I need help simplifying this expression?: cos^2 (x) /(1+sin(x))

Recall that #cos^2x = 1- sin^2x#.

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

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

#=1 - sinx#

Hopefully this helps!

#"using the "color(blue)"trigonometric identity"#

#•color(white)(x)sin^2x+cos^2x=1#

#rArrcos^2x=1-sin^2x#

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

#1-sin^2x" is a "color(blue)"difference of squares"#

#[•color(white)(x)a^2-b^2=(a-b)(a+b)]#

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

Simplifying #cos^2x/(1+sinx#:

Because of the identity #sin^2x+cos^2x=1# (based on the Pythagorean Theorem), you can move the #sin^2x# to the other side of the equation and you get
#cos^2x=1-sin^2x#.

Substitute it in the equation, and you will get #(1-sin^2x)/(1+sinx#.

You can think of #1-sin^2x# as #1^2 - (sinx)^2#.

Based on the rule #(x-y)(x+y)=x^2-y^2#, you can therefore write #1-sin^2x# as #(1-sinx)(1+sinx)#.

Substitute it into the equation and you get #((1-sinx)(1+sinx))/(1+sinx#.

However, in both the numerator and denominator there is a #1+sinx#. These can cancel out, and, thus, you will get #1-sinx# as your final answer.

Hope this helps!