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) 
$$(\sec^2x + \tan^2x)(\csc^2x + \cot^2x)$$
=>
$$(\sec^2x\csc^2x + \tan^2x\csc^2x + \sec^2x\cot^2x + \tan^2x\cot^2x)$$
=>
$$(\sec^2x\csc^2x+ \frac{\sin^2x}{\cos^2x\sin^2x} + \frac{\cos^2x}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{1}{\cos^2x} + \frac{1}{\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{\sin^2x + \cos^2x}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{1}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \sec^2x\csc^2x + 1)$$
=>
$$(1 + 2\sec^2x\csc^2x)$$
ii) $$\frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x}$$
=>
 $$\frac{\cos x(1 - \cot x) + \sin x(1 - \tan x)}{(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{\frac{\cos^2x \sin x - \cos x\cos^2x + \sin^2 x\cos x - \sin x\sin^2x}{\sin x \cos x}}{(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{\cos^2 x(\sin x - \cos x) +  \sin^2 x(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{(\sin^2 x - \cos^2 x)(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{(\sin^2 x - \cos^2x)(\cos x - \sin x)}{(\cos x - \sin x)(\sin x - \cos x)}$$
=>
 $$\frac{(\sin^2 x - cos^2 x)}{(\sin x - \cos x)}$$
=>
 $$\frac{(\sin x  + \cos x)(\sin x - \cos x)}{(\sin x - \cos x)}$$
=>
 $$\sin x  + \cos x$$
