Simplify (1- cos theta +sin theta)/(1+ cos theta + sin theta)?

1 Answer
MattyMatty
Feb 11, 2018

#=sin(theta)/(1+cos(theta))#

Explanation:

#(1-cos(theta)+sin(theta))/(1+cos(theta)+sin(theta))#
#=(1-cos(theta)+sin(theta))*(1+cos(theta)+sin(theta))/(1+cos(theta)+sin(theta))^2#
#=((1+sin(theta))^2-cos^2(theta))/(1+cos^2(theta)+sin^2(theta)+2 sin(theta)+2 cos(theta) + 2 sin(theta) cos(theta))#
#=((1+sin(theta))^2-cos^2(theta))/(2+2 sin(theta)+2 cos(theta) + 2 sin(theta) cos(theta))#
#=((1+sin(theta))^2-cos^2(theta))/(2(1+cos(theta)) + 2 sin(theta)(1+cos(theta))#
#=(1/2)((1+sin(theta))^2-cos^2(theta))/((1+cos(theta))(1+sin(theta))#
#=(1/2)(1+sin(theta))/(1+cos(theta))-(1/2)(cos^2(theta))/((1+cos(theta))(1+sin(theta)))#

#=(1/2)(1+sin(theta))/(1+cos(theta))-(1/2)(1-sin^2(theta))/((1+cos(theta))(1+sin(theta)))#
#=(1/2)(1+sin(theta))/(1+cos(theta))-(1/2)((1-sin(theta))*(1+sin(theta)))/((1+cos(theta))(1+sin(theta)))#

#=(1/2)(1+sin(theta))/(1+cos(theta))-(1/2)(1-sin(theta))/(1+cos(theta))#
#=sin(theta)/(1+cos(theta))#

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