How can this be reduced to the simplest form?

#sinx/(1+cosx)# + #cos/sinx#
Please include a detailed explanation for the process of simplifying it.

2 Answers
VNVDVI
Mar 2, 2018

#cscx#

Explanation:

First, let's take a common denominator so that we can add our two rational expressions (fractions).

#sinx(1+cosx)# will be our common denominator; thus, we will multiply #sinx/(1+cosx)# by #sinx/sinx# and #cosx/sinx# by #(1+cosx)/(1+cosx)#. Both of these actions are the same as multiplying by #1,# we're just trying to get our common denominator:

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

Since we have a common denominator, we can combine everything into one fraction:

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

Recall the identity

#sin^2x+cos^2x=1#

We can apply this in our numerator, as #sin^2x+cos^2x# shows up:

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

#(1+cosx)# cancels out:

#(cancel(1+cosx))/(sinxcancel(1+cosx))=1/sinx=cscx#

Shwetank Mauria
Mar 2, 2018

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

Explanation:

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

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

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

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

= #1/sinx#

= #cscx#

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