Упр.7.13 ГДЗ Мордкович 10-11 класс (Алгебра)
- а) $$\frac{\sin t}{1+\cos t}+\frac{\sin t}{1-\cos t}$$; б) $$\operatorname{ctg}^2(t)\left(\cos^2(t)-1\right)+1$$; в) $$\frac{\cos t}{1+\sin t}+\frac{\cos t}{1-\sin t}$$; г) $$\frac{\operatorname{tg} t+1}{1+\operatorname{ctg} t}$$.
а)
$$\frac{\sin t}{1+\cos t}+\frac{\sin t}{1-\cos t} = \frac{\sin t(1-\cos t)+\sin t(1+\cos t)}{(1+\cos t)(1-\cos t)}$$
$$= \frac{\sin t-\sin t\cos t+\sin t+\sin t\cos t}{1-\cos^2 t} = \frac{2\sin t}{\sin^2 t} = \frac{2}{\sin t}$$
Ответ: $$\frac{2}{\sin t}$$
б)
$$\ctg^2 t\cdot(\cos^2 t-1)+1 = -\ctg^2 t\cdot(1-\cos^2 t)+1$$
$$= -\frac{\cos^2 t}{\sin^2 t}\cdot \sin^2 t+1 = -\cos^2 t+1 = \sin^2 t$$
Ответ: $$\sin^2 t$$
в)
$$\frac{\cos t}{1+\sin t}+\frac{\cos t}{1-\sin t} = \frac{\cos t(1-\sin t)+\cos t(1+\sin t)}{(1+\sin t)(1-\sin t)}$$
$$= \frac{\cos t-\sin t\cos t+\cos t+\sin t\cos t}{1-\sin^2 t} = \frac{2\cos t}{\cos^2 t} = \frac{2}{\cos t}$$
Ответ: $$\frac{2}{\cos t}$$
г)
$$\frac{\tg t+1}{1+\ctg t} = \frac{\frac{\sin t}{\cos t}+1}{1+\frac{\cos t}{\sin t}} = \frac{\frac{\sin t+\cos t}{\cos t}}{\frac{\sin t+\cos t}{\sin t}} = \frac{\sin t}{\cos t} = \tg t$$
Ответ: $$\tg t$$
