How do you simplify #sinx/(1+cosx)+sinx/(1-cosx)#?

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Answer 1 · 2016-11-23T08:46:03

#2cscx#

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

providing #cosx!=+-1#

using the rules of fractions .

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

#((sinx-sinxcosx+sinx+sinxcosx))/(1-cos^2x)#

#((sinxcancel(-sinxcosx)+sinx+cancel(sinxcosx)))/sin^2x#

#=(2sinx)/sin^2x#

#=(2cancel(sinx))/cancel(sin^2x)^sinx#

#=2/sinx=2cscx#