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

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(B) We use the identity $	an A = \frac{\sin 2A}{1 + \cos 2A}$.Setting $A = 7\frac{1}{2}^\circ$,we get $2A = 15^\circ$.Thus,$	an 7\frac{1}{2}^\circ =

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

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(B) We use the identity $	an A = \frac{\sin 2A}{1 + \cos 2A}$.Setting $A = 7\frac{1}{2}^\circ$,we get $2A = 15^\circ$.Thus,$	an 7\frac{1}{2}^\circ = \frac{\sin 15^\circ}{1 + \cos 15^\circ}$.We know that $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$.Substituting these values: $	an 7\frac{1}{2}^\circ = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{1 + \frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{4 + \sqrt{6} + \sqrt{2}}$.Multiplying the numerator and denominator by the conjugate $(4 + \sqrt{2}) - \sqrt{6}$,we simplify to get $\sqrt{6} - \sqrt{3} + \sqrt{2} - 2$.

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

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