Question

Докажите тождество. 100б

Answer 1

Доказать тождество:

4.

$\frac{cos\alpha +ctg\alpha }{1+sin\alpha }=ctg\alpha\\\\\frac{\cos (\alpha)+\frac{\cos (\alpha)}{\sin (\alpha)}}{1+\sin (\alpha)}=ctg\alpha\\\\\frac{\frac{\cos (\alpha)\sin (\alpha)+\cos (\alpha)}{\sin (\alpha)}}{1+\sin (\alpha)}=ctg\alpha\\\\\frac{\cos (\alpha)\sin (\alpha)+\cos (\alpha)}{\sin (\alpha)(1+\sin (\alpha))}=ctg\alpha\\\\\frac{\cos (\alpha)(\sin (\alpha)+1)}{\sin (\alpha)\left(1+\sin (\alpha)\right)}=ctg\alpha\\\\\frac{\cos (\alpha)}{\sin (\alpha)} = ctg\alpha\\\\ctg\alpha=ctg\alpha$

5.

$\frac{cos^2\alpha -sin^2\alpha }{ctg\alpha \:-tg\alpha \:}=sin\alpha \cdot \:cos\alpha \\\\\frac{\cos ^2(\alpha)-\sin ^2(\alpha)}{\frac{\cos (\alpha)}{\sin (\alpha)}-\frac{\sin (\alpha)}{\cos (\alpha)}} =sin\alpha \cdot \:cos\alpha \\\\\frac{\cos ^2(\alpha)-\sin ^2(\alpha)}{\frac{\cos ^2(\alpha)-\sin ^2(\alpha)}{\sin (\alpha)\cos (\alpha)}}=sin\alpha \cdot \:cos\alpha$

$\frac{ (\cos ^2 (\alpha )-\sin ^2 (\alpha ) )\sin (\alpha )\cos (\alpha )}{\cos ^2 (\alpha )-\sin ^2 (\alpha )} = sin\alpha \cdot \:cos\alpha \\\\sin\alpha \cdot \:cos\alpha = sin\alpha \cdot \:cos\alpha$