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.
You are on the right track.
writing $\tan\theta$ as$ \dfrac {\sin\theta}{\cos\theta}$ and $\cot\theta$ as $ \dfrac {\cos\theta}{\sin\theta} $, we get
$ \dfrac {\frac {\sin\theta}{\cos\theta} }{1-\frac {\cos\theta}{\sin\theta} }+\dfrac {\frac {\cos\theta}{\sin\theta} }{1-\frac {\sin\theta}{\cos\theta} }$
$= \dfrac {\sin^2\theta}{cos\theta\cdot(\sin\theta-\cos\theta)} + \dfrac {\cos^2\theta}{\sin\theta\cdot(\cos\theta-\sin\theta)}$ (how?)
$= \dfrac {\sin^2\theta}{\cos\theta\cdot(\sin\theta-\cos\theta)} - \dfrac {\cos^2\theta}{\sin\theta\cdot(\sin\theta-\cos\theta)}$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^2\theta}{\cos\theta}-\dfrac {\cos^2\theta}{\sin\theta})$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^3\theta-\cos^3\theta}{\sin\theta\cdot\cos\theta})$
$=\dfrac{\sin\theta-\cos\theta}{\sin\theta-\cos\theta}\dfrac{\big(\sin^2\theta+\sin\theta\cdot\cos\theta+\cos^2\theta)}{\sin\theta\cdot\cos\theta}$(how?)
$=1\cdot \dfrac{1+\sin\theta\cdot\cos\theta}{\sin\theta\cdot\cos\theta}$ (why?)
which is
$1+\sec\theta\cdot\csc\theta$
QED.
Best Answer
That is exactly correct! Just two things: First, $\tan,\sin,\cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $\tan x$, $\cos x$ etc rather than just $\tan$ or $\cos$.
Finally, you could save time on your proof by noticing on the fourth step that $$ \frac{1}{\cos x\sin x}=\frac{1}{\cos x}\frac{1}{\sin x}=\sec x \csc x $$