Question

ПОМОГИТЕ РЕШИТЬ СРОЧНО ПРОШУ

Answer 1

Ответ:

$\displaystyle 1a)\ \ sina=-\dfrac{2}{3}\\\\\frac{3\pi}{2}<a<2\pi \ \ \ \Rightarrow \ \ \ cosa>0\ ,\ tga<0\ ,\ ctga<0\\\\cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt5}{3}\\\\tga=\frac{sina}{cosa}=-\frac{2}{\sqrt5}\ \ ,\ \ \ ctga=\frac{cosa}{sina}=-\frac{\sqrt5}{2}$

$\displaystyle 1b)\ \ cosa=-\dfrac{7}{25}\\\\\pi <a<\frac{3\pi}{2}\ \ \ \Rightarrow \ \ \ sina<0\ ,\ tga>0\ ,\ ctga>0\\\\sina=-\sqrt{1-cos^2a}=-\sqrt{1-\dfrac{49}{625}}=-\sqrt{\frac{576}{625}}=-\frac{24}{25}\\\\tga=\frac{sina}{cosa}=\frac{24}{7}\ \ ,\ \ \ ctga=\frac{cosa}{sina}=\frac{7}{24}$

$2)\ \ tgx+\dfrac{cosx}{1+sinx}=\dfrac{sinx}{cosx}+\dfrac{cosx}{1+sinx}=\dfrac{sinx+sin^2x+cos^2x}{cosx\, (1+sinx)}=\\\\\\=\dfrac{sinx+1}{cosx(1+sinx)}=\dfrac{1}{cosx}\\\\\\3)\ \ sin^2a-cos^2a=sin^4a-cos^4a\ \ \ (!)\\\\sin^2a-cos^2a=(sin^2a-cos^2a)\cdot 1=(sin^2a-cos^2a)(\underbrace{sin^2a+cos^2a}_{1})=\\\\=sin^4a-cos^4a\ ;\\\\sin^4a-cos^4a=sin^4a-cos^4a$