\(tan(\frac\pi4 + \theta) + tan(\frac\pi4 - \theta) = x\)

⇒ \(\cfrac{tan\frac\pi4 + tan\theta}{1- tan\frac\pi4 \,tan\theta} + \cfrac{tan\frac\pi4 - tan\theta}{1 + tan\frac\pi4\,tan\theta} = x\)

⇒ \(\frac{1 + tan\theta}{1 - tan\theta} + \frac{1-tan\theta}{1 + tan\theta} =x\)

⇒ \(\frac{(1 + tan\theta)^2 + (1 - tan\theta)^2}{(1 - tan\theta)(1 + tan\theta)} = x\)

⇒ \(x = \frac{2(1 + tan^2\theta)}{1 - tan^2\theta}\)     ......(1)

Now, \(tan^2(\frac\pi4 + \theta) + tan^2(\frac\pi4 - \theta)\)

\(= \left(\frac{1 + tan\theta}{1 - tan\theta}\right)^2 + \left(\frac{1 - tan\theta}{1 + tan\theta}\right)^2\)

\(= \frac{(1 + tan\theta)^4 + (1 - \tan\theta)^4}{(1 - tan\theta)^2(1 + tan\theta)^2}\)

\(= \frac{2(1 + 6tan^2\theta + tan^4\theta)}{(1 - tan^2\theta)^2}\)    ......(2)

From (1), 

\(x^2 -2 = \frac{4(1 + tan^2\theta)^2}{(1 - tan^2\theta)^2} - 2 \)

\(= \frac{4(1 +2tan^2\theta + tan^4\theta)}{(1 - tan^2\theta)^2} - 2\)

\(= \frac{4(1 + 2tan^2\theta + tan^4\theta) - 2(1 - 2tan^2\theta + tan^4\theta)}{(1 -tan^2\theta)^2}\)

\(= \frac{2 + 12tan^2\theta + 2tan^4\theta}{(1 - tan^2\theta )^2}\)

\(= \frac{2(1 + 6tan^2\theta + tan^4\theta)}{( 1 - tan^2\theta )^2}\)

\(= tan^2(\frac\pi4 + \theta) + tan^2(\frac\pi4 -\theta)\)    (From(2))

Hence proved.