<s>[INST] Verify trigonometry equation $\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A$ Sorry for asking so many of these type of questions. 
How would I verify the following trigonometry identity:
$$\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A.$$
My work is
$$\frac{\sin A + \frac{\sin A}{\cos A}}{\frac{\cos A}{\sin A} + \frac{1}{\sin A}}.$$
Do I have to use a common denominator between the sin and tan to solve the identity?
 [/INST] For what it's worth (at this late date)...
Since cotangent and cosecant are linked by the Pythagorean Identity, the "conjugate factor" method is helpful:
$$
\frac{\sin A + \tan A}{\cot A + \csc A} \cdot \frac{\cot A - \csc A}{\cot A - \csc A}  =  \frac{(\sin A + \tan A)(\cot A - \csc A)}{\cot^2 A - \csc^2 A}$$
$$=  \frac{\sin A \cot A - \sin A \csc A + \tan A \cot A - \tan A \csc A}{-1}  =  - (\cos A - 1 + 1 - \frac{1}{\cos A} )$$
$$= \frac{1 - \cos^2 A}{\cos A}  =  \frac{\sin^2 A}{\cos A}  =  \sin A \tan A  .$$
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