Ответ:

Значение   [tex]\bf sin\dfrac{\pi }{4}=cos\dfrac{\pi }{4}=\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}[/tex]   .  

[tex]\bf \sqrt{\Big(1+2sin\dfrac{\pi }{4}\Big)^2}-\sqrt{\Big(1-2cos\dfrac{\pi }{4}\Big)^2}=\sqrt{\Big(1+2\cdot \dfrac{\sqrt2}{2}\Big)^2}-\sqrt{\Big(1-2\cdot \dfrac{\sqrt2}{2}\Big)^2}=\\\\\\=\sqrt{(1+\sqrt2)^2}-\sqrt{(1-\sqrt2)^2}=|1+\sqrt2|-|1-\sqrt{2}|=1+\sqrt2-(\sqrt2-1)=2[/tex]

Или, возможно, такое было условие :

[tex]\bf \sqrt{1+2sin^2\dfrac{\pi }{4}}-\sqrt{1-2cos^2\dfrac{\pi }{4}}=\sqrt{1+2\cdot \dfrac{1}{2}}-\sqrt{1-2\cdot \dfrac{1}{2}}=\\\\\\=\sqrt{2}-\sqrt{0}=\sqrt{2}[/tex]