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 $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$
 A: (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) $$\frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x}$$
=>
 $$\frac{\cos x(1 - \cot x) + \sin x(1 - \tan x)}{(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{\frac{\cos^2x \sin x - \cos x\cos^2x + \sin^2 x\cos x - \sin x\sin^2x}{\sin x \cos x}}{(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{\cos^2 x(\sin x - \cos x) +  \sin^2 x(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{(\sin^2 x - \cos^2 x)(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$
=>
 $$\frac{(\sin^2 x - \cos^2x)(\cos x - \sin x)}{(\cos x - \sin x)(\sin x - \cos x)}$$
=>
 $$\frac{(\sin^2 x - cos^2 x)}{(\sin x - \cos x)}$$
=>
 $$\frac{(\sin x  + \cos x)(\sin x - \cos x)}{(\sin x - \cos x)}$$
=>
 $$\sin x  + \cos x$$
A: $$(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)$$
$$=(2\sec^2x-1)(2\csc^2x-1)$$
$$=4\sec^2x\csc^2x-2(\sec^2x+\csc^2x)+1$$
Now use $\sec^2x+\csc^2x=\cdots=\sec^2x\csc^2x$
The second one has been solved by Taussig
A: (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*}
A: (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
