Question #88b3e

1 Answer
Feb 13, 2018

#1/cosx#

Explanation:

Convert everything to #sinx# and #cosx#:

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

Dividing by #1/sinx# is the same as multiplying by the reciprocal #sinx#

#(sinx)(sinx/cosx+cosx/sinx)#

Distribute

#sin^2x/cosx+(sinxcosx)/sinx#

Simplify

#sin^2x/cosx+cosx/1#

The common denominator is #cosx#

#sin^2x/cosx+cos^2x/cosx#

Add fractions

#(sin^2x+cos^2x)/cosx#

#sin^2x+cos^2x=1# (Pythagorean Identity) Sooooo:

#(sin^2x+cos^2x)/cosx=1/cosx#

That's your answer in terms of #cosx#. Most people would convert #1/cosx# to #secx#, though.