Question

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

Answer 1

Hint: 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, \[tan2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\]