Question

Solve the given question in a detail manner:
Prove that:
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$

Answer 1

Take the left-hand side and the right-hand side of the given equation separately. Consider the right-hand side and by replacing the $\tan x$ in terms of $\sin x$ and $\cos x$ simplify the terms until the right-hand side value is obtained.

Complete step-by-step solution:
Given,
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Let us take the two sides individually. Here, first let us take the right-hand side and simplify it to get the left-hand side;
Right-hand side $= \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Let us replace the $\tan x$ in the terms of $\sin x$ and $\cos x$
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Now, substituting the $\tan x$ in the right-hand side we get;
$\Rightarrow \dfrac{{2\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 - \left( {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$
Simplifying the denominator, we get;
$\Rightarrow \dfrac{{\dfrac{{2\sin x}}{{\cos x}}}}{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}}}$
In the trigonometric relations, we have;
${\cos ^2}x - {\sin ^2}x = \cos 2x$
So, substituting this in the expression, we get;
$\Rightarrow \dfrac{{\dfrac{{2\sin x}}{{\cos x}}}}{{\dfrac{{\cos 2x}}{{{{\cos }^2}x}}}}$
By rearranging the expression, we get;
$\Rightarrow \dfrac{{2\sin x{{\cos }^2}x}}{{\cos x\cos 2x}}$
Cancelling out the common terms in the numerator and the denominator, we get;
$\Rightarrow \dfrac{{2\sin x\cos x}}{{\cos 2x}}$
Now, in the trigonometric relations, we have;
$2\sin x\cos x = \sin 2x$
Substituting this above value in the obtained expression, we get;
$\Rightarrow \dfrac{{\sin 2x}}{{\cos 2x}}$
This can be written as;
$\tan 2x$
The left-hand side$= \tan 2x$
We have the right-hand side$= \tan 2x$
And left-hand side$= \tan 2x$
Since the right-hand side and the left-hand side are equal, we have that;
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Hence proved.

Note: Here, we can prove the values of $\sin 2x$ and $\cos 2x$.
$\sin 2x$ can be written as $\sin \left( {x + x} \right)$
$\Rightarrow \sin 2x = \sin \left( {x + x} \right)$
We have the formula:
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
Using this formula, we substitute $x$ in the place of $A$ and $B$ since both the angles are the same.
$\sin \left( {x + x} \right) = \sin x\cos x + \cos x\sin x$
Now, since both the terms are equal to each other, we can add the terms.
$\Rightarrow \sin 2x = 2\sin x\cos x$
$\cos 2x$ can be written as $\cos \left( {x + x} \right)$
We have the formula:
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Using this formula, we substitute $x$ in the place of $A$ and $B$ since both the angles are the same.
$\cos \left( {x + x} \right) = \cos x\cos x - \sin x\sin x$
The like terms are multiplied and the simplified equation, we get;
$\cos 2x = {\cos ^2}x - {\sin ^2}x$