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 $

Question

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 $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$

Answer 1

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

Answer 2

(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*}

Answer 3

$$(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)$$

$$=(2\sec^2x-1)(2\csc^2x-1)$$

$$=4\sec^2x\csc^2x-2(\sec^2x+\csc^2x)+1$$

Now use $\sec^2x+\csc^2x=\cdots=\sec^2x\csc^2x$

The second one has been solved by Taussig

Answer 4

(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