Petrovcic2x

2022-06-20

Find the value of $\left(\left({\mathrm{tan}}^{2}\frac{7\pi }{24}-{\mathrm{tan}}^{2}\frac{\pi }{24}\right):\left(1-{\mathrm{tan}}^{2}\frac{7\pi }{24}{\mathrm{tan}}^{2}\frac{\pi }{24}\right){\right)}^{2}$

### Answer & Explanation

last99erib

Hint: ${A}^{2}-{B}^{2}=\left(A-B\right)\left(A+B\right)$
Further hint:
Use the hint to rewrite the expression as ${\left(\frac{\mathrm{tan}\left(7\pi /24\right)-\mathrm{tan}\left(\pi /24\right)}{1+\mathrm{tan}\left(7\pi /24\right)\mathrm{tan}\left(\pi /24\right)}\cdot \frac{\mathrm{tan}\left(7\pi /24\right)+\mathrm{tan}\left(\pi /24\right)}{1-\mathrm{tan}\left(7\pi /24\right)\mathrm{tan}\left(\pi /24\right)}\right)}^{2}$
The addition formula for $\mathrm{tan}$ then gives you the answer:
${\left(\mathrm{tan}\frac{6\pi }{24}\mathrm{tan}\frac{8\pi }{24}\right)}^{2}={\left(\mathrm{tan}\frac{\pi }{4}\mathrm{tan}\frac{\pi }{3}\right)}^{2}={\left(1\cdot \sqrt{3}\right)}^{2}=3$