How do you prove #(1+sinx)/(1-sinx)+(sinx-1)/(1+sinx)#?

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Answer 1 · 2016-10-23T02:01:56

The expression simplifies to #4sinxsec^2x#.

Put on a common denominator.

#=((1 + sinx)(1 + sinx))/((1 -sinx)(1 + sinx)) + ((sinx - 1)(1 - sinx))/((1 + sinx)(1 - sinx))#

#=(1 + 2sinx + sin^2x - sin^2x + sinx + sinx - 1)/(1 - sin^2x)#

Apply the identity #sin^2theta + cos^2theta = 1 -> cos^2theta = 1 - sin^2theta# to the numerator and simply the denominator.

#=(4sinx)/cos^2x#

Apply the identity #1/cosbeta = secbeta#.

#=4sinxsec^2x#

Hopefully this helps!