How do you prove that \[ - \cot 2x = \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}\]?

Hint: We directly use the formula of \[\tan 2x\] after we convert the cotangent function in left hand side of the equation to tangent function using the formula \[\cot x = \dfrac{1}{{\tan x}}\]. Substitute the value of \[\tan 2x\] in the denominator and multiply negative sign wherever required.
* \[\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\]

Complete step-by-step solution:
We have to prove that \[ - \cot 2x = \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}\]
We take left hand side of the equation and convert the cotangent function into tangent function
Left hand side of the equation is \[ - \cot 2x\]
We use the conversion of \[\cot x = \dfrac{1}{{\tan x}}\] to write the left hand side of the equation in terms of tangent
\[ \Rightarrow - \cot 2x = \dfrac{{ - 1}}{{\tan 2x}}\]
Now we know the formula for \[\tan 2x = \dfrac{{1 - {{\tan }^2}x}}{{2\tan x}}\]
\[ \Rightarrow - \cot 2x = \dfrac{{ - (1 - {{\tan }^2}x)}}{{2\tan x}}\]
Multiply negative sign outside the bracket with terms inside the bracket
\[ \Rightarrow - \cot 2x = \dfrac{{ - 1 - ( - {{\tan }^2}x)}}{{2\tan x}}\]
Use the concept that when a negative number is multiplied by a negative number, we get the result as a positive number.
\[ \Rightarrow - \cot 2x = \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}\]
Since the term on right hand side of the equation is same as right hand side of the equation, we can write LHS \[ = \] RHS
Hence Proved

Note: Alternate method:
We can convert the terms given on the right hand side of the equation in terms of sine and cosine using the formula \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]. Later use the formula \[{\cos ^2}x - {\sin ^2}x = \cos 2x\] and \[2\sin x\cos = \sin 2x\] to convert the angle from x to 2x.
Right hand side of the equation is \[\dfrac{{{{\tan }^2}x - 1}}{{2\tan x}}\]
Substitute the value of \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] in both numerator and denominator of the equation on right hand side
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1}}{{\dfrac{{2\sin x}}{{\cos x}}}}\]
Take LCM of the terms in numerator of the equation
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}}}}{{\dfrac{{2\sin x}}{{\cos x}}}}\]
Write the fraction in simpler form
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}} \times \dfrac{{\cos x}}{{2\sin x}}\]
Cancel same factors from numerator and denominator of the fraction
 \[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\cos x}} \times \dfrac{1}{{2\sin x}}\]
Now we can write the product of two fractions as one fraction
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{2\sin x\cos x}}\]
Take -1 common from numerator of the fraction
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{ - ({{\cos }^2}x - {{\sin }^2}x)}}{{2\sin x\cos x}}\]
Now we know that \[{\cos ^2}x - {\sin ^2}x = \cos 2x\] and \[2\sin x\cos = \sin 2x\].
Substitute the value of \[{\cos ^2}x - {\sin ^2}x = \cos 2x\] in numerator of the fraction and \[2\sin x\cos = \sin 2x\] in the denominator of the fraction.
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = \dfrac{{ - \cos 2x}}{{\sin 2x}}\]
We know that \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] then changing the angle from x to 2x we can write \[\cot 2x = \dfrac{{\cos 2x}}{{\sin 2x}}\]
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 1}}{{2\tan x}} = - \cot 2x\]
Since the term on right hand side of the equation is same as left hand side of the equation, we can write LHS \[ = \] RHS
Hence Proved