<s>[INST] Find the value of $\Biggl(\Biggl(\tan^2\frac{7\pi}{24}-\tan^2\frac{\pi}{24}\Biggr):\Biggl(1-\tan^2\frac{7\pi}{24}\tan^2\frac{\pi}{24}\Biggr)\Biggr)^2$ This problem reminds me of the formula $\tan(x\pm y)=\frac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}.$ But two minus signs in the problem interfere in applying the formula. I can't think of other ways. How to apply this formula?
 [/INST] Hint: $A^{2}-B^{2} = (A-B)(A+B)$
Further hint:

 Use the hint to rewrite the expression as $\left(\frac{\tan{(7\pi/24)}-\tan{(\pi/24)}}{1+\tan{(7\pi/24)}\tan{(\pi/24)}}\cdot\frac{\tan{(7\pi/24)}+\tan{(\pi/24)}}{1-\tan{(7\pi/24)}\tan{(\pi/24)}}\right)^{2}$

Answer:

 The addition formula for $\tan$ then gives you the answer: $\left(\tan{\frac{6\pi}{24}}\tan{\frac{8\pi}{24}}\right)^{2} = \left(\tan{\frac{\pi}{4}}\tan{\frac{\pi}{3}}\right)^{2} = \left(1\cdot\sqrt{3}\right)^{2} = 3$

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