Question

Ayudemen porfavor es urgente!! ​

Answer 1

Respuesta:

Explicación paso a paso:

2)

$\frac{(1+cosx)(1-cosx)}{tan^{2}x}= cos^{2}x$

$\frac{1-cos^{2}x}{\frac{sen^{2}x}{cos^{2}x}}=cos^{2}x$

$\frac{\frac{sen^{2}x}{1}}{\frac{sen^{2}x}{cos^{2}x}}=cos^{2}x$

$\frac{sen^{2}xcos^{2}x}{sen^{2}x}=cos^{2}x$

cos²x = cos²x

3)

$\frac{tanx+cotx}{secx}=cscx$

$\frac{\frac{senx}{cosx}+\frac{cosx}{senx}}{\frac{1}{cosx}}=cscx$

$\frac{\frac{sen^{2}x+cos^{2}x}{senxcosx}}{\frac{1}{cosx}}=cscx$

$\frac{\frac{1}{senxcosx}}{\frac{1}{cosx}}=cscx$

$\frac{cosx}{senxcosx}=cscx$

1/senx = cscx

cscx = cscx

4)

$\frac{cscx-secx}{cotx}=\frac{cosx-senx}{cos^{2}x}$

$\frac{\frac{1}{senx}-\frac{1}{cosx}}{\frac{cosx}{senx}}=\frac{cosx-senx}{cos^{2}x}$

$\frac{\frac{cosx-senx}{senxcosx}}{\frac{cosx}{senx}}=\frac{cosx-senx}{cos^{2}x}$

$\frac{senx(cosx-senx)}{senxcos^{2}x}=\frac{cosx-senx}{cos^{2}x}$

$\frac{cosx-senx}{cos^{2}x}=\frac{cosx-senx}{cos^{2}x}$

5)

$\frac{(senx-cosx)^{2}}{1-2senxcosx}=1$

$\frac{sen^{2}x-2senxcosx+cos^{2}x}{1-2senxcosx}=1$

$\frac{1-2senxcosx}{1-2senxcosx}=1$

1 = 1

6)

cscx(tanx - senx) = secx - 1

$\frac{1}{senx}(\frac{senx}{cosx}-senx)=secx-1$

$\frac{1}{cosx}-1=secx-1$

secx-1=secx-1