พิสูจน์ตรีโกณ
 

1.$\frac{tan?a-sin?a}{sin?a?sec?a}= tan\frac{a}{2}sin?a$

2.$\frac{1-cosx-tan^2\frac{x}{2}}{sin^2\frac{x}{2}}=\frac{2cosx}{1+cosx}$

แนวคิด
 

ข้อที่ 1

$\qquad \begin{array}{l}
LHS. = \dfrac{{\dfrac{{\sin a}}{{\cos a}} - \sin a}}{{\sin a\dfrac{1}{{\cos a}}}} = \sin a\left( {\dfrac{1}{{\cos a}} - 1} \right) \cdot \dfrac{{\cos a}}{{\sin a}}\\
= \sin a\left( {\dfrac{1}{{\sin a}} - \dfrac{{\cos a}}{{\sin a}}} \right)\quad \quad \quad from\quad \tan a = \dfrac{{2\tan \dfrac{a}{2}}}{{1 - {{\tan }^2}\dfrac{a}{2}}},\sin a = \dfrac{{2\tan \dfrac{a}{2}}}{{1 + {{\tan }^2}\dfrac{a}{2}}},\cos a = \dfrac{{1 - {{\tan }^2}\dfrac{a}{2}}}{{1 + {{\tan }^2}\dfrac{a}{2}}}\\
= \sin a\left( {\dfrac{{1 + {{\tan }^2}\dfrac{a}{2}}}{{2\tan \dfrac{a}{2}}} - \dfrac{{1 - {{\tan }^2}\dfrac{a}{2}}}{{2\tan \dfrac{a}{2}}}} \right)\\
= \sin a\left( {\dfrac{{1 + {{\tan }^2}\dfrac{a}{2} - 1 + {{\tan }^2}\dfrac{a}{2}}}{{2\tan \dfrac{a}{2}}}} \right)\\
= \sin a \cdot \tan \dfrac{a}{2}
\end{array}$

ข้อ 2

$\begin{array}{l}
LHS. = \dfrac{{1 - \cos x - {{\tan }^2}\dfrac{x}{2}}}{{{{\sin }^2}\dfrac{x}{2}}}\quad \quad from\quad {\tan ^2}\dfrac{x}{2} = \dfrac{{{{\sin }^2}\dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}} = \dfrac{{1 - \cos x}}{{1 + \cos x}}\\
= \dfrac{{1 - \cos x - \dfrac{{1 - \cos x}}{{1 + \cos x}}}}{{\dfrac{{1 - \cos x}}{2}}} = \dfrac{{\left[ {\left( {1 + \cos x} \right) - \cos x\left( {1 + \cos x} \right) - \left( {1 - \cos x} \right)} \right]}}{{1 + \cos x}} \cdot \dfrac{2}{{1 - \cos x}}\\
= \dfrac{{1 + \cos x - \cos x - {{\cos }^2}x - 1 + \cos x}}{{1 + \cos x}} \cdot \dfrac{2}{{1 - \cos x}}\\
= \dfrac{{\cos x - {{\cos }^2}x}}{{1 + \cos x}} \cdot \dfrac{2}{{1 - \cos x}}\\
= \dfrac{{\cos x\left( {1 - \cos x} \right)}}{{1 + \cos x}} \cdot \dfrac{2}{{1 - \cos x}}\\
= \dfrac{{2\cos x}}{{1 + \cos x}}
\end{array}$