Find the value of tan 105^circ. | vedclass

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(C) Step $1$: Express $105^\circ$ as $(60^\circ + 45^\circ)$.Step $2$: Use the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Step

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$\frac{1 + \sqrt{3}}{1 - \sqrt{3}}$

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(C) Step $1$: Express $105^\circ$ as $(60^\circ + 45^\circ)$.Step $2$: Use the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Step $3$: Substitute $A = 60^\circ$ and $B = 45^\circ$: $	an(60^\circ + 45^\circ) = \frac{	an 60^\circ + 	an 45^\circ}{1 - 	an 60^\circ 	an 45^\circ}$.Step $4$: Substitute values $	an 60^\circ = \sqrt{3}$ and $	an 45^\circ = 1$: $\frac{\sqrt{3} + 1}{1 - \sqrt{3} 	imes 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$.Step $5$: Rationalize the denominator: $\frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{3 + 1 + 2\sqrt{3}}{-2} = \frac{4 + 2\sqrt{3}}{-2} = -(2 + \sqrt{3})$.Step $6$: Comparing with options, option $C$ is $\frac{1 + \sqrt{3}}{1 - \sqrt{3}}$, which is equivalent to our result in Step $4$.

Find the value of $\tan 105^\circ$.

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