EvoN1020v Posted January 7, 2006 Share Posted January 7, 2006 [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] Link to comment Share on other sites More sharing options...
timo Posted January 7, 2006 Share Posted January 7, 2006 [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). Link to comment Share on other sites More sharing options...
EvoN1020v Posted January 7, 2006 Author Share Posted January 7, 2006 Why? Link to comment Share on other sites More sharing options...
timo Posted January 7, 2006 Share Posted January 7, 2006 Calculate out the (...)² as (...)*(...) by hand and see why . Link to comment Share on other sites More sharing options...
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