Question

How do you prove $\tan (2x) = \dfrac{{2\tan (x)}}{{1 - {{\tan }^2}x}}$ ?

Answer 1

Hint:
Whenever they ask to prove something we need to consider the Left hand side function and we need to make necessary changes to arrive at the right hand side function. So first take $\tan (2x)$ we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ , so we can make use of this as $\tan (2x) = \dfrac{{\sin 2x}}{{\cos 2x}}$. Now make use of trigonometric identities to reduce this to the function which is equal to the right hand side.

Complete step by step solution:
Here they have asked to prove something, so we need to consider the Left hand side function and we need to make necessary changes to arrive at the right hand side function.
So first take $\tan (2x)$ we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ , so we can make use of this and write the given function as below.
 $L.H.S. = \tan (2x) = \dfrac{{\sin 2x}}{{\cos 2x}}$.
Now make use of trigonometric double angle identities to reduce the above function which is equal to right hand side.
We know that $\sin 2x = 2\sin x\cos x$ and $\cos 2x = {\cos ^2}x - {\sin ^2}x$ . by making use of these formulas in the above expression, we get
$ \Rightarrow L.H.S. = \tan (2x) = \dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$
Now divide both numerator and denominator by ${\cos ^2}x$
$ \Rightarrow L.H.S. = \tan (2x) = \dfrac{{\left( {\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x}}} \right)}}{{\left( {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$
$ \Rightarrow L.H.S. = \tan (2x) = \dfrac{{\left( {\dfrac{{2\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$
On simplifying the above expression, we get
$ \Rightarrow L.H.S. = \tan (2x) = \dfrac{{\left( {\dfrac{{2\sin x}}{{\cos x}}} \right)}}{{\left( {1 - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$
Now we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ also remember that ${\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}$ . So by substituting these in the above expression we get,
$ \Rightarrow L.H.S. = \tan (2x) = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} = R.H.S.$
Hence we proved that the left hand side function is equal to the right hand side function.

Note:
This problem can also be solved in another way that is by rewriting the L.H.S. $\tan (2x)$ as $\tan (x + x)$ which is of the form $\tan (A + B)$ . So by using the formula $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ we can prove the given function.