Question

помогите пожалуйста
А) (1 - sin a)(1 + sin a)

б) 1/cos^2a - tg^2a

в) cos^a/1-sin^2a

г) 1-cos^2a+sin^2

д) cos^4 a + sin^2a cos^2a

е) sin a/1+cos a + 1+cos a/sin a

Answer 1

Решение:

а) (1-sinα)(1+sinα) = 1+sinα-sinα-sinα×sinα = 1+sinα-sinα-sin²α = 1-sin²α = cos²α

б) $\frac{1}{cos^{2}\alpha}-tg^{2}\alpha = \frac{1}{cos^{2}\alpha}-\frac{sin^{2}\alpha}{cos^{2}\alpha} = \frac{1-sin^{2}\alpha}{cos^{2}\alpha} = \frac{cos^{2}\alpha}{cos^{2}\alpha} = 1$

в) $\frac{cos^{2}\alpha}{1-sin^{2}\alpha} =\frac{cos^{2}\alpha}{cos^{2}\alpha}=1$

г)  1-cos²α+sin²α = sin²α+sin²α = 2sin²α

д) cos⁴α + sin²α×cos²α = (cos²α + sin²α)×cos²α = 1×cos²α = cos²α

е) $\frac{sin\alpha}{1+cos\alpha}+\frac{1+cos\alpha}{sin\alpha} = \frac{sin^{2}\alpha+(1+cos\alpha)^{2}}{(1+cos\alpha)*sin\alpha} = \frac{sin^{2}\alpha + 1+2cos\alpha +cos^{2}\alpha}{(1+cos\alpha)*sin\alpha} = \frac{1+1+2cos\alpha}{(1+cos\alpha)*sin\alpha} = \frac{2+2cos\alpha}{(1+cos\alpha)*sin\alpha} = \frac{2(1+cos\alpha)}{(1+cos\alpha)*sin\alpha} = \frac{2}{sin\alpha} = 2csc\alpha$