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Explanation:

In △OBD, ∠OBD = 90 - 75 = 15° 
In △EBC, ∠ECB = 90 - 15 = 75° 
Similarly, ∠B = 75° and ∠A = 180 - 75 - 75 = 30°

Area of △ABC = $\frac{1}{2}\times \mathrm{AD}\times \mathrm{BC}$ = $\frac{1}{2}\times \mathrm{BE}\times \mathrm{AC}$

⇒ $\frac{\mathrm{AD}}{\mathrm{BE}}$ = $\frac{\mathrm{AC}}{\mathrm{BC}}$   ...(2)

Area of △ABC = $\frac{1}{2}\times \mathrm{AB}\times \mathrm{AC}\times \mathrm{Sin} 30°$ = $\frac{1}{2}\times A\mathrm{B}\times \mathrm{BC}\times \mathrm{Sin} 75°$

⇒ $\frac{\mathrm{AC}}{\mathrm{BC}}$ = $\frac{Sin 75°}{Sin 30°}$ = 2 × Sin 75° = 2 × Cos (90 - 15)° = 2 × Cos 15°

⇒ $\frac{\mathrm{AD}}{\mathrm{BE}}$ = $\frac{\mathrm{AC}}{\mathrm{BC}}$ = 2 × Cos 15°

Hence, option (a).