Question

Va rog ajutati-ma cu un ex. la mate.: Aratati ca:
tg²x-sin²x=tg²x*sin²x
1+tg²x=1/cos²*x
1+ctg²x=1/sin²x

Answer 1

$1+ctg^{2}x=\frac{1}{sin^{2}x} \Leftrightarrow$
$\Leftrightarrow 1+ \frac{cos^{2}x}{sin^{2}x}=\frac{1}{sin^{2}x} \Leftrightarrow$
$\Leftrightarrow \frac{cos^{2}x+sin^{2}x}{sin^{2}x}=\frac{1}{sin^{2}x}$
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?