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. 
 A: Solution 1:
$$\dfrac{\dfrac{\sin{A} + \cos{A}}{\sin{A}  \cos{A}}}{\dfrac{\sin^2{A} + \cos^2{A}}{\sin{A}  \cos{A}}}$$
$$ = \frac{\sin{A} + \cos{A}}{\sin^2{A} + \cos^2{A}}$$
$$ = \sin{A} + \cos{A}$$
Solution 2:
$$\frac{\cos{A}(1 + \cos{A}) + \sin^2{A}}{\sin{A}  (1 + \cos{A})}$$
$$= \frac{\color{red}{\cos{A} + 1}}{\sin{A}  (\color{red}{\cos{A} + 1})}$$
$$= \frac{1}{\sin{A}} = \csc{A}$$
PS: I don't know how to put those cross-marks(cancellations) on fractions, if someone knows, please comment it, I'll edit it.
A: For the first problem, use $\sec A = \frac{1}{\cos A}$, and $\csc A = \frac{1}{\sin A}$, then put them over a common denominator.  Then convert $\tan$ and $\cot$ to $\sin$ and $\cos$ as you have done and put them over the same common denominator.  Cancel these denominators and use $\sin^2 A + \cos^2 A = 1$ and you're done.
For the second one, begin as you did with $\frac{\cos}{\sin} + \frac{\sin}{1+\cos}$ and put them each over a denominator of $\sin^2$.  Hint: You can convert $1+\cos$ to $\sin^2$ by multiplying it by $1-\cos$ and using the fact that $1-\cos^2=\sin^2$.  The rest is just algebra.
A: 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 :)
