Question

Tylko podpunkt f)
Proszę niech ktoś pomoże

Answer 1

Odpowiedź:

[tex]\displaystyle \frac{1-sin13^{\circ}}{cos13^{\circ}} -\frac{cos13^{\circ}}{1+sin13^{\circ}} =\frac{(1-sin13^{\circ})(1+sin13^{\circ})-cos^213^{\circ}}{(1+sin13^{\circ})cos13^{\circ}} =\\=\frac{1-sin^213^{\circ}-cos^213^{\circ}}{(1+sin13^{\circ})cos13^{\circ}} =\frac{cos^213^{\circ}-cos^213^{\circ}}{(1+sin13^{\circ})cos13^{\circ}} =\frac{0}{(1+sin13^{\circ})cos13^{\circ}} =0[/tex]

Answer 2

Verified answer

Odpowiedź:

[tex]\huge\boxed{f) \ \frac{1-sin13^{o}}{cos13^{o}}-\frac{cos13^{o}}{1+sin13^{o}} = 0}[/tex]

Szczegółowe wyjaśnienie:

Korzystamy z jedynki trygonometrycznej:

[tex]sin^{2}\alpha + cos^{2}\alpha = 1[/tex]

oraz ze wzoru skróconego mnożenia:

[tex](a-b)(a+b) = a^{2}-b^{2}[/tex]

[tex]f) \ \frac{1-sin13^{o}}{cos13^{o}} - \frac{cos13^{o}}{1+sin13^{o}}=\frac{(1-sin13^{o})(1+sin13^{o})-cos^{2}13^{o}}{cos13^{o}(1+sin13^{o})}=\frac{1-sin^{2}13^{o}-cos^{2}13^{o}}{cos13^{o}(1+sin13^{o})}=\\\\\\=\frac{sin^{2}13^{o}+cos^{2}13^{o}-sin^{2}13^{o}-cos^{2}sin13^{o}}{cos13^{o}(1+sin13^{o}} = \frac{0}{cos13^{o}(1+sin13^{o}} = \boxed{0}[/tex]