Упр.7.12 ГДЗ Мордкович 10-11 класс (Алгебра)

a) ctg t — (cos t — 1) / sin t;
6) ctg^2(t) — (sin^(-2)(t) — 1);
в) cos^2(t) — (ctg^2(t) + 1) * sin^2(t);
г) (sin^2(t) — 1) / (cos^2(t) — 1) + tg t *ctg t.

1. $$\ctg t-\frac{\cos t-1}{\sin t}=\frac{\cos t}{\sin t}-\frac{\cos t-1}{\sin t}=\frac{\cos t-\cos t+1}{\sin t}=\frac{1}{\sin t}.$$

2. $$\ctg^2 t-\left(\sin^{-2} t-1\right)=\frac{\cos^2 t}{\sin^2 t}-\left(\frac{1}{\sin^2 t}-1\right)=\frac{\cos^2 t-1+\sin^2 t}{\sin^2 t}=\frac{0}{\sin^2 t}=0.$$

3. $$\cos^2 t-\left(\ctg^2 t+1\right)\sin^2 t=\cos^2 t-\left(\frac{\cos^2 t}{\sin^2 t}+1\right)\sin^2 t$$

   $$=\cos^2 t-\left(\cos^2 t+\sin^2 t\right)=\cos^2 t-1=-\sin^2 t.$$

4. $$\frac{\sin^2 t-1}{\cos^2 t-1}+\tg t\cdot\ctg t=\frac{-(1-\sin^2 t)}{-(1-\cos^2 t)}+1=\frac{\cos^2 t}{\sin^2 t}+1$$

   $$=\frac{\cos^2 t+\sin^2 t}{\sin^2 t}=\frac{1}{\sin^2 t}.$$

Ответ

а) $$\frac{1}{\sin t}$$; б) $$0$$; в) $$-\sin^2 t$$; г) $$\frac{1}{\sin^2 t}$$.