The value of tan 7frac{1}{2}^circ is equal to | vedclass

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(B) We know the formula $	an 	heta = \frac{1 - \cos 2	heta}{\sin 2	heta}$.Setting $	heta = 7\frac{1}{2}^\circ$,we have $2	heta = 15^\circ$.$	an

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$\sqrt{6} - \sqrt{3} + \sqrt{2} - 2$

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(B) We know the formula $	an 	heta = \frac{1 - \cos 2	heta}{\sin 2	heta}$.Setting $	heta = 7\frac{1}{2}^\circ$,we have $2	heta = 15^\circ$.$	an 7\frac{1}{2}^\circ = \frac{1 - \cos 15^\circ}{\sin 15^\circ}$.We know $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.And $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.Substituting these values:$	an 7\frac{1}{2}^\circ = \frac{1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} = \frac{2\sqrt{2} - \sqrt{3} - 1}{\sqrt{3} - 1}$.Rationalizing the denominator:$= \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2\sqrt{6} + 2\sqrt{2} - 3 - \sqrt{3} - \sqrt{3} - 1}{3 - 1} = \frac{2\sqrt{6} + 2\sqrt{2} - 2\sqrt{3} - 4}{2} = \sqrt{6} + \sqrt{2} - \sqrt{3} - 2$.

The value of $\tan 7\frac{1}{2}^\circ$ is equal to

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