<s>[INST] 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?
 [/INST] Another possibility, just continuing to get common denominators:
$$\sin A \left(1 + {\sin A\over \cos A}\right) + \cos A\left(1 + {\cos A\over \sin A}\right) \\ = \sin A \left({\cos A + \sin A\over \cos A}\right) + \cos A \left({\sin A + \cos A\over \sin A}\right) \\ = {\sin A(\cos A + \sin A)\over \cos A} + {\cos A (\sin A + \cos A)\over \sin A}\\= {\sin^2 A(\cos A + \sin A) + \cos^2 A(\sin A + \cos A)\over \cos A \sin A} \\ = {(\sin^2 A + \cos^2 A)(\cos A + \sin A)\over \cos A \sin A} \\ = {\cos A + \sin A\over \cos A \sin A} \\ = {1\over \sin A} + {1\over \cos A}$$
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