ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2555 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากสูตร}\hspace{0.17em} {\sec }^2\theta =1+{\tan }^2\theta \\ \mathrm{กำหนดให้} \mathrm{\theta }={\tan }^{-1}\sqrt{2}\to \mathrm{tan\theta }=\sqrt{2} \\ \mathrm{จะได้} {\sec }^2(2 {\tan }^{-1}\sqrt{2})=1+{\tan }^2(2 {\tan }^{-1}\sqrt{2})\to \boxed{1} \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} {\sec }^2(2 {\tan }^{-1}\sqrt{2})=1+{\tan }^2\left(2\mathrm{\theta }\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta } \mathrm{จะได้}; \\ {\sec }^2(2 {\tan }^{-1}\sqrt{2})=1+{\left(\frac{2\mathrm{tan\theta }}{1-{\tan }^2\mathrm{\theta }}\right)}^2 \\ =1+{\left(\frac{2\sqrt{2}}{1-{\left(\sqrt{2}\right)}^2}\right)}^2 \\ =1+{\left(-2\sqrt{2}\right)}^2 \\ =1+8 \\ =9 \end{gathered}$$