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[kapka1a]

1)
\(\displaystyle{ \frac{tgx}{sinx}}\)

2)
\(\displaystyle{ cosx+cosx tg^2x}\)

3)
\(\displaystyle{ sinx \sqrt{1+ctg^2x}}\) przy \(\displaystyle{ x (\pi , \frac{3}{2} \pi)}\)

[bolo]

1.

\(\displaystyle{ \frac{tgx}{sin x}= \\ =\frac{\frac{sin x}{cos x}}{sin x}= \\ =\frac{1}{cos x}}\)

2.

\(\displaystyle{ cos x+cos x\cdot tg^{2}x=\\=cos x+\frac{sin^{2}x}{cos x}=\\=\frac{cos^{2}x+sin^{2}x}{cos x}=\\=\frac{1}{cos x}}\)

3.

\(\displaystyle{ sin x\cdot\sqrt{1+ctg^{2}x}=\\=sin x\cdot\sqrt{\frac{{cos^{2}}x+sin^{2}x}{sin^{2}x}}=\\=sin x\cdot\sqrt{\frac{1}{sin^{2}}}=\\=\frac{sin x}{|sin x|}=-1}\)

[kapka1a]

ale jak \(\displaystyle{ cosx+\frac{sin^2x}{cosx}}\)
daje \(\displaystyle{ \frac{cos^2x+sin^2x}{cosx}}\)

Moim zdaniem powinno być \(\displaystyle{ \frac{cosx+sin^2x}{cosx}}\)

bo wspólnym mianownikiem jest \(\displaystyle{ cosx}\)

[bolo]

Ale: \(\displaystyle{ \frac{cos x+sin^{2}x}{cos x}=\frac{cos x}{cos x}+\frac{sin^{2}x}{cos x}=1+\frac{sin^{2}x}{cos x}}\). Więc w liczniku musi być \(\displaystyle{ cos^{2}x}\).

[kapka1a]

nie rozumie naprawde chyba jestem za głupi

[bolo]

\(\displaystyle{ cos x+\frac{sin^{2}x}{cos x}=\frac{cos^{2}x}{cos x}+\frac{sin^{2}x}{cos x}=\frac{cos^{2}x+sin^{2}x}{cos x}}\)