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# Solve $f\left( x \right)=\dfrac{1-2\tan x}{1+2\tan x}$

Last updated date: 13th Jun 2024
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Hint: We don’t have any information about what to do in the problem, they have only asked to solve the function which is in terms of $\tan x$ . So, we will use the basic trigonometric equation that we have $\tan x=\dfrac{\sin x}{\cos x}$ in the given functions and simplify the equation by taking LCM then we will get the result.

Given that,
$f\left( x \right)=\dfrac{1-2\tan x}{1+2\tan x}$
We have the basic trigonometric formula $\tan x=\dfrac{\sin x}{\cos x}$ , substituting this value in given equation, then we will get
\begin{align} & f\left( x \right)=\dfrac{1-2\tan x}{1+2\tan x} \\ & \Rightarrow f\left( x \right)=\dfrac{1-2\left( \dfrac{\sin x}{\cos x} \right)}{1+2\left( \dfrac{\sin x}{\cos x} \right)} \\ \end{align}
Multiplying $2$ in to the parenthesis.
$f\left( x \right)=\dfrac{1-\dfrac{2\sin x}{\cos x}}{1+\dfrac{2\sin x}{\cos x}}$
Now take the numerator and denominator individually, then
Numerator $=1-\dfrac{2\sin x}{\cos x}$
Doing LCM in the above equation, then we will get
\begin{align} & 1-\dfrac{2\sin x}{\cos x}=\dfrac{1\times \cos x-2\sin x}{\cos x} \\ & \Rightarrow 1-\dfrac{2\sin x}{\cos x}=\dfrac{\cos x-2\sin x}{\cos x} \\ \end{align}
Considering the denominator,
Denominator $=1+\dfrac{2\sin x}{\cos x}$
Doing LCM in the above equation, then we will get
\begin{align} & 1+\dfrac{2\sin x}{\cos x}=\dfrac{1\times \cos x+2\sin x}{\cos x} \\ & \Rightarrow 1+\dfrac{2\sin x}{\cos x}=\dfrac{\cos x+2\sin x}{\cos x} \\ \end{align}
Substituting the both the values of numerator and denominator in the given equation, then we will get
$f\left( x \right)=\dfrac{1-\dfrac{2\sin x}{\cos x}}{1+\dfrac{2\sin x}{\cos x}}$
Writing the denominator in the form multiplication i.e., $\dfrac{a}{b}=a\times \dfrac{1}{b}$ , then we will get
$\Rightarrow f\left( x \right)=\left( 1-\dfrac{2\sin x}{\cos x} \right)\times \dfrac{1}{\left( 1+\dfrac{2\sin x}{\cos x} \right)}$
We have the value of $1+\dfrac{2\sin x}{\cos x}=\dfrac{\cos x+2\sin x}{\cos x}$ , then the value of $\dfrac{1}{1+\dfrac{2\sin x}{\cos x}}$ can be written as
\begin{align} & \dfrac{1}{1+\dfrac{2\sin x}{\cos x}}=\dfrac{1}{\dfrac{\cos x+2\sin x}{\cos x}} \\ & \Rightarrow \dfrac{1}{1+\dfrac{2\sin x}{\cos x}}=\dfrac{\cos x}{\cos x+2\sin x} \\ \end{align}
Now substituting the above value in the given function, then we will get
$\Rightarrow f\left( x \right)=\dfrac{\left( \cos x-2\sin x \right)}{\cos x}\times \dfrac{\cos x}{\cos x+2\sin x}$
Rearranging the terms in the above equation, then we will get
$\Rightarrow f\left( x \right)=\dfrac{\cos x-2\sin x}{\cos x+2\sin x}\times \dfrac{\cos x}{\cos x}$
We know that $\dfrac{a}{a}=1$ , then we will get
$\Rightarrow f\left( x \right)=\dfrac{\cos x-2\sin x}{\cos x+2\sin x}$
Hence the above equation is simplified equation of above equation.

Note:
We can directly go for the simplification without considering the numerator and denominator individually. But students who are not comfortable with the fractions may do mistakes while simplifying, so we have considered the numerator and denominator individually and then we will simplify both and substituted in the given equation.