How do you simplify # [(1 + sin theta)/cos theta] + [cos theta/(1 + sin theta)]#?

2 Answers
Apr 25, 2016

#2sectheta#

Explanation:

Multiply the fractions to achieve a common denominator.

#=[(1+sintheta)/costheta][(1+sintheta)/(1+sintheta)]+[costheta/(1+sintheta)][costheta/costheta]#

#=(1+2sintheta+sin^2theta)/(costheta(1+sintheta))+cos^2theta/(costheta(1+sintheta))#

#=(1+2sintheta+(sin^2theta+cos^2theta))/(costheta(1+sintheta))#

Recall the Pythagorean Identity #sin^2theta+cos^2theta=1#.

#=(2+2sintheta)/(costheta(1+sintheta))#

#=(2(1+sintheta))/(costheta(1+sintheta))#

#=2/costheta#

#=2sectheta#

Apr 25, 2016

# 2sectheta #

Explanation:

Begin by writing the fractions as a single fraction by extracting the lowest common denominator
In this case # costheta(1 + sintheta)#

#rArr( (1 +sintheta)^2 + cos^2 theta)/(costheta(1+sintheta)#

Expanding the numerator

#( (1+2sintheta+sin^2theta) + cos^2theta)/(costheta(1+sintheta))#

using the identity #(sin^2theta+cos^2theta=1)#

then # (1+2sintheta+1)/(costheta(1+sintheta)) =(2+2sintheta)/(costheta(1+sintheta))#

#=(2(1+sintheta))/(costheta(1+sintheta))=(2cancel((1+sintheta)))/(costhetacancel((1+sintheta)))#

#= 2/costheta=2sectheta#