Question

помогите решить
заранее спасибо!

Answer 1

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: $cos^2x$