Question

Prove that:
$tan4x = \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$

Answer 1

To solve this problem we are to use a trigonometric result of $tan2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$ to get through the result. First we write $tan4x = \tan [2(2x)]$ so, that the formula of $\tan 2x$ can be used, now again we substitute the value of $tan2x$ , and simplify to find the solution.

Complete step by step Answer:

We are given to prove $tan4x = \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$
We have our L.H.S as,
$tan4x = \tan [2(2x)]$
Now as per the formula of $tan2x$ we get, $tan2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Then we can write,
$= \dfrac{{2\tan 2x}}{{1 - {{\tan }^2}2x}}$
Again using the same formula $tan2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$, we get,
$= \dfrac{{2\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}}{{1 - {{\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}^2}}}$
On Simplifying, we get,
$= \dfrac{{\left( {\dfrac{{4\tan x}}{{1 - {{\tan }^2}x}}} \right)}}{{1 - \dfrac{{4{{\tan }^2}x}}{{{{(1 - {{\tan }^2}x)}^2}}}}}$
On taking LCM of the denominator, we get,
$= \dfrac{{\left( {\dfrac{{4\tan x}}{{1 - {{\tan }^2}x}}} \right)}}{{\dfrac{{{{(1 - {{\tan }^2}x)}^2} - 4{{\tan }^2}x}}{{{{(1 - {{\tan }^2}x)}^2}}}}}$
Transforming division into multiplication, we get,
$= \dfrac{{4\tan x}}{{1 - {{\tan }^2}x}} \times \dfrac{{{{(1 - {{\tan }^2}x)}^2}}}{{{{(1 - {{\tan }^2}x)}^2} - 4{{\tan }^2}x}}$
On Cancelling out common terms we get,
$= \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{{{(1 - {{\tan }^2}x)}^2} - 4{{\tan }^2}x}}$
On Elaborating, we get,
$= \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 2{{\tan }^2}x + {{\tan }^4}x - 4{{\tan }^2}x}}$
$= \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$(R.H.S)
So, we have, L.H.S = R.H.S.
i.e., $tan4x = \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$
Hence, our result is proved.

Note: The result $tan2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$ can be proved in the following way,
Use
$tanx = \dfrac{{sinx}}{{\cos x}}$,$sin2x = 2sinxcosx$ and $cos2x = co{s^2}x - si{n^2}x$, for the right hand side expression
Explanation:
$\dfrac{{2tanx}}{{1 - {{\tan }^2}x}}$
On substituting the value of tanx we get,
$= \dfrac{{2\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 - {{\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}^2}}}$
On simplification we get,
$= \dfrac{{\dfrac{{2\sin x}}{{\cos x}}}}{{1 - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}$
On taking LCM in the denominator we get,
$= \dfrac{{\dfrac{{2\sin x}}{{\cos x}}}}{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}}}$
On further simplification we get,
$= \dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$
On using $sin2x = 2sinxcosx$ and $cos2x = co{s^2}x - si{n^2}x$, we get,
$= \dfrac{{\sin 2x}}{{\cos 2x}}$
On using, $tanx = \dfrac{{sinx}}{{\cos x}}$, we get,
$= \tan 2x$
Hence, $tan2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$