Answer 1

Since, 50° = 10° + 40°

∴ tan50° = tan (10° + 40°)

∴ tan50° = `(tan10^circ + tan40^circ)/(1 - tan10^circ tan40^circ)`

∴ tan50° (1 – tan10° tan40°) = tan10° + tan40°

∴ tan50° – tan10° tan40° tan50° = tan10° + tan40°

∴ tan50° – tan10° tan40° tan (90° – 40°) = tan10° + tan40°

∴ tan50° – tan10° tan40° cot40° = tan10° + tan40° ...[∵ tan(90° – θ) = cot θ]

∴ `tan50^circ - tan10^circ tan40^circ*1/tan40^circ` = tan10° + tan40° 

∴ tan50° – tan10° . 1 = tan10° + tan40°

∴ tan50° = tan40° + 2 tan10°