,

กระจายกำลังสองสมบูรณ์

\begin{eqnarray*}
2 \cot \frac{\theta}{2} &=& (1 + \cot \theta)^2\\
2 \cot \frac{\theta}{2} &=& 1 + 2 \cot \theta + \cot^2 \theta\\
2 \cot \frac{\theta}{2} - 2 \cot \theta &=& 1 + \cot^2 \theta
\end{eqnarray*}

ใช้เอกลักษณ์ $\displaystyle \operatorname{cosec}^2 \theta - \cot^2 \theta = 1$ จะได้ว่า $\displaystyle \operatorname{cosec}^2 \theta = 1 + \cot^2 \theta$

\begin{eqnarray*}
2 \cot \frac{\theta}{2} - 2 \cot \theta &=& \operatorname{cosec}^2 \theta\\
2 \left( \cot \frac{\theta}{2} - \cot \theta \right) &=& \operatorname{cosec}^2 \theta
\end{eqnarray*}

กระจายฟังก์ชัน $\displaystyle \cot \theta = \frac{\cos \theta}{\sin \theta}$ จะได้

\begin{eqnarray*}
2 \left( \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} - \frac{\cos \theta}{\sin \theta} \right) &=& \operatorname{cosec}^2 \theta\\
2 \left( \frac{\cos\frac{\theta}{2} \sin \theta - \sin \frac{\theta}{2} \cos \theta}{\sin \frac{\theta}{2} \sin \theta} \right) &=& \operatorname{cosec}^2 \theta      \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)
\end{eqnarray*}

พิจารณา $\displaystyle \cos\frac{\theta}{2} \sin \theta - \sin \frac{\theta}{2} \cos \theta$

ใช้สูตร $\displaystyle \sin$ ของผลต่างของมุม คือ $\displaystyle \sin(A-B) = \sin A \cos B - \cos A \sin B$ จะได้ว่า

\begin{eqnarray*}
\cos\frac{\theta}{2} \sin \theta - \sin \frac{\theta}{2} \cos \theta &=& \sin \theta \cos\frac{\theta}{2} - \cos \theta \sin \frac{\theta}{2}\\
&=& \sin \left( \theta - \frac{\theta}{2} \right)\\
&=& \sin \frac{\theta}{2}
\end{eqnarray*}

แทนกลับในสมการ $(1)$ จะได้

\begin{eqnarray*}
2 \left( \frac{\sin \frac{\theta}{2}}{\sin \frac{\theta}{2} \sin \theta} \right) &=& \operatorname{cosec}^2 \theta\\
2 \left( \frac{\cancel{\sin \frac{\theta}{2}}}{\cancel{\sin \frac{\theta}{2}} \sin \theta} \right) &=& \operatorname{cosec}^2 \theta\\
2 \left( \frac{1}{\sin \theta} \right) &=& \operatorname{cosec}^2 \theta
\end{eqnarray*}

เปลี่ยน $\displaystyle \operatorname{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ จะได้

\begin{eqnarray*}
2 \left( \frac{1}{\sin \theta} \right) &=& \frac{1}{\sin^2 \theta}\\
\sin^2 \theta \left( \frac{1}{\sin \theta} \right) &=& \frac{1}{2}\\
\sin \theta &=& \frac{1}{2}
\end{eqnarray*}

นั่นคือ $\displaystyle \theta = 30^{\circ}$

,

แทนค่า $\displaystyle \theta = 30^{\circ}$

\begin{eqnarray*}
\frac{(1 + \sin \theta) \sec^2 \theta}{\cos 2 \theta} &=& \frac{(1 + \sin 30^{\circ}) \sec^2 30^{\circ}}{\cos 2 (30^{\circ})}\\
&=& \frac{\left( 1 + \frac{1}{2} \right) \left( \frac{2}{\sqrt{3}} \right)^2}{\frac{1}{2}}\\
&=& \left( \frac{\cancel{3}}{\bcancel{2}} \right)\left( \frac{4}{\cancel{3}} \right) (\bcancel{2})\\
&=& 4
\end{eqnarray*}