Тригонометрия,срочно,пожалуйста!!!

Question 1

Question 2

Решите задачу:

$tg\alpha=\frac12;\ tg\beta=\frac13;\ \ \pi<\alpha+\beta<2\pi;\\ \alpha+\beta-?;\\ tg(\alpha+\beta)=\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}=\\ =\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}\cdot\frac{\frac{1}{\cos\alpha\cos\beta}}{\frac{1}{\cos\alpha\cos\beta}}=\\$
$=\frac{\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{\cos\alpha\cos\beta-\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}=\frac{tg\alpha+\tg\beta}{1-tg\alpha\cdot tg\beta}=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\\ =\frac{\frac{3+2}{2\cdot3}}{1-\frac16}=\frac{\frac56}{\frac56}=1;\\ tg(\alpha+\beta)=1;\ \ \pi<\alpha+\beta<2\pi;\\ \alpha+\beta=\pi+\frac\pi4=\frac{5\pi}{4}$

$\left \{ {{\sin x\sin y=\frac14} \atop {ctg x\cdot ctg y=3}} \right.;\ \ \\ \cos(x-y)-?;\\ D(f): x,y\neq \pi k,\ \ k\in Z;\\ ctgx\cdot ctgy=\frac{\cos x\cdot\cos y}{\sin x\cdot\sin y}=\frac{\cos x\cos y-\sin x\sin y+\sin x\sin y}{\sin x\sin y}=\\$
$\frac{(\cos x\cos y+\sin x\sin y)-\sin x\sin y}{\sin x\sin y}=\frac{\cos(x-y)-\sin x\sin y}{\sin x\sin y}=ctgx\cdot ctgy;\\ \cos(x-y)=ctgx\cdot ctgy\cdot\sin x\cdot\sin y+\sin x\sin y=\\ =\sin x\cdot\sin y\cdot(ctgx\cdot ctgy+1)=\frac14\cdot(3+1)=\frac14\cdot4=1;\\ \cos(x-y)=1$

Question 3

Громадное спасибо!

Question 4

1)tg(a+b)=(tga+tgb)/(1-tgatgb)=(1/2+1/3):(1-1/2*1/3)=(3/6+2/6):(1-1/6)=5/6:5/6=1
a+b=π/4+πn U π2)sinxsiny=1/4
ctgx*ctgy=cosxcosy/sinxsiny=3⇒cosxcosy=3*1/4=3/4
cos(x-y)=cosxcosy+sinxsiny=3/4+1/4=1