Помогите решить заранее спасибо!

1)
$\frac{cos(a+125^{\circ})}{sin(a+35^{\circ})} = \frac{cos((a+35^{\circ})+90^{\circ})}{sin(a+35^{\circ})} = \frac{-sin(a+35^{\circ})}{sin(a+35^{\circ})} =-1$
Ответ: -1
2)
$b_n=b_1*q^{n-1}$
$b_3=b_1*q^2 \\b_5=b_1*q^4 \\b_7=b_1*q^6 \\ \left \{ {{b_1*q^4+b_1*q^2=60} \atop {b_1*q^6-b_1*q^4=144}} \right. \\b_1(q^4+q^2)=60 \\b_1= \frac{60}{q^4+q^2} \\ \frac{60q^6}{q^4+q^2} - \frac{60q^4}{q^4+q^2} =144 \\ \frac{60q^4}{q^2+1} - \frac{60q^2}{q^2+1} =144 \\ \frac{60q^4-60q^2}{q^2+1} =144 \\q^2=x,\ x \geq 0 \\ \frac{60x^2-60x}{x+1} =144 \\x \neq 1 \\60x^2-60x=144(x+1) \\60x^2-60x=144x+144 \\60x^2-204x-144=0 \\5x^2-17x-12=0 \\D=289+240=529=23^2 \\x_1= \frac{17+23}{10}= \frac{40}{10}=4$
$x_2= \frac{17-23}{10} \ \textless \ 0 \\q^2=4 \\q=\pm2 \\b_1= \frac{60}{q^4+q^2}= \frac{60}{16+4} =3 \\b_2=\pm 3*2=\pm6 \\b_4=\pm3*8=\pm24 \\b_2*b_4=\pm6*\pm24=144$
Ответ: 144
4)
$tg(-x)*ctg(-x)-sin^2(-x)=-tgx*(-ctgx)-(-sinx)^2= \\=tgx*ctgx-sin^2x=1-sin^2x=cos^2x$
Ответ: $cos^2x$