Question

How do you simplify $\tan \left( {2x} \right) \times \tan \left( {2x} \right)$?

Answer 1

This problem deals with solving the given equation with trigonometric identities and compound sum angles of trigonometric functions. A compound angle formula or addition formula is a trigonometric identity which expresses a trigonometric function of $\left( {A + B} \right)$ or $\left( {A - B} \right)$ in terms of trigonometric functions of $A$ and $B$.

Formula Used:
$\Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Here when $A = B,$ then $\tan \left( {A + B} \right) = \tan 2A$, which is given by:
$\Rightarrow \tan \left( {2A} \right) = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Complete step-by-step answer:
Given an expression of trigonometric tangent expression functions.
The given expression is $\tan \left( {2x} \right) \times \tan \left( {2x} \right)$, consider this as given below:
$\Rightarrow \tan \left( {2x} \right) \times \tan \left( {2x} \right)$
Consider $\tan \left( {2x} \right)$, we know that from the tangent compound angle formula, it can be written as given below:
$\Rightarrow \tan \left( {2x} \right) = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Hence substituting this in the given expression $\tan \left( {2x} \right) \times \tan \left( {2x} \right)$, as shown below:
$\Rightarrow \tan \left( {2x} \right) \times \tan \left( {2x} \right)$
$\Rightarrow \left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)$
Now simplifying the above and solving as given below:
$\Rightarrow \dfrac{{4{{\tan }^2}x}}{{\left( {1 - {{\tan }^2}x} \right)\left( {1 - {{\tan }^2}x} \right)}}$
Now simplifying the denominator, by multiplying both the expressions with each other, as shown below:
$\Rightarrow \dfrac{{4{{\tan }^2}x}}{{{{\left( {1 - {{\tan }^2}x} \right)}^2}}}$
As it can be written as the square of the expression, as both the expressions are one and the same, as shown in the above expression.
Now simplifying the square of the given expression, as shown below:
$\Rightarrow \dfrac{{4{{\tan }^2}x}}{{1 - 2{{\tan }^2}x + {{\tan }^4}x}}$
So the simplification of the product of the two expressions can be written as:
$\therefore \tan \left( {2x} \right) \times \tan \left( {2x} \right) = \dfrac{{4{{\tan }^2}x}}{{1 - 2{{\tan }^2}x + {{\tan }^4}x}}$

Final Answer: The expression $\tan \left( {2x} \right) \times \tan \left( {2x} \right)$ is equal to $\dfrac{{4{{\tan }^2}x}}{{1 - 2{{\tan }^2}x + {{\tan }^4}x}}$.

Note:
Please note that the formula of cosine compound angles formula is used to solve this problem. Here instead of simplifying the product of the expressions of $\tan \left( {2x} \right)$, we can directly square them first as given the product of $\tan \left( {2x} \right)$ and itself, and then hence simplify the expression of the square, which gives the same final answer. But there are a few other trigonometric compound angle formulas of sine, cosine and tangent, which are shown below:
$\Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\Rightarrow \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
$\Rightarrow \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
$\Rightarrow \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$\Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\Rightarrow \tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$