Trigonometry Problem

Prove the Identity  4cot(4x)= 2cot(2x)-2tan(2x)

$$\\\small{\text{ If $ t = \tan{(2x)} $ than $ \cot(4x)=\frac{1-t^2}{2t} $ and $ \cot{(2x)}=\frac{1}{t}$ . } $\\$ \small{\text{ So $ 4* ( \frac{1-t^2}{2t} ) = 2\frac{1}{t} - 2t $ or $ \frac{1-t^2}{t} = \frac{1}{t} - t = \frac{1-t^2}{t} $ }$$

Since  tan(2x) = [2tan(x)] / [1 - tan²(x)]  

--->  cot(2x)  =  [1 - tan²(x)]  / [2tan(x)]

--->  cot(4x)  =  [1 - tan²(2x)]  / [2tan(2x)]

--->  4cot(4x)  =  2[1 - tan²(2x)]  / [tan(2x)]

                      =  2[ 1 - ( 2tan(x) / [1 - tan²(x)] )² ] / [ 2tan(x) / (1 - tan²(x) ) ]  

                      = 2 / [ 2tan(x) / ( 1 - tan²(x) ) ]   -   2( 2tan(x) / [1 - tan²(x)] )² /[ 2tan(x) / ( 1 - tan²(x) ) ]  

                       =  2 · ( 1 - tan²(x) ) / 2tan(x)  -  2( 2tan(x) / [1 - tan²(x)] )² · ( 1 - tan²(x) ) / ( 2tan(x) )

                      =  ( 1 - tan²(x) ) / tan(x) - 2( 2tan(x) / [1 - tan²(x)] )

                      = ( 1 - tan²(x) ) / tan(x) - ( 4tan(x) / [1 - tan²(x)] )

                      = 2cot(2x) - 2tan(2x)

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Because:

2cot(2x)  =  [1 - tan²(x)]  / [tan(x)]

2tan(2x)  =  [4tan(x)] / [1 - tan²(x)]  

4cot(4x)= 2cot(2x)-2tan(2x) 

4cos(4x)/sin(4x) =

4[cos²(2x) - sin²(2x) ] / [(2cos(2x)sin(2x)] =

2 [( cos²(2x)/cos(2x)sin(2x) ) - sin²(2x)/cos(2x)sin(2x) ] =

2[ cos(2x)/sin(2x) - sin(2x)/cos(2x)]

2 [cot(2x) - tan(2x) ] =

2cot(2x) - 2tan(2x)

And the left side = the right side