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.
Best Answer
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.