Prove that:

$
\frac{\tan\left( \frac{\pi}{4} + x \right)}{\tan\left( \frac{\pi}{4} - x \right)} = \left( \frac{1 + \tan x}{1 - \tan x} \right)^2
$

Q: Prove that: $ \frac{\tan\left( \frac{\pi}{4} + x \right)}{\tan\left( \frac{\pi}{4} - x \right)} = \left( \frac{1 + \tan x}{1 - \tan x} \right)^2 $

Explanation: \begin{align*} \text{LHS} &= \frac{\tan\left( \frac{\pi}{4} + x \right)}{\tan\left( \frac{\pi}{4} - x \right)} \\ &= \frac{ \frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4} \cdot \tan x} }{ \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4} \cdot \tan x} } \\ &= \frac{ \frac{1 + \tan x}{1 - \tan x} }{ \frac{1 - \tan x}{1 + \tan x} } \\ &= \left( \frac{1 + \tan x}{1 - \tan x} \right)^2 = \text{RHS} \qquad \textbf{Hence proved.} \end{align*}

Question

Prove that:

$
\frac{\tan\left( \frac{\pi}{4} + x \right)}{\tan\left( \frac{\pi}{4} - x \right)} = \left( \frac{1 + \tan x}{1 - \tan x} \right)^2
$

Accepted Answer

Explanation: \begin{align*}
	ext{LHS} &= \frac{	an\left( \frac{\pi}{4} + x ight)}{	an\left( \frac{\pi}{4} - x ight)} \
&= \frac{ \frac{	an\frac{\pi}{4} + 	an x}{1 - 	an\frac{\pi}{4} \cdot 	an x} }{ \frac{	an\frac{\pi}{4} - 	an x}{1 + 	an\frac{\pi}{4} \cdot 	an x} } \
&= \frac{ \frac{1 + 	an x}{1 - 	an x} }{ \frac{1 - 	an x}{1 + 	an x} } \
&= \left( \frac{1 + 	an x}{1 - 	an x} ight)^2 = 	ext{RHS}  \qquad 	extbf{Hence proved.}
\end{align*}

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