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Решите задачу:

$1)\quad \frac{(1-sin^2a)+ctg^2a\cdot sin ^2a}{cos^2a} =\frac{cos^2a+\frac{cos^2a}{sin^2a}\cdot sin^2a}{cos^2a} = \frac{2cos^2a}{cos^2a} =2\\\\2)\quad \frac{1+ctg^2a}{1+tg^2a} = \frac{\frac{1}{sin^2a}}{\frac{1}{cos^2a}} =\frac{cos^2a}{sin^2a} =ctg^2a\\\\3)\quad \frac{tga}{sina} - \frac{sina}{ctga}=\frac{\frac{sina}{cosa}}{sina} - \frac{sina}{\frac{cosa}{sina}} =\frac{1}{cosa}-\frac{sin^2a}{cosa}= \frac{1-sin^2a}{cosa}=\frac{cos^2a}{cosa}=cosa$