Prove that:tan4x = 4tanx(1-tan²x) / 1-6tan²x + tan⁴xPlz...No fake

1.	Prove that:tan4x = 4tanx(1-tan²x) / 1-6tan²x + tan⁴xPlz...No fake answersIt's urgent...
Answer» ANSWER: Taking LHS tan 4x we know that tan 2x = $\frac{2tanx}{1 - {tan \: }^{2} x}$ replacing x with 2x tan (2 × 2x)= $\frac{2tan \: 2x}{1 - {tan}^{2} 2x}$ USING tan2x = $\frac{2tanx}{1 - {tan}^{2}x }$ $2 \times \frac{2tanx}{1 - {tan}^{2} x} \div 1 - \frac{2tanx}{1 - {tan}^{2}x } {}^{2}$ $\frac{4tanx}{1} \times \frac{1 - {tan}^{2}x }{1 - {{tan }^{2}x} ^{2} } - {4tan}^{2}x$ using (a-b)= ${a}^{2} + {b}^{2} - 2ab$ $\frac{4tanx(1 - {tan}^{2}x) }{( {1}^{2} + ( {tan}^{2} {x})^{2} - 2 \times 1 \times {tan}^{2}x } - 4 {tan}^{2}x$ $\frac{4tanx(1 - {tan}^{2}x) }{1 + {tan}^{4}x - 6 {tan}^{4}x }$ = RHS LHS=RHS