If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
(a) a2 + 1
(b) a2 + 2
(c) a2 − 2
(d) None of these
(c) a2 − 2
Given:
$\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)=a$
$\Rightarrow\left[\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)\right]^{2}=a^{2}$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)+2 \tan \left(\frac{\pi}{4}-x\right) \tan \left(\frac{\pi}{4}+x\right)=a^{2}$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)=a^{2}-2 \tan \left(\frac{\pi}{4}-x\right) \tan \left(\frac{\pi}{4}+x\right)$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)=a^{2}-2\left[\frac{\tan 45^{\circ}-\tan x}{1+\tan 45^{\circ} \tan x} \times \frac{\tan 45^{\circ}+\tan x}{1-\tan 45^{\circ} \tan x}\right]$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)=a^{2}-2\left[\frac{1^{\circ}-\tan x}{1+\tan x} \times \frac{1+\tan x}{1-\tan x}\right]$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)=a^{2}-2\left(\frac{1-\tan ^{2} x}{1-\tan ^{2} x}\right)$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)=a^{2}-2$
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