Question

tan A / 1- cot A + cot A / 1- tan A = 1+ sec A cosec A.

Answer 1

(tan A)/(1 - cot A) + cot A /(1 - tan A)

= (tan A)/[(1 - (1/tan A)] + cot A /(1 - tan A)

= (tan² A)/[(tan A - 1)] + cot A /(1 - tan A)

= (tan² A)/[(tan A - 1)] - cot A /(tan A - 1)

= (tan² A - cot A) / (tan A - 1) = (tan2 A - 1/tan A) / (tan A - 1)

= (tan³ A - 1) / [tan A (tan A - 1)]

= (tan A - 1)(tan² A + tan A + 1) / [tan A (tan A - 1)]

= (tan² A + tan A + 1) / tan A

= 1 + tan A + cot A

= 1 + [(sin A/cosA) + (cos A/sin A)]

= 1 + [(sin² A + cos² A) / sin A cos A]

= 1 + [1 / (sin A cos A)]

= 1 + (sec A × cos A)

Hence proved