Question

Prove the following:
$\dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)}={{\left( \dfrac{1+\tan x}{1-\tan x} \right)}^{2}}$

Answer 1

Hint: For solving this question we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas like $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ and $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Given:
We have to prove the following equation:
$\dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)}={{\left( \dfrac{1+\tan x}{1-\tan x} \right)}^{2}}$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}................\left( 1 \right) \\
 & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}................\left( 2 \right) \\
 & \tan \dfrac{\pi }{4}=1..............................................\left( 3 \right) \\
\end{align}$
Now, we will use the above three formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)}$ .
Now, we will use the formula from the equation (1) to write $\tan \left( \dfrac{\pi }{4}+x \right)=\dfrac{\tan \dfrac{\pi }{4}+\tan x}{1-\tan \dfrac{\pi }{4}\tan x}$ and formula from the equation (2) to write $\tan \left( \dfrac{\pi }{4}-x \right)=\dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x}$ in the term on the left-hand side. Then,
\[\begin{align}
  & \dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)} \\
 & \Rightarrow \dfrac{\left( \dfrac{\tan \dfrac{\pi }{4}+\tan x}{1-\tan \dfrac{\pi }{4}\tan x} \right)}{\left( \dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x} \right)} \\
 & \Rightarrow \left( \dfrac{\tan \dfrac{\pi }{4}+\tan x}{1-\tan \dfrac{\pi }{4}\tan x} \right)\times \left( \dfrac{1+\tan \dfrac{\pi }{4}\tan x}{\tan \dfrac{\pi }{4}-\tan x} \right) \\
\end{align}\]
Now, we will use the formula from the equation (3) to write $\tan \dfrac{\pi }{4}=1$ in the above expression. Then,
\[\begin{align}
  & \left( \dfrac{\tan \dfrac{\pi }{4}+\tan x}{1-\tan \dfrac{\pi }{4}\tan x} \right)\times \left( \dfrac{1+\tan \dfrac{\pi }{4}\tan x}{\tan \dfrac{\pi }{4}-\tan x} \right) \\
 & \Rightarrow \left( \dfrac{1+\tan x}{1-1\times \tan x} \right)\times \left( \dfrac{1+1\times \tan x}{1-\tan x} \right) \\
 & \Rightarrow \left( \dfrac{1+\tan x}{1-\tan x} \right)\times \left( \dfrac{1+\tan x}{1-\tan x} \right) \\
 & \Rightarrow \dfrac{\left( 1+\tan x \right)\times \left( 1+\tan x \right)}{\left( 1-\tan x \right)\times \left( 1-\tan x \right)} \\
 & \Rightarrow \dfrac{{{\left( 1+\tan x \right)}^{2}}}{{{\left( 1-\tan x \right)}^{2}}} \\
 & \Rightarrow {{\left( \dfrac{1+\tan x}{1-\tan x} \right)}^{2}} \\
\end{align}\]
Now, from the above result, we conclude that the value of the expression $\dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)}$ will be equal to the value of the expression ${{\left( \dfrac{1+\tan x}{1-\tan x} \right)}^{2}}$ . Then,
$\dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)}={{\left( \dfrac{1+\tan x}{1-\tan x} \right)}^{2}}$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\dfrac{\tan \left( \dfrac{\pi }{4}+x \right)}{\tan \left( \dfrac{\pi }{4}-x \right)}={{\left( \dfrac{1+\tan x}{1-\tan x} \right)}^{2}}$ .
Hence, proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to prove the desired result quickly. . After that, we should proceed in a stepwise manner and apply trigonometric formulas of $\tan \left( A+B \right)$ and $\tan \left( A-B \right)$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.