[math]\frac{1-sinx}{1+sinx} = (secx - tanx)^2[/math]

 

The solution by solving the right side:

[math]=(\frac{1}{cosx} - \frac{sinx}{cosx})^2[/math]

 

[math]=\frac{1}{cos^{2}x} - 2\frac{sinx}{cosx} + \frac{sin^2{x}}{cos^{2}x}[/math] This is the question: How did I get to join all the [math]cos^{2}x[/math]? Because the middle one is only [math]cosx[/math]? Can anyone help me here?

 

[math]=\frac{1-2sinx+sin^{2}x}{cos^{2}x}[/math]

 

[math]=\frac{1-2sinx+sin^{2}x}{1-sin^{2}x}[/math]

 

[math]=\frac{(sinx-1)(sinx-1)}{(1-sinx)(1+sinx)}[/math]

 

[math]= \frac{1-sinx}{1+sinx}[/math]

[math]=(\frac{1}{cosx} - \frac{sinx}{cosx})^2=\frac{1}{cos^{2}x} - 2\frac{sinx}{cosx} + \frac{sin^2{x}}{cos^{2}x}[/math]

 

This is the question: How did I get to join all the [math]cos^{2}x[/math]? Because the middle one is only [math]cosx[/math]? Can anyone help me here?

Yes. You made an error in above step. The denominator in the second term actually is cos²(x).

Why?

Calculate out the (...)² as (...)*(...) by hand and see why .