How do you simplify #tanx + cosx/(1 + sinx)#?

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Answer 1 · 2016-05-09T02:02:39

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

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

Applying the Pythagorean identity #sin^2x + cos^2x = 1#:

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

#1/cosx#

Applying the reciprocal identity #1/cosx = secx#:

#secx#

Hopefully this helps!