Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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

#### 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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