How do you prove $(tan u + cot u)(cos u + sin u) = csc u + sec u$ ?

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Answer 1 · 2015-09-01T21:08:57

First, expand the Left Hand Side $("LHS")$

$"LHS"=(tan u + cot u)(cos u + sin u)$

$=tanucosu+tanusinu+cotucosu+cotusinu$

Recall that : $tanu=sinu/cosu$
and $" "cotu=cosu/sinu$

$=>"LHS"=sinu/cancel(cosu)*cancel(cosu)+sinu/cosu*sinu+cosu/sinu*cosu+cosu/cancel(sinu)*cancel(sinu)$

$=sinu+sin^2u/cosu+cos^2u/sinu+cosu$

$=sinu+cos^2u/sinu+cosu+sin^2u/cosu$

$=sin^2u/sinu+cos^2u/sinu+cos^2u/cosu+sin^2u/cosu$

$=(sin^2u+cos^2u)/sinu+(cos^2u+sin^2u)/cosu$

Recall again that : $sin^2u + cos^2u=1$

$=>"LHS"=1/sinu+1/cosu$

Recall that : $1/sinu=cscu$
and also $" "1/cosu=secu$

$=>"LHS"=color(blue)(cscu+secu)$

Quad Era Demonstrandum

How do you prove (tan u + cot u)(cos u + sin u) = csc u + sec u(tan u + cot u)(cos u + sin u) = csc u + sec u?