Răspuns:
Explicație pas cu pas:
Avem formula
[tex]\tan\dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{1+\cos x}}, \ x\in\left[0,\pi\right)[/tex]
Atunci
[tex]\tan\dfrac{\pi}{8}=\tan\dfrac{\frac{\pi}{4}}{2}=\sqrt{\dfrac{1-\cos\frac{\pi}{4}}{1+\cos\frac{\pi}{4}}}=\sqrt{\dfrac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}=\sqrt{\dfrac{2-\sqrt{2}}{2+\sqrt{2}}}=\\=\dfrac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1[/tex]
Atunci
[tex]\cot\dfrac{\pi}{8}=\dfrac{1}{\sqrt{2}-1}[/tex]
Rezultă
[tex]\cot\dfrac{\pi}{8}-\tan\dfrac{\pi}{8}=\dfrac{1}{\sqrt{2}-1}-(\sqrt{2}-1)=\dfrac{1-(\sqrt{2}-1)^2}{\sqrt{2}-1}=\dfrac{2(\sqrt{2}-1)}{\sqrt{2}-1}=2[/tex]
