Question

How do you simplify $\dfrac{{{\tan }^{2}}\left( x \right)}{1+{{\tan }^{2}}\left( x \right)}$ ?

Answer 1

Hint: We are give term as $\dfrac{{{\tan }^{2}}\left( x \right)}{1+{{\tan }^{2}}\left( x \right)}$, we are asked to simplify it, in order to simplify it, we will try to reduce it to the simple form of trigonometric ratio, to do so, reduce this into simpler form, we will use that $\tan x=\dfrac{\sin x}{\cos x}$ , we change tan x into the form of $\dfrac{\sin x}{\cos x}$ , then we will also use that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , using these relation between sin, cos and tan we will reach to our answer.

Complete step by step answer:
We are given term as $\dfrac{{{\tan }^{2}}x}{1-{{\tan }^{2}}x}$ , we have to simplify it, simplifying here means reducing it to the much easier trigonometric form or we say much less complex trigonometric ratios.
To do so, we will learn how the ratio is connected to each other.
We start by learning how the ratio is connected to $\tan \theta $ .
We know that sin and cos are connected to tan.
We have that –
$\tan x=\dfrac{\sin x}{\cos x}$ .
We also have relation between sin x and cos x they are connected as –
${{\sin }^{2}}x+{{\cos }^{2}}x=1$ .
Now we will use these relations to solve and simplify our problem.
Now we have $\dfrac{{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$
As we know that $\tan x=\dfrac{\sin x}{\cos x}$ .
So, using this we get –
$\Rightarrow \dfrac{{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}}{1+{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}}$
By simplifying, we get –
$=\dfrac{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{1+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}$
By simplifying denominator, we get –
$=\dfrac{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x}}$
As we know that $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}$ can be written as $\dfrac{a\times d}{b\times c}$ .
So, our above equation will be written as –
$=\dfrac{{{\sin }^{2}}x\times {{\cos }^{2}}x}{{{\cos }^{2}}x\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)}$ .
By cancelling like terms, we get –
$=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x}$
Now as ${{\cos }^{2}}x+{{\sin }^{2}}x$ is 1. So, we get –
$\begin{align}
  & =\dfrac{{{\sin }^{2}}x}{1} \\
 & \Rightarrow {{\sin }^{2}}x \\
\end{align}$
So, we get the simplification value of $\dfrac{{{\tan }^{2}}\left( x \right)}{1+{{\tan }^{2}}\left( x \right)}={{\sin }^{2}}x$ .

Note:
As the ratios are connected to one another is more than 1 way, so there are more than one way to solve.
So, we can also solve our problem as –
$\dfrac{{{\tan }^{2}}\left( x \right)}{1+{{\tan }^{2}}\left( x \right)}=\dfrac{{{\tan }^{2}}x}{{{\sec }^{2}}x}$ (as $1+{{\tan }^{2}}x={{\sec }^{2}}x$ )
Now as $\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$ .
So, $=\dfrac{{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}}{{{\left( \dfrac{1}{\cos } \right)}^{2}}}$
Opening bracket, we get –
$=\dfrac{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{\dfrac{1}{{{\cos }^{2}}x}}$
Cancelling like term, we get –
$={{\sin }^{2}}x$
Hence,
$\Rightarrow \dfrac{{{\tan }^{2}}\left( x \right)}{1+{{\tan }^{2}}\left( x \right)}={{\sin }^{2}}x$