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ID:
Class: 11
Subject: Math
Topic: Trigonometric Functions
Type: Short
Question:
Prove that: \( \frac{\tan\left( \frac{\pi}{4} + x \right)}{\tan\left( \frac{\pi}{4} - x \right)} = \left( \frac{1 + \tan x}{1 - \tan x} \right)^2 \)
Official Solution
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*}
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