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Question
If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
Options
a2 + 1
a2 + 2
a2 − 2
None of these
Solution
Given:
\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = a\]
\[ \Rightarrow \left[ \tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) \right]^2 = a^2 \]
\[ \Rightarrow \tan^2 \left( \frac{\pi}{4} + x \right) + \tan^2 \left( \frac{\pi}{4} - x \right) + 2 \tan\left( \frac{\pi}{4} - x \right) \tan\left( \frac{\pi}{4} + x \right) = a^2 \]
\[ \Rightarrow \tan^2 \left( \frac{\pi}{4} + x \right) + \tan^2 \left( \frac{\pi}{4} - x \right) = a^2 - 2 \tan\left( \frac{\pi}{4} - x \right) \tan\left( \frac{\pi}{4} + x \right)\]
\[ \Rightarrow \tan^2 \left( \frac{\pi}{4} + x \right) + \tan^2 \left( \frac{\pi}{4} - x \right) = a^2 - 2\left[ \frac{\tan45^\circ - \tan x}{1 + \tan45^\circ \tan x} \times \frac{\tan45^\circ + \tan x}{1 - \tan45^\circ \tan x} \right] \]
\[ \Rightarrow \tan^2 \left( \frac{\pi}{4} + x \right) + \tan^2 \left( \frac{\pi}{4} - x \right) = a^2 - 2\left[ \frac{1^\circ - \tan x}{1 + \tan x} \times \frac{1 + \tan x}{1 - \tan x} \right]\]
\[ \Rightarrow \tan^2 \left( \frac{\pi}{4} + x \right) + \tan^2 \left( \frac{\pi}{4} - x \right) = a^2 - 2\left( \frac{1 - \tan^2 x}{1 - \tan^2 x} \right)\]
\[ \Rightarrow \tan^2 \left( \frac{\pi}{4} + x \right) + \tan^2 \left( \frac{\pi}{4} - x \right) = a^2 - 2\]
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