Помогите с алгеброй пример e,f,g

Question 1

Question 2

Решите задачу:

$1)\; \; ctga=-7\; ,\\\\\frac{3\pi}{2}\ \textless \ a\ \textless \ 2\pi \; \; \Rightarrow \; \; \; sina\ \textless \ 0\; ,\; \; cosa\ \textgreater \ 0\; ,\; \; tga\ \textless \ 0\\\\1+ctg^2a=\frac{1}{sin^2a}\; \; \Rightarrow \; \; \; sin^2a= \frac{1}{1+ctg^2a} =\frac{1}{1+49}=\frac{1}{50}=0,02\\\\sina=-\sqrt{0,02}=-\frac{\sqrt2}{10}\\\\cosa=+\sqrt{1-sin^2a}=\sqrt{1-0,02}=\sqrt{0,98}=\frac{\sqrt{98}}{10}=\frac{7\sqrt2}{10}\\\\tga=\frac{sina}{cosa}=-\frac{\sqrt2}{10}:\frac{7\sqrt2}{10}=-\frac{1}{7}$

$2)\; \; sina=-\frac{2}{3}\\\\\frac{3\pi}{2}\ \textless \ a\ \textless \ 2\pi \; \; \Rightarrow \; \; \; cosa\ \textgreater \ 0,\; tga\ \textless \ 0\; ,ctga\ \textless \ 0$

$cosa=+\sqrt{1-sin^2a}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt5}{3}\\\\tga=\frac{sina}{cosa}=-\frac{2}{3}:\frac{\sqrt5}{3}==-\frac{2}{3}\cdot \frac{3}{\sqrt5}=- \frac{2}{\sqrt5}\\\\ctg=\frac{1}{tga}=-\frac{\sqrt5}{2}$

$3)\; \; tga=-2\; ,\; \; 0\ \textless \ a\ \textless \ \frac{3\pi}{2}\\\\Tak\; kak\; tga\ \textless \ 0\; ,\; to\; \; \frac{\pi}{2}\ \textless \ a\ \textless \ \pi \; \Rightarrow \; \; cosa\ \textless \ 0,\; sina\ \textgreater \ 0,\; ctga\ \textless \ 0.\\\\1+tg^2a=\frac{1}{cos^2a}\; \; \Rightarrow \; \; \; cos^2a=\frac{1}{1+tg^2a}=\frac{1}{1+4}=\frac{1}{5}\\\\cosa=- \frac{1}{\sqrt5}\\\\sina=+\sqrt{1-cos^2a}=\sqrt{1-\frac{1}{5}}=\frac{2}{\sqrt5}\\\\ctga=\frac{1}{tga}=-\frac{1}{2}$