Question

Prove the trigonometric relation
$\tan 4x = \dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$

Answer 1

Hint – In this question use the basic trigonometric formula that $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$ on the left hand side of the given equation. Resolve tan4A in terms of tan2A and use basic algebraic identities to get the proof.

Complete step-by-step answer:

Given trigonometric equation
$\tan 4x = \dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$
Proof –
Consider L.H.S
$ \Rightarrow \tan 4x$
Now as we know that $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
So use this property in above equation we have,
$ \Rightarrow \tan 4x = \tan 2\left( {2x} \right) = \dfrac{{2\tan 2x}}{{1 - {{\tan }^2}2x}}$....................... (1)
Now again apply the property we have,
$ \Rightarrow \tan 4x = \dfrac{{2\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}}}{{1 - {{\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}^2}}}$
Now simplify the above equation we have,
$ \Rightarrow \tan 4x = \dfrac{{\dfrac{{4\tan x}}{{1 - {{\tan }^2}x}}}}{{1 - \dfrac{{4{{\tan }^2}x}}{{{{\left( {1 - {{\tan }^2}x} \right)}^2}}}}}$
$ \Rightarrow \tan 4x = \dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{{{\left( {1 - {{\tan }^2}x} \right)}^2} - 4{{\tan }^2}x}}$
Now open the denominator square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we have,
$ \Rightarrow \tan 4x = \dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 + {{\tan }^4}x - 2{{\tan }^2}x - 4{{\tan }^2}x}}$
$ \Rightarrow \tan 4x = \dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}$
= R.H.S
Hence Proved.

Note – These problems are based upon direct trigonometric formula, identity and algebraic identities. It is advised to grasp all these formulas although mugging up them will be difficult therefore practice can help getting things on the right track.
We can also prove this by using $\tan 4x = \tan \left( {2x + 2x} \right)$
And we all know that $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Therefore, $ \Rightarrow \tan 4x = \tan \left( {2x + 2x} \right) = \dfrac{{\tan 2x + \tan 2x}}{{1 - {{\tan }^2}2x}} = \dfrac{{2\tan 2x}}{{1 - {{\tan }^2}2x}}$
Now as we see that this equation is the same as equation (1).
Now we further apply the property of tan in this equation and simplify we will get the same answer as above.