Hello, jimbo10!
This is a particularly tricky one . . .
Prove: \(\displaystyle \,\sec\theta(1\,+\,\sin\theta) \:=\:\tan\theta(1\,+\,\csc\theta)\)
The right side is: \(\displaystyle \L\,\tan\theta(1\,+\,\csc\theta) \;= \;\frac{\sin\theta}{\cos\theta}\left(1\,+\,\frac{1}{\sin\theta}\right)\)
Multiply: \(\displaystyle \L\,\frac{\sin\theta}{\cos\theta}\,+\,\frac{1}{\cos\theta}\)
Since \(\displaystyle \frac{1}{\cos\theta}\,=\,\sec\theta\), we have: \(\displaystyle \L\,\sin\theta\cdot\sec\theta\,+\,\sec\theta\)
Factor: \(\displaystyle \L\,\sec\theta(\sin\theta\,+\,1)\)
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I tried it "the other way", but it took Olympic-level gymnastics.
On the left, we have: \(\displaystyle \,\sec\theta(1\,+\,\sin\theta) \;= \;\L\frac{1}{\cos\theta}\,\)\(\displaystyle (1\,+\,\sin\theta)\)
Multiply: \(\displaystyle \L\,\frac{1}{\cos\theta}\,+\,\frac{\sin\theta}{\cos\theta}\)
Multiply the first fraction by \(\displaystyle \frac{\sin\theta}{\sin\theta}\):
\(\displaystyle \L\;\;\frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\sin\theta}\,+\,\frac{\sin\theta}{\cos\theta} \:=\;\frac{\sin\theta}{\cos\theta}\cdot\frac{1}{\sin\theta}\,+\,\frac{\sin\theta}{\cos\theta} \;=\)\(\displaystyle \;\tan\theta\cdot\csc\theta\,+\,\tan\theta\)
Factor: \(\displaystyle \,\tan\theta(\csc\theta\,+\,1)\)
