Trigonometric identities in one angle

Using the following relation, we can find the cosine of the above mentioned angle: $$1+\tan^2(\alpha)=\dfrac{1}{\cos^2(\alpha)} \Rightarrow \cos^2(\alpha)=\dfrac{1}{1+\tan^2(\alpha)}$$ By substituting, we obtain: $$\cos^2(\alpha)=\dfrac{1}{1+\tan^2(\alpha)}=\dfrac{1}{1+2^2}=\dfrac{1}{5} \Rightarrow \cos(\alpha)=\pm\sqrt{\dfrac{1}{5}}=\pm\dfrac{\sqrt{5}}{5}$$

From the following relation $$\sin^2(\alpha)+\cos^2(\alpha)=1 \Rightarrow \sin^2(\alpha)=1-\cos^2(\alpha)$$ By substituting we then have: $$\sin^2(\alpha)=1-\cos^2(\alpha)=1-\dfrac{1}{5}=\dfrac{5-1}{5}=\dfrac{4}{5} \Rightarrow \sin(\alpha)=\pm\sqrt{\dfrac{4}{5}}=\pm\dfrac{2}{\sqrt{5}}=\pm\dfrac{2\sqrt{5}}{5}$$ Bearing in mind that $0 < \alpha < 90^\circ$, the cosine and the sine take positive values. Therefore, the correct solution is the result of taking the positive determinations of these angles.