Question

A)sin²2+cos²2 + tg²2 - 1/cos²2
b)(tg2×ctg2-cos²2)×1/sin²2

Answer 1

а) sin²2+cos²2 + tg²2 - 1/cos²2 =0

b) (tg2×ctg2-cos²2)×1/sin²2 = 1

Объяснение:

а) sin²2+cos²2 + tg²2 - 1/cos²2 =0

$\sin^22+\cos^22 + \tg^22 - \frac{1}{cos^22} = \\ (\sin^22+\cos^22) + \frac{\sin^22}{ \cos^22}- \frac{1}{\cos^22 } = \\ = 1 + \frac{\sin^22 - 1}{\cos^22} =...$

Заменим единицу в числителе по основному триг.тождеству на:

${{ [1 = sin^22+cos^22]}}\\$

Получим:

$...= 1 + \frac{\sin^22 - (\sin^22+\cos^22)}{\cos^22} = \\ = 1 + \frac{ - \cos^22}{\cos^22} = 1 - \frac{ \cos^22}{\cos^22} = \\ = 1 - \frac{ \cancel{\cos^22}}{\cancel{\cos^22}} = 1 - 1 = 0$

b) (tg2×ctg2-cos²2)×1/sin²2 = 1

$( \tg2 \cdot{ \ctg2}- \cos^22) \cdot \dfrac{1}{\sin^22} = \\ = ( \frac{\cancel{\sin2}}{\cancel{\cos2}} \cdot \frac{\cancel{\cos2}}{\cancel{\sin2}} - \cos^22) \cdot \dfrac{1}{\sin^22} = \\ = (1- \cos^22) \cdot \dfrac{1}{\sin^22} = \\ = ( \sin^{2} 2)\cdot \dfrac{1}{\sin^22} = \dfrac{\cancel{\sin^22}}{\cancel{\sin^22}} = 1 \\$