Accepted Answer

$2 - \sqrt{3}$

Vedclass · Answer

(C) We know that $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.$	an 15^\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 \cdot (1/\sqrt{3})} = \frac{(\sqrt{3} - 1)/\sqrt{3}}{(\sqrt{3} + 1)/\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$Rationalizing the denominator:$= \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

$\tan 15^\circ = $

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