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Prove that $\tan 45^{\circ} = 1$ geometrically.
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Consider an isosceles right $\Delta ABC$
Using angle sum property $\angle A = \angle C = 45^{\circ}$
Clearly, $\tan 45^{\circ} = \frac{AB}{BC} = \frac{x}{x} = 1$
Using angle sum property $\angle A = \angle C = 45^{\circ}$
Clearly, $\tan 45^{\circ} = \frac{AB}{BC} = \frac{x}{x} = 1$