Verify the given identity

1/(1-sinx)+1/(1+sinx)=2sec(^2)x

1/(1-sinx) * (1+sinx) + 1/(1+sinx) * (1-sinx)


[1+sinx+1-sinx]/(1-sin^2x)

2/cos^2x


=2sec^2x.

Start out by combining fractions on the left side to get:
LHS = 1/[1 - sin(x)] + 1/[1 + sin(x)]
= [1 + sin(x)]/{[1 + sin(x)][1 - sin(x)]} + [1 - sin(x)]/{[1 + sin(x)][1 - sin(x)]}
(Note that the LCD is [1 + sin(x)][1 - sin(x)].)
= [1 + sin(x) + 1 - sin(x)]/{[1 + sin(x)][1 - sin(x)]}, by adding
= 2/[1 - sin^2(x)], by simplifying the numerator and denominator
= 2/cos^2(x), since cos^2(x) + sin^2(x) = 1
= 2sec^2(x), since cos(x) = 1/sec(x)
= RHS.

I hope this helps!