How do you simplify the expression #(sinx+cosx)/(sinxcosx)#?

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Answer 1 · 2018-04-03T03:46:06

#(sinx+cosx)/(sinxcosx)=color(red)(secx+cscx)#

(see simplification process below)

#(sinx+cosx)/(sinxcosx)#

#rArrsinx/(sinxcosx)+cosx/(sinxcosx)#

#rArrcancelsinx/(cancelsinxcosx)+cancelcosx/(sinxcancelcosx)#

#rArr1/cosx+1/sinx#

#rArrcolor(red)(secx+cscx)#

Answer 2 · 2018-04-03T03:49:19

sec x + csc x

#(sin x + cos x)/(sin x.cos x) = #
#= (sin x)/(sin x.cos x) + (cos x)/(sin x.cos x)= = sec x + csc x#