10. Найдите: tg²α, если 13sin²α + 5cos²α = 11.

Дано: $$13\sin^2{\alpha} + 5\cos^2{\alpha} = 11$$. Найти: $$\tan^2{\alpha}$$. Преобразуем данное уравнение: $$13\sin^2{\alpha} + 5\cos^2{\alpha} = 11$$ $$13\sin^2{\alpha} + 5(1 - \sin^2{\alpha}) = 11$$ $$13\sin^2{\alpha} + 5 - 5\sin^2{\alpha} = 11$$ $$8\sin^2{\alpha} = 6$$ $$\sin^2{\alpha} = \frac{6}{8} = \frac{3}{4}$$ Тогда $$\cos^2{\alpha} = 1 - \sin^2{\alpha} = 1 - \frac{3}{4} = \frac{1}{4}$$ $$\tan^2{\alpha} = \frac{\sin^2{\alpha}}{\cos^2{\alpha}} = \frac{\frac{3}{4}}{\frac{1}{4}} = \frac{3}{4} \cdot \frac{4}{1} = 3$$ Ответ: 3