Alas, poor Whoppie, you are making it harder than it needs to be.
Remember, \(\displaystyle cscA=\frac{1}{sinA}\).
\(\displaystyle \L\\\frac{\frac{1}{sinA}(sin^{2}A+cos^{2}AtanA)}{sinA+cosA}\)
\(\displaystyle \L\\\frac{\frac{sin^{2}A}{sinA}+\frac{cos^{2}AtanA}{sinA}}{sinA+cosA}\)
Now, simplify the top and you should see it.
Remember what \(\displaystyle \L\\\frac{tanA}{sinA}\) is equivalent to?.