Question

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

Answer 1

Hint: 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;
\[\cos 2x = {\cos ^2}x - {\sin ^2}x\]