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Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that tan 4x=4 tanx(1−tan2x)1−6 tan2x+tan4xtan\,4x=\frac{4\,tanx(1-tan^2x)}{1-6\,tan^2x+tan^4x}.

Answer

It is known that tan 2A=2 tanA1−tan2A.tan\,2A=\frac{2\,tanA}{1-tan^2A}.

∴L.H.S. = tan 4x = tan 2(2x).

=2 tan2x1−tan2(2x)=\frac{2\,tan2x}{1-tan^2(2x)}

=2( tan x1−tan2x)−1(2 tan x1−tan2x)2=\frac{2(\frac{\,tan\,x}{1-tan^2x})}{-1(\frac{2\,tan\,x}{1-tan^2x})^2}

=( 4tan2 x(1−tan2x)[−14 tan2 x(1−tan2x)2]=\frac{(\frac{\,4tan^2\,x}{(1-tan^2x})}{[-1\frac{4\,tan^2\,x}{(1-tan^2x)^2}]}

=( 4tan2 x(1−tan2x)[(1 tan2 x)2−4 tan2 x(1−tan2x)2]=\frac{(\frac{\,4tan^2\,x}{(1-tan^2x})}{[\frac{(1\,tan^2\,x)^2-4\,tan^2\,x}{(1-tan^2x)^2}]}

=4 tan x(1−tan2x)(1−tan2 x)2−4 tan2x=\frac{4\,tan\,x(1-tan^2x)}{(1-tan^2\,x)^2-4\,tan^2x}

=4 tan x(1−tan2x)1+tan4x−2 tan2−4 tan2=\frac{4\,tan\,x(1-tan^2x)}{1+tan^4x-2\,tan^2-4\,tan^2}

=4 tan x(1−tan2 x)1−6 tan2x+tan4x=R.H.S.= \frac{4\,tan\,x(1-tan^2\,x)}{1-6\,tan^2x+tan^4x}=R.H.S.