Find the value of: tan 15^{circ} | vedclass

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(A) We know that $	an 15^{\circ} = 	an (45^{\circ} - 30^{\circ})$.Using the formula $	an (x-y) = \frac{	an x - 	an y}{1 + 	an x 	an y}$:$	an 1

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$2-\sqrt{3}$

Vedclass · Answer

(A) We know that $	an 15^{\circ} = 	an (45^{\circ} - 30^{\circ})$.Using the formula $	an (x-y) = \frac{	an x - 	an y}{1 + 	an x 	an y}$:$	an 15^{\circ} = \frac{	an 45^{\circ} - 	an 30^{\circ}}{1 + 	an 45^{\circ} 	an 30^{\circ}}$$= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$$= \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 + 1 - 2\sqrt{3}}{3 - 1}$$= \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

Find the value of: $\tan 15^{\circ}$

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