Prove that (secx+tanx)^2=1+sinx/1-sinx

Question

Thanks in advance.

Answer 1

(secx + tanx)^2 = [(1 + sinx)/cosx]^2
= [1 + sinx]^2/[1 - sin^2x]
= (1 + sinx)(1 + sinx)/[(1 + sinx)(1-sinx)]
= (1+sinx)/(1-sinx)