Помогите.............................................

Задание 1.
ctg a = -√2; pi/2 < a < pi
Такой промежуток означает 2 четверть, то есть sin a > 0, cos a < 0.
tg a = 1/ctg a = -1/√2 = -√2/2
1 + ctg^2 a = 1/sin^2 a
$sin (a)= \frac{1}{ \sqrt{1+ctg^2(a)} } = \frac{1}{ \sqrt{1+2} } = \frac{1}{ \sqrt{3} }= \frac{ \sqrt{3} }{3}$
$cos(a)=- \sqrt{1-sin^2(a)} =- \sqrt{1- \frac{1}{3} } =- \sqrt{ \frac{2}{3} } =- \frac{ \sqrt{6} }{3}$

Задание 2.
1) sin^2 (2a) + cos^2 (2a) + ctg^2 (5a) = 1 + ctg^2 (5a) = 1/sin^2 (5a)
2) $\frac{ctg(a)}{tg(a)+ctg(a)} = \frac{cos(a)}{sin(a)} :( \frac{sin(a)}{cos(a)}+\frac{cos(a)}{sin(a)})=\frac{cos(a)}{sin(a)}: \frac{sin^2(a)+cos^2(a)}{sin(a)*cos(a)} = \\ = \frac{cos(a)}{sin(a)}*sin(a)*cos(a)=cos^2(a)$