Помогите пожалуйста 1) ctg(arcsin1/5) 2) sin²(1/2arcsin1/4)-cos²(1/2arcsin1/4) 3)...

Решите задачу:

$1) ctg(arcsin1/5)=\frac{cos(arcsin1/5)}{sin(arcsin1/5)}=\frac{\sqrt{1-sin^2(arcsin1/5)}}{1/5}=\frac{\sqrt{1-(1/5)^2}}{1/5}=\\= \frac{\sqrt{24}}{5}:\frac{1}{5}=\sqrt{24}$

$2) sin^2(1/2arcsin1/4)-cos^2(1/2arcsin1/4)=[arcsin1/4=\alpha]=\\= sin^2\frac{\alpha}{2}-cos^2\frac{\alpha}{2}=-cos\alpha=-cos(arcsin1/4)=\\=-\sqrt{1-sin^2(arcsin1/4)}=-\sqrt{1-1/16}=-\frac{\sqrt{15}}{4}$

$3) cos(arccos3/4-arcsin1/3)=[arccos3/4=\alpha, arcsin1/3=\beta]=\\=cos(\alpha-\beta)=cos\alpha*cos\beta+sin\alpha*sin\beta=\\= cos(arccos3/4)*cos(arcsin1/3)+sin(arccos3/4)*sin(arcsin1/3)=\\= 3/4*\sqrt{1-1/9}+\sqrt{1-9/16}*1/3=\frac{3}{4}*\frac{\sqrt8}{3}+\frac{\sqrt7}{4}*\frac{1}{3}=\\=\frac{\sqrt2}{2}+\frac{\sqrt7}{12}=\frac{6\sqrt2+\sqrt7}{12}$