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

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