How do you prove that #(1 + secx)/secx = sin^2x/(1 - cosx)#?

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Answer 1 · 2017-08-26T16:20:38

We have:

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

This means that

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

We can use #sin^2x + cos^2x = 1# to rewrite the right hand side.

#(1 + 1/cosx)/(1/cosx) = (1 - cos^2x)/(1 - cosx)#

Now it's time to simplify. We can use the difference of squares formula to say that #1 - cos^2x= (1 + cosx)(1 - cosx)#.

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

#(cosx + 1)/cosx * cosx = 1 + cosx#

#cosx + 1 = cosx +1#

This is clearly true for all values of #x#, therefore, the identity is true.

Hopefully this helps!