Question

Prove that \(tan\,4x=\frac{4\,tanx(1-tan^2x)}{1-6\,tan^2x+tan^4x}\).

Answer 1

It is known that \(tan\,2A=\frac{2\,tanA}{1-tan^2A}.\)

∴L.H.S. = tan 4x = tan 2(2x).

\(=\frac{2\,tan2x}{1-tan^2(2x)}\)

\(=\frac{2(\frac{\,tan\,x}{1-tan^2x})}{-1(\frac{2\,tan\,x}{1-tan^2x})^2}\)

\(=\frac{(\frac{\,4tan^2\,x}{(1-tan^2x})}{[-1\frac{4\,tan^2\,x}{(1-tan^2x)^2}]}\)

\(=\frac{(\frac{\,4tan^2\,x}{(1-tan^2x})}{[\frac{(1\,tan^2\,x)^2-4\,tan^2\,x}{(1-tan^2x)^2}]}\)

\(=\frac{4\,tan\,x(1-tan^2x)}{(1-tan^2\,x)^2-4\,tan^2x}\)

\(=\frac{4\,tan\,x(1-tan^2x)}{1+tan^4x-2\,tan^2-4\,tan^2}\)

\(= \frac{4\,tan\,x(1-tan^2\,x)}{1-6\,tan^2x+tan^4x}=R.H.S.\)