How do you prove #(1+sec[theta])/(tan[theta]+sin[theta])= csc[theta]#?

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Answer 1 · 2016-05-03T20:02:43

see below

Left Side:#=(1+sec theta)/(tan theta +sin theta)#

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

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

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

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

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

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

#=1/sin theta#

#=csc theta#

#=#Right Side