Cho góc α thỏa mãn 0 ∘ < α < 90 ∘ . Chứng minh rằng:

Xét \(\Delta ABC\) vuông tại \(A\) có \(\widehat {B\,} = \alpha \).

Do \(\widehat {B\,}\) là góc nhọn nên \(0^\circ  < \widehat {B\,} < 90^\circ \) hay \[0^\circ  < \alpha  < 90^\circ .\]

Ta có: \(\sin \alpha  = \frac{{AC}}{{BC}}\) và \(\cos \alpha  = \frac{{AB}}{{BC}}.\)

\(B{C^2} = A{B^2} + A{C^2}\) (định lí Pythagore).

Khi đó:

\[{\sin ^2}\alpha  + {\cos ^2}\alpha  = {\left( {\frac{{AC}}{{BC}}} \right)^2} + {\left( {\frac{{AB}}{{BC}}} \right)^2} = \frac{{A{C^2}}}{{B{C^2}}} + \frac{{A{B^2}}}{{B{C^2}}} = \frac{{A{C^2} + A{B^2}}}{{B{C^2}}} = \frac{{B{C^2}}}{{B{C^2}}} = 1.\]

Với \[0^\circ  < \alpha  < 90^\circ \] thì \[1 - \cos \alpha  \ne 0\] và \[\sin \alpha  - \cos \alpha  + 1 \ne 0\].

Ta có: \[\frac{{\sin \alpha  + \cos \alpha  - 1}}{{1 - \cos \alpha }} - \frac{{2\cos \alpha }}{{\sin \alpha  - \cos \alpha  + 1}}\]

       \[ = \frac{{\left( {\sin \alpha  + \cos \alpha  - 1} \right)\left( {\sin \alpha  - \cos \alpha  + 1} \right) - 2\cos \alpha \left( {1 - \cos \alpha } \right)}}{{\left( {1 - \cos \alpha } \right)\left( {\sin \alpha  - \cos \alpha  + 1} \right)}}\]

       \[ = \frac{{\left[ {\sin \alpha  + \left( {\cos \alpha  - 1} \right)} \right]\left[ {\sin \alpha  - \left( {\cos \alpha  - 1} \right)} \right] - 2\cos \alpha \left( {1 - \cos \alpha } \right)}}{{\left( {1 - \cos \alpha } \right)\left( {\sin \alpha  - \cos \alpha  + 1} \right)}}\]

       \[ = \frac{{{{\sin }^2}\alpha  - {{\left( {\cos \alpha  - 1} \right)}^2} - 2\cos \alpha  + 2{{\cos }^2}\alpha }}{{\left( {1 - \cos \alpha } \right)\left( {\sin \alpha  - \cos \alpha  + 1} \right)}}\]

       \[ = \frac{{{{\sin }^2}\alpha  - \left( {{{\cos }^2}\alpha  - 2\cos \alpha  + 1} \right) - 2\cos \alpha  + 2{{\cos }^2}\alpha }}{{\left( {1 - \cos \alpha } \right)\left( {\sin \alpha  - \cos \alpha  + 1} \right)}}\]

       \[ = \frac{{{{\sin }^2}\alpha  - {{\cos }^2}\alpha  + 2\cos \alpha  - 1 - 2\cos \alpha  + 2{{\cos }^2}\alpha }}{{\left( {1 - \cos \alpha } \right)\left( {\sin \alpha  - \cos \alpha  + 1} \right)}}\]

       \[ = \frac{{{{\sin }^2}x + {{\cos }^2}x - 1}}{{\left( {1 - \cos x} \right)\left( {\sin x - \cos x + 1} \right)}}.\]

       \[ = \frac{{1 - 1}}{{\left( {1 - \cos x} \right)\left( {\sin x - \cos x + 1} \right)}} = 0\] (vì \[1 - \cos x \ne 0\] và \[\sin x - \cos x + 1 \ne 0)\]

     Vậy \[\frac{{\sin x + \cos x - 1}}{{1 - \cos x}} = \frac{{2\cos x}}{{\sin x - \cos x + 1}}.\]