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

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Answer 1 · 2017-09-14T16:40:57

#(sin^2x+cos^2)/(sin x + cos x)=color(red)(1/(sin x +cos x)#

Based on the Pythagorean Theorem and basic definitions of #sin# and #cos#:
#sin^2(x)+cos^2(x)=1#

Answer 2 · 2017-09-15T08:54:21

#(sin^2x+cos^2x)/(sinx+cosx)=1/sqrt2csc(x+pi/4)#

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

= #1/(sqrt2(sinx xx 1/sqrt2+cosx xx 1/sqrt2)#

= #1/(sqrt2(sinx xx cos(pi/4)+cosx xx sin(pi/4))#

= #1/(sqrt2sin(x+pi/4))#

= #1/sqrt2csc(x+pi/4)#