ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากสูตร} {\sec }^2\mathrm{x}-{\tan }^2\mathrm{x}=1\to {\sec }^2\mathrm{x}=1+{\tan }^2\mathrm{x} \\ \mathrm{จากโจทย์} \left(1+ {\tan }^2 \mathrm{x}\right)\cot \mathrm{x} = \frac{\mathrm{a}}{\mathrm{b}} \\ {\sec }^2\mathrm{x} \cot \mathrm{x}=\frac{\mathrm{a}}{\mathrm{b}} \\ \frac{1}{{\cos }^2\mathrm{x}}·\frac{\cos \mathrm{x}}{\sin \mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}} \end{gathered}$$

$$\begin{gathered} \frac{1}{\sin \mathrm{x}·\cos \mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}}\to \boxed{1} \\ \mathrm{จากโจทย์} \sin \mathrm{x}+\cos \mathrm{x}=\frac{4}{3} \\ {(\sin \mathrm{x}+\cos \mathrm{x})}^2={\left(\frac{4}{3}\right)}^2 \end{gathered}$$

$$\begin{gathered} {\sin }^2\mathrm{x}+2\sin \mathrm{x} \cos \mathrm{x}+{\cos }^2\mathrm{x}=\frac{16}{9} \\ \left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)+2\sin \mathrm{x} \cos \mathrm{x}=\frac{16}{9} \\ 1+2\sin \mathrm{x} \cos \mathrm{x}=\frac{16}{9} \\ 2\sin \mathrm{x} \cos \mathrm{x}=\frac{7}{9} \\ \sin \mathrm{x} \cos \mathrm{x}=\frac{7}{18} ; \mathrm{แทนใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะได้} \frac{1}{{\displaystyle \frac{7}{18}}}=\frac{\mathrm{a}}{\mathrm{b}} \\ \frac{\mathrm{a}}{\mathrm{b}}=\frac{18}{7} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{ห}.\mathrm{ร}.\mathrm{ม}. \mathrm{ของ} \mathrm{a} \mathrm{และ} \mathrm{b} \mathrm{เท่ากับ} 1 \\ \mathrm{แสดงว่า} \mathrm{a}=18 , \mathrm{b}=7 \\ \mathrm{ดังนั้น} {\mathrm{a}}^2+{\mathrm{b}}^2={18}^2+7^2 \\ =324+49 \\ =373 \end{gathered}$$