How do you rationalize the numerator for #sqrt((1+siny)/(1-siny))#?

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Answer 1 · 2016-09-03T14:42:12

The expression can be simplified to #(1 + siny)/cosy#.

#sqrt(1 + siny)/sqrt(1 - siny)#

Multiply by the entire expression by the numerator to cancel the square root:

#=sqrt(1 + siny)/sqrt(1 - siny) xx sqrt(1 + siny)/sqrt(1 + siny)#

#=(1 + siny)/sqrt(1 - sin^2y)#

This can be simplified further. Apply the pythagorean identity #1 - sin^2theta = cos^2theta#.

#=(1 + siny)/(sqrt(cos^2y)#

#=(1 + siny)/(sqrt(cosy xx cosy))#

#=(1 + siny)/cosy#

Hopefully this helps!