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)
\begin{align*}
(\sec^2x + \tan^2x)(\csc^2x + \cot^2x) & = (1 + \tan^2x + \tan^2x)(1 + \cot^2x + \cot^2x)\\
& = (1 + 2\tan^2x)(1 + 2\cot^2x)\\
& = 1 + 2\cot^2x + 2\tan^2x + 4\\
& = 5 + 2(\csc^2x - 1) + 2(\sec^2x - 1)\\
& = 5 + 2\csc^2x - 2 + 2\sec^2x - 2\\
& = 1 + 2\csc^2x + 2\sec^2x\\
& = 1 + 2\left(\frac{1}{\sin^2x} + \frac{1}{\cos^2x}\right)\\
& = 1 + 2\left(\frac{\cos^2x + \sin^2x}{\sin^2x\cos^2x}\right)\\
& = 1 + \frac{2}{\cos^2x\sin^2x}\\
& = 1 + 2\sec^2x\csc^2x
\end{align*}
(ii) 
\begin{align*}
\frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x} & = \frac{\cos x}{1 - \tan x} \cdot \frac{\cos x}{\cos x} + \frac{\sin x}{1 - \cot x} \cdot \frac{\sin x}{\sin x}\\
& = \frac{\cos^2x}{\cos x - \sin x} + \frac{\sin^2x}{\sin x - \cos x}\\
& = \frac{\cos^2x}{\cos x - \sin x} - \frac{\sin^2x}{\cos x - \sin x}\\
& = \frac{\cos^2x - \sin^2x}{\cos x - \sin x}\\
& = \frac{(\cos x + \sin x)(\cos x - \sin x)}{\cos x - \sin x}\\
& = \cos x + \sin x
\end{align*}
