<s>[INST] Prove $\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$ and $\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$ Can anyone help me solve the following trig equations.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$$
My work thus far 
$$\frac{\frac{1}{\cos{A}}+\frac{1}{\sin{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$
$$\frac{\frac{\sin{A} + \cos{A}}{\sin{A} * \cos{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$
But how would I continue?
My second question is 
$$\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$
My work is 
$$\frac{\cos{A}}{\sin{A}} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$
I think I know how to solve this one by using a common denominator but I am not sure. 
 [/INST] Try waiting until the last minute before converting to sines and cosines.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}$$
Recall that $\tan A\cot A = 1$ and $\tan A = \frac{\sec A}{\csc A}$
$$=\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}\cdot\frac{\tan A}{\tan A} $$
$$=\frac{\sec{A}\tan A+\csc{A}\tan A}{\tan^2{A} + 1} $$
$$=\frac{\sec{A}\tan A+\csc{A}\cdot\frac{\sec A}{\csc A}}{\sec^2 A} $$
$$=\frac{\sec{A}\tan A+\sec A}{\sec^2 A} $$
$$=\frac{\sec A (\tan A+1)}{\sec^2 A} $$
$$=\frac{\tan A+1}{\sec A} $$
$$=\frac{\frac{\sec A}{\csc A}}{\sec A} + \frac{1}{\sec A} $$
$$=\frac{1}{\csc A} + \frac{1}{\sec A} $$
$$=\sin A + \cos A $$
But I do like hjpotter92's answer better :)
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