Question

Помогите пожалуйста только б,в.​

Answer 1

б) приведем к общему знаменателю,

tgα+(cosα/(sinα-1))=(sinα/cosα)+cosα/(sinα-1)=

(sin²α-sinα+cos²α)/(cosα*(sinα-1))=(1-sinα)/(-(1-sinα)*сosα)=-1/cosα;

-1/cos(-3π/4)=-1/cos(3π/4)= -1/(сos(π-(π/4))=-1/(-сosπ/4))=-1/(-1/√2)=√2

d) ctg²α*(cos²α-1)+2cos²α=cos²α*(((cos²α-1)/sin²α)+2)=

cos²α*((cos²α-1+2sin²α))/sin²α=cos²α*((-sin²α+2sin²α))/sin²α=

cos²α*(sin²α)/sin²α=cos²α; cos²(7π/3)=cos²(2π+π/3)=cos²(π/3)=(1/2)²=1/4

ПОЯСНЕНИЕ К РЕШЕНИЮ

использовал основное тригонометрическое тождество sin²α+cos²α=1; tgα=sinα/cosα; формулу приведения

(сos(π-(π/4))=-сosπ/4;  четность косинуса, т.е. cos(-3π/4)=cos(3π/4); сtg²α=cos²α/sin²α; cos²(2π+π/3)- можно отбрасывать период косинуса 2π; (cos²α-1)=-(1-cos²α)=-sin²α

Answer 2

Ответ:

Формулы:  $1=sin^2a+cos^2a\ \ ,\ \ a^2-b^2=(a-b)(a+b)$

$\displaystyle 1)\ \ a=\frac{\pi }{6}\ \ \ ,\ \ \ \frac{1-2sina\cdot cosa}{sin^2a-cos^2a}=\frac{sin^2a+cos^2a-2sina\cdot cosa}{(sina-cosa)(sina+cosa)}=\\\\\\=\frac{(sina-cosa)^2}{(sina-cosa)(sina+cosa)}=\frac{sina-cosa}{sina+cosa}=\frac{sin\frac{\pi}{6}-cos\frac{\pi}{6}}{sin\frac{\pi}{6}+cos\frac{\pi}{6}}=\frac{\frac{1}{2}-\frac{\sqrt3}{2}}{\frac{1}{2}+\frac{\sqrt3}{2}}=\\\\\\=\frac{1-\sqrt3}{1+\sqrt3}=\frac{(1-\sqrt3)^2}{(1+\sqrt3)(1-\sqrt3)}=\frac{1-2\sqrt3+3}{1-3}=\frac{4-2\sqrt3}{-2}=\sqrt3-2$

Формулы:    $1-cosx=2sin^2\frac{x}{2}\ \ ,\ \ sin(\frac{\pi}{2}+x)=cosx\ .$

$\displaystyle 2)\ \ a=-\dfrac{3\pi}{4}\ \ ,\ \ tga+\frac{cosa}{sina-1}=tga-\frac{cosa}{1-sina}=tga-\frac{cosa}{1-cos(\frac{\pi}{2}-a)}=\\\\\\=tga-\frac{cosa}{2sin^2(\frac{\pi}{4}-\frac{a}{2})}=tg(-\frac{3\pi}{4})-\frac{cos(-\frac{3\pi}{4})}{2sin^2(\frac{\pi}{4}+\frac{3\pi}{8})}=-tg\frac{3\pi}{4}-\frac{cos\frac{3\pi}{4}}{2sin^2\frac{5\pi}{8}}=$

$\displaystyle =-(-1)-\frac{-\frac{\sqrt2}{2}}{2sin^2(\frac{\pi}{2}+\frac{\pi}{8})}=1+\frac{\sqrt2}{4cos^2\frac{\pi}{8}}=1+\frac{\sqrt2}{4\cdot \frac{2+\sqrt2}{4}}=1+\frac{\sqrt2}{2+\sqrt2}=\\\\\\=\frac{2+2\sqrt2}{2+\sqrt2}=\frac{2\cdot (1+\sqrt2)}{\sqrt2\, (\sqrt2+1)}=\frac{2}{\sqrt2}=\sqrt2\\\\\\\star \ \ cos^2\frac{\pi}{8}=cos^2(\frac{\pi}{4}:2)=\frac{1+cos\frac{\pi}{4}}{2}=\frac{1+\frac{\sqrt2}{2}}{2}=\frac{2+\sqrt2}{4}\ \ \star$  

Формулы:  $ctga=\dfrac{cosa}{sina}\ \ ,\ \ sin^2a+cos^2a=1\ \ \Rightarrow \ \ sin^2a=1-cos^2a$

$\displaystyle 3)\ \ a=\frac{7\pi }{3}\ \ ,\ \ ctg^2a\cdot (cos^2a-1)+2cos^2a=\frac{cos^2a}{sin^2a}\cdot (-sin^2a)+2cos^2a=\\\\\\=-cos^2a+2cos^2a=cos^2a=cos^2\frac{7\pi }{3}=cos^2(2\pi +\frac{\pi}{3})=cos^2\frac{\pi}{3}=\Big(\frac{1}{2}\Big)^2=\frac{1}{4}$