Harder Trigonometry Identity ($\sec A+\csc A$) How do I prove:
$\sin A (1 + \tan A) + \cos A (1 + \cot A) = \sec A + \csc A$
I've tried expanding the brackets by multiplying sin A and cos A to the left hand side but to no avail. Where should I start from?

$\tan A=\sin A/\cos A$, and $\cot A=\cos A/\sin A$, so that making this substitution, we have \begin{align} 
\sin A\left(1+\frac{\sin A}{\cos A}\right)+\cos A\left(1+\frac{\cos A}{\sin A}\right) & =\sin A + \frac{\sin^2 A}{\cos A}+\cos A +\frac{\cos^2 A}{\sin A} \\
& =\sin A+\cos A+\frac{1-\cos^2 A}{\cos A}+\frac{1-\sin^2 A}{\sin A} \\
& = \sin A+\cos A+\sec A-\cos A+\csc A-\sin A \\
& = \sec A +\csc A
\end{align}
Where I also made use of $\cos^2 A+\sin^2 A=1$.
