Prove that: 4x = 4 x(1 - ^2 x) 1 - 6 ^2 x + ^4 x

Question

Prove that: $\tan 4x = \frac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$

✓

Answer

We have L.H.S. = tan 4x $ = \frac{{2\tan 2x}}{{1 - {{\tan }^2}2x}}\left[ {\because \tan 2A = \frac{{2\tan A}}{{1 - {{\tan }^2}A}}} \right]$
$ = \frac{{2 \cdot \frac{{2\tan x}}{{1 - {{\tan }^2}x}}}}{{1 - {{\left( {\frac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}^2}}}$$ = \frac{{\frac{{4\tan x}}{{1 - {{\tan }^2}x}}}}{{\frac{{{{(1 - {{\tan }^2}x)}^2} - 4{{\tan }^2}x}}{{{{(1 - {{\tan }^2}x)}^2}}}}}$
$ = \frac{{4\tan x}}{{1 - {{\tan }^2}x}} \times $$\frac{{{{(1 - {{\tan }^2}x)}^2}}}{{1 + {{\tan }^4}x - 2{{\tan }^2}x - 4{{\tan }^2}x}}$
$ = \frac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$ = R.H.S.

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