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Cel mai inteligent răspuns!
2014-02-19T22:01:32+02:00
1+ctg^{2}x=\frac{1}{sin^{2}x} \Leftrightarrow
\Leftrightarrow 1+ \frac{cos^{2}x}{sin^{2}x}=\frac{1}{sin^{2}x} \Leftrightarrow
\Leftrightarrow \frac{cos^{2}x+sin^{2}x}{sin^{2}x}=\frac{1}{sin^{2}x}
De unde egalitatea.
1+tg^{2}x=\frac{1}{cos^{2}x}<- Se rezolva exact ca exemplul de mai sus.
tg^{2}x-sin^{2}x=tg^{2}x*sin^{2}x \Leftrightarrow
\Leftrightarrow \frac{sin^{2}x}{cos^{2}x}-\frac{sin^{4}x}{cos^{2}x}=\frac{sin^{4}x}{cos^{2}x} \Leftrightarrow
\Leftrightarrow sin^{2}x*(1-sin^{2}x)=sin^{4}x
1-sin^{2}x=sin^{2}x
Sigur e bine exercitiul?
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