The set of all possible values of α in [−π,π] such that 1−sin⁡α1+sin⁡α is equal to sec⁡α−tan⁡α, is

Clearly, sec⁡α−tan⁡α is not defined for α=±π/2 Now,  1−sin⁡α1+sin⁡α=(1−sin⁡α)2cos2⁡α⇒ 1−sin⁡α1+sin⁡α=1−sin⁡α|cos⁡α|=sec⁡α−tan⁡α,ifcos⁡α>0Clearly, cos⁡α>0⇒α∈(−π/2,π/2)