Odpowiedź :
5. tgα = 2 + √3
tg(90° - α) = ctgα
ctgα = 1/tgα
ctgα = 1/(2 + √3) = (2 - √3)/[2 + √3)(2 - √3) = (2 - √3)/(4 - 3) =
= (2 - √3)/1 = 2 - √3
5.
[tex]tg\alpha = 2+\sqrt{3}\\\\tg(90^{o}-\alpha) = ctg\alpha\\\\ctg\alpha = \frac{1}{tg\alpha} = \frac{1}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{4-3} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}\\\\Odp. \ A.[/tex]
6.
[tex]cos\alpha = 0,96\\sin\alpha = ?\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha = 1 - cos^{2}\alpha = 1 - 0,96^{2} = 1 - 0,9216 = 0,0784\\\\sin^{2}\alpha = 0,0784\\\\sin\alpha = \sqrt{0,0784}\\\\sin\alpha = 0,28\\\\Odp. \ D.[/tex]
7.
[tex]2cos\alpha\cdot(sin\alpha\cdot tg\alpha+cos\alpha) = 2cos\alpha\cdot(sin\alpha\cdot\frac{sin\alpha}{cos\alpha} + cos\alpha) =2cos\alpha(\frac{sin^{2}\alpha}{cos\alpha}+cos\alpha) =\\\\=2sin^{2}\alpha + 2cos^{2}\alpha = 2(sin^{2}\alpha + cos^{2}\alpha) = 2\cdot1 = 2\\\\Odp. \ D.[/tex]