trigonometry!!!!!!!!!!!!!!!!!!!

Question

show that this is an identity
(sec β/cos β) - (tan β/ cot β) = 1

Answer 1

LHS = (secβ / cosβ) - (tanβ / cotβ)

         = (1 / cos^2β) -  tan^2β                      [ Since secA = 1/cosA and 1/cotA = tanA ]

         = (1 / cos^2β) - sin^2β / cos^2β          [ Since tanA = sinA / cosA ]

        = 1/cos^2β- sin^2β / cos^2β             

        = (1 - sin^2β) / cos^2β                         [ By taking 1/cos^2A as common ]

        = cos^2β/ cos^2β                                 [ Since sin^2A + cos^2A = 1 ]

        = 1

        = RHS

Hence it is proved that (secβ / cosβ) - (tanβ / cotβ) = 1