ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \mathbf{สูตร} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1\to {\sec }^2\mathrm{\theta }=1+{\tan }^2\mathrm{\theta }\to \boxed{\mathrm{a}} \\ \hspace{0.17em} \hspace{0.17em}\tan \left(\mathrm{A}+\mathrm{B}\right)=\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A} \tan \mathrm{B}}\to \boxed{\mathrm{b}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} {\sec }^2 \left(2 \arctan \frac{1}{3}+\arctan \frac{1}{7}\right) \\ \mathrm{จาก} \boxed{\mathrm{a}} \mathrm{จะได้} =1+{\tan }^2\mathrm{\theta }\left(2 \arctan \frac{1}{3}+\arctan \frac{1}{7}\right)\to \boxed{1} \\ \mathrm{กำหนดให้} \mathrm{A}=\arctan \frac{1}{3} \mathrm{B}=\arctan \frac{1}{7} \\ \tan \mathrm{A}=\frac{1}{3} \tan \mathrm{B}=\frac{1}{7} \\ -\mathrm{วาดรูปสามเหลี่ยม} \end{gathered}$$

$$\begin{gathered} \tan 2\mathrm{A}=\frac{2 \tan \mathrm{A}}{1-{\tan }^2\mathrm{A}} \\ \tan 2\mathrm{A}=\frac{2\left({\displaystyle \frac{1}{3}}\right)}{1-{\left(\frac{1}{3}\right)}^2}=\frac{{\displaystyle \frac{2}{3}}}{1-{\displaystyle \frac{1}{9}}}=\frac{2}{3}\left(\frac{9}{8}\right) \\ \tan 2\mathrm{A}=\frac{3}{4} \\ \mathrm{จาก} \boxed{1} {\sec }^2 \left(2 \arctan \frac{1}{3}+\arctan \frac{1}{7}\right) \\ =1+{\tan }^2\left(2 \arctan \frac{1}{3}+\arctan \frac{1}{7}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{แทนค่า} \mathrm{A} , \mathrm{B} \\ \mathrm{จะได้} =1+{\tan }^2\left(2\mathrm{A}+\mathrm{B}\right) \\ =1+{\left[\tan \left(2\mathrm{A}+\mathrm{B}\right)\right]}^2 \\ \mathrm{จาก} \boxed{\mathrm{b}} \mathrm{จะได้} =1+ \frac{\tan 2\mathrm{A}+\tan \mathrm{B}}{1-\tan 2\mathrm{A} \tan \mathrm{B}} \\ =1+\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{7}}}{1-\left({\displaystyle \frac{3}{4}}\right)\left({\displaystyle \frac{1}{7}}\right)}\right) \end{gathered}$$

$$\begin{gathered} =1+{\left(\frac{{\displaystyle \frac{25}{28}}}{\frac{25}{28}}\right)}^2 \\ =1+1^2 \\ =2 \\ \mathrm{ดังนั้น} {\sec }^2 \left(2 \arctan \frac{1}{3}+\arctan \frac{1}{7}\right)=2 \end{gathered}$$