Question #38190

Rule number two or three for identities (depending on the math teacher): when you see fractions, add them. It's an extremely useful step that reveals more information than what you have to begin with.

Our common denominator for the addition here is #(sin)(1+cos)#. For the first term:
#sin/(1+cos)#

If we want #(sin)(1+cos)# on the bottom, then we need to multiply the bottom by #sin#; and what we do to the bottom, we do to the top. Therefore, this becomes:
#((sin)(sin))/((1+cos)(sin))#
#=((sin)(sin))/((1+cos)(sin))#

Note that I'm not in a hurry to simplify anything further. Leave everything by itself (for instance, don't combine #(1+cos)(sin)# to make #sin+(sin)(cos)# because this makes it more complicated than it needs to be).

Likewise, we multiply the second term by #(1+cos)/(1+cos)#, to get:
#((1+cos)(1+cos))/((sin)(1+cos))#

We now have:
#((sin)(sin))/((1+cos)(sin))+((1+cos)(1+cos))/((sin)(1+cos))#

We now add to get:
#((sin)(sin)+(1+cos)(1+cos))/((sin)(1+cos))#

Now we can multiply everything out, since we're obviously getting nowhere with the above expression:
#(sin^2+2cos+cos^2+1)/((sin)(1+cos))#

Recognize a Pythagorean Identity in here? I think this is the most challenging part of identities - at least, from what I saw from my classmates in precalc. It's kind of hard spotting identities in expressions like these, and it takes tons of practice. Recall that #sin^2+cos^2=1#; this is probably the most important identity in trig. It's also hiding in the numerator, right here:
#(color(red)(sin^2)+2cos+color(red)(cos^2)+1)/((sin)(1+cos))#

We can replace that with #1#, since it equals #1#:
#(color(red)(1)+2cos+1)/((sin)(1+cos))#

And add the two ones in the numerator to get #2#:
#(2+2cos)/((sin)(1+cos))#

Since both terms in the numerator have a common factor of two, we might try factoring out the two:
#(2(1+cos))/((sin)(1+cos))#

And what do you know, the #1+cos# cancels:
#(2cancel((1+cos)))/((sin)cancel((1+cos)))#
#=2/sin)=2csc->#since #1/sin)=csc#

And boom, we're done. Also, I've been using #sin# and #cos# to make the answer easier on the eyes, but we really should be using #sinx# and #cosx#. Doesn't matter to me, but it does matter to the people giving out the grades.