Question

am nevoie de ajutor, cine ma poate ajuta va rog

Answer 1



1. a). 1/2 + (√6)(√2/2)(√3/2) - √3/3(-√3) + 1 =

1/2 + √36/2 + √9/3 + 1 =

1/2 + 6/2 + 3/3 + 1 =

1/2 + 3 + 1 + 1 = 11/2

b). cos 60° - √2 sin 135° + √3 tg 60° =

1/2 + √2(√2/2) + √3(√3) =

1/2 + √4/2 + √9 =

1/2 + 1 + 3 = 9/2

$2. a). \frac{1 - cos^{2}\alpha}{(\sqrt{1 - cos^{2}\alpha)^{2}}}} = \frac{1 - cos^{2}\alpha}{1 - cos^{2}\alpha}}} = 1$

b. $sin (3 \pi +\alpha) + cos (\pi - \alpha) - sin(- \alpha) + cos (- \alpha) =$$- sin \alpha - cos \alpha + sin \alpha +cos \alpha = 0$

3. $(sin^{2} \alpha - cos^{2} \alpha)^{2} + 4sin^{2}\alpha cos^{2} \alpha =$

$sin^{4} \alpha - 2sin^{2}\alpha cos^{2} \alpha + cos^{4} \alpha+ 4sin^{2}\alpha cos^{2} \alpha =$

$sin^{4} \alpha + 2sin^{2}\alpha cos^{2} \alpha + cos^{4} \alpha$

$= [(sin \alpha)^{2} + (cos\alpha)^{2}]^{2} = 1 ^{2} = 1$]

b. $tg \alpha + ctg \alpha = \frac{sin \alpha }{cos \alpha} + \frac{cos \alpha}{sin \alpha} =$

$\frac{sin^{2} \alpha + cos^{2} \alpha }{sin \alpha cos \alpha} = \frac{1}{0.3} = \frac{10}{3} = 3.(3)$