Prove that:

tan(4x) = [4tan(x) × (1 − tan²(x))] / [1 − 6tan²(x) + tan⁴(x)]

L.H.S. = tan(4x) = tan(2(2x))

= (2tan(2x)) / (1 − tan²(2x))

= 2 × (tan(x)/(1 − tan²(x))) ÷ [1 − (2tan(x)/(1 − tan²(x)))²]

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

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

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

= (4tan(x)(1 − tan²(x))) / (1 + tan⁴(x) − 2tan²(x) − 4tan²(x))

= (4tan(x)(1 − tan²(x))) / (1 − 6tan²(x) + tan⁴(x)) = R.H.S.