Question #d1e3c

1 Answer
Noah G
Dec 4, 2016

Put on a common denominator and use the identity #sectheta = 1/costheta#.

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

Use the identity #sin^2theta + cos^2theta = 1#.

#(1 + 2sinx + sin^2x + cos^2x)/(cosx + cosxsinx) = 2/cosx#

#(1 + 2sinx + 1)/(cosx + cosxsinx) = 2/cosx#

#(2 + 2sinx)/(cosx + cosxsinx) = 2/cosx#

#(2(1 + sinx))/(cosx(1 + sinx)) = 2/cosx#

#2/cosx = 2/cosx#

Hopefully this helps!

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