I've been given a trigonometric identity worksheet. We need to use conjugates to solve these identities. One sided proofs. I've spent some time on it but just can't crack it. Here it is, help appreciated. Any tips (such as which side to start on etc) is appreciated.
$$\frac{(\cos x\cot x)}{\cot x-\cos x}=\frac{\cot x+\cos x}{(\cos x\cot x)}$$
On
For $\sin x\neq0$,
$\dfrac{(\cos x\cot x)}{\cot x-\cos x}=\dfrac{(\cos x\cot x)}{\cos x(\csc x-1)}=\dfrac{\cot x}{\csc x-1}\cdot\dfrac{\csc x+1}{\csc x+1}=\dfrac{\csc x\cot x+\cot x}{\cot^2x}=\dfrac{\dfrac{\cos x}{\sin^2 x}+\dfrac{\cos x}{\sin x}}{\dfrac{\cos^2x}{\sin^2 x}}=\dfrac{\dfrac{\cos x+\sin x\cos x}{\sin^2 x}}{\dfrac{\cos^2x}{\sin^2 x}}={\dfrac{\cos x+\sin x\cos x}{\sin^2 x}}\cdot{\dfrac{\sin^2 x}{\cos^2 x}}={\dfrac{\cos x+\sin x\cos x}{\sin x}}\cdot{\dfrac{\sin x}{\cos^2 x}}=\left(\cot x+\cos x\right)\dfrac{1}{\cos x\cot x}=\dfrac{\cot x+\cos x}{(\cos x\cot x)}$
On
$$\dfrac{\cos x\cot x}{\cot x-\cos x}=\cdots=\dfrac{\cos x}{1-\sin x}$$
Similarly simplify the right hand side
Finally use $$\cos^2x=(1-\sin x)(1+\sin x)$$
$$\implies\dfrac{\cos x}{1-\sin x}=?$$
Alternatively, $$\dfrac{\cot x-\cos x}{\cot x\cos x}=\dfrac1{\cos x}-\dfrac1{\cot x}=\sec x-\tan x$$
which is $=\dfrac1{\sec x+\tan x}=?$
As $\cos x=\cot x\cdot\sin x$ for $\sin x\ne0,$
$$(\cot x+\cos x)(\cot x-\cos x)=\cot^2x\left(1-\sin^2x\right)=?$$