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For $A = 60^{\circ} and $B = 30^{°}, verify that $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
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(b) LHS = $\tan (60^{\circ} - 30^{\circ}) = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ (1/2 Mark)
RHS = $\frac{\tan 60^{\circ} - \tan 30^{\circ}}{1 + \tan 60^{\circ} \tan 30^{\circ}} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} = \frac{\frac{3-1}{\sqrt{3}}}{1+1} = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}}$ (1/2 Mark)
Hence verified
RHS = $\frac{\tan 60^{\circ} - \tan 30^{\circ}}{1 + \tan 60^{\circ} \tan 30^{\circ}} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} = \frac{\frac{3-1}{\sqrt{3}}}{1+1} = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}}$ (1/2 Mark)
Hence verified