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

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(A) We use the half-angle formula: $	an A = \frac{1 - \cos 2A}{\sin 2A}$.Setting $A = 7 \frac{1}{2}^{\circ}$,we get $2A = 15^{\circ}$.Thus,$	an 7 \f

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$\frac{2 \sqrt{2}-(1+\sqrt{3})}{\sqrt{3}-1}$

Vedclass · Answer

(A) We use the half-angle formula: $	an A = \frac{1 - \cos 2A}{\sin 2A}$.Setting $A = 7 \frac{1}{2}^{\circ}$,we get $2A = 15^{\circ}$.Thus,$	an 7 \frac{1}{2}^{\circ} = \frac{1 - \cos 15^{\circ}}{\sin 15^{\circ}}$.We know that $\cos 15^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ and $\sin 15^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$.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{\frac{2 \sqrt{2} - (\sqrt{3} + 1)}{2 \sqrt{2}}}{\frac{\sqrt{3} - 1}{2 \sqrt{2}}} = \frac{2 \sqrt{2} - (1 + \sqrt{3})}{\sqrt{3} - 1}$.

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

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