Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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If Tan (π/4 + X) + Tan (π/4 − X) = A, Then Tan2 (π/4 + X) + Tan2 (π/4 − X) = - Mathematics

MCQ

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

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Solution

\[a^2 - 2\]

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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Q 17 | Page 28
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