Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $ 
Prove the following identities: 
  $$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x 
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\tag {ii}$$

For $(\mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1. 
I then tried just multiplying out the brackets and got as far as $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$

(i) Let $C=\cos(2x), S=\sin(2x)=2\sin(x)\cos(x)$ 
$S^2*RHS= S^2(1+\frac{2}{(S/2)^2})= S^2+8=9-C^2$ 
$\begin{align}S^2*LHS
&=(2+2\sin^2(x))(2+2\cos^2(x)) \cr
&=(2+(1-C))(2+(1+C))\cr
&=9-C^2 =S^2*RHS
\end{align}$
QED
(ii) Let $t=\tan(x/2),\text{then }\sin(x)=\frac{2t}{1+t^2}\text{ , }\cos(x)=\frac{1-t^2}{1+t^2}$ 
$1-\tan(x)=1-\frac{2t}{1-t^2}=\frac{1-2t-t^2}{1-t^2}$
$1-\cot(x)=1-\frac{1-t^2}{2t}=\frac{1-2t-t^2}{-2t}$ 
$\begin{align}
\frac{\cos(x)}{1-\tan(x)} + \frac{\sin(x)}{1-\cot(x)}
&=\frac{(1-t^2)^2 - (2t)^2}{(1-2t-t^2)(1+t^2)} \cr
&=\frac{(1-2t-t^2)(1+2t-t^2)}{(1-2t-t^2)(1+t^2)} \cr 
&= \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = \sin(x) + \cos(x)
\end{align}$
QED
