Question

ПОМОГИТЕ С ТЕСТОМ
С 13 ПО15

Answer 1

$13) \: {tg}^{2} x - {sin}^{2} x - {tg}^{2} x {sin}^{2} x = \\ = {tg}^{2} x(1 - {sin}^{2} x) - {sin}^{2} x = \\ = \frac{ {sin}^{2} x}{ {cos}^{2} x} \times {cos}^{2} x - {sin}^{2} x = \\ = {sin}^{2} x - {sin}^{2} x = 0 \\ 14) \: 2 {sin}^{2} x - 1 + {cos}^{2} x = \\ = 2 {sin}^{2} x - {sin}^{2} x - {cos}^{2} x + {cos}^{2} x = \\ = {sin}^{2} x \\ 15) \: {(tg \alpha + ctg \alpha )}^{2} - 2 = \\ {tg}^{2} \alpha + 2tg \alpha ctg \alpha + {ctg}^{2} \alpha - 2 = \\ = {tg}^{2} \alpha + {ctg}^{2} \alpha = \\ = \lbrace\alpha = \frac{\pi}{4} \rbrace = \\ = {tg}^{2} \frac{\pi}{4} + {ctg}^{2} \frac{\pi}{4} = {1}^{2} + {1}^{2} = 2$