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Question:
differentiate w.r.t.x : y = squre root of 1 +sinx/1-sinx
Answer:

Given y = √{(1 + sinx)/(1 - sinx)}

=> y = √[{(1 + sinx)*(1 + sinx)}/{(1 - sinx)*(1 + sinx)}]    (Multiply numerator and denominator by (1+sinx))

=> y=  √[{(1 + sinx)2 }/{(1 - sin2 x)}]

=> y=  √[{(1 + sinx)2 }/cos2 x]

=> y=  (1 + sinx)/cosx

Now 

dy/dx = {cosx*cosx  - (1 + sinx)(-sinx)}/cos2 x

         = (cos2 x + sinx + sin2 x)/cos2 x

         = (1 + sinx)/cos2 x  (Since cos2 x + sin2 x = 1 )

 So derivative of √{(1 + sinx)/(1 - sinx)} = (1 + sinx)/cos2 x

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