ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \cot \left(\frac{\mathrm{\pi }}{2}+\mathrm{arcsec} \mathrm{x}\right)\to \boxed{1} \\ \mathrm{กำหนดให้} \mathrm{arcsec} \mathrm{x}=\mathrm{\theta } \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \\ \cot \left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right)=\cot \left(\frac{\mathrm{\pi }}{2}-\left(-\mathrm{\theta }\right)\right)=\tan \left(-\mathrm{\theta }\right)=-\mathrm{tan\theta } \\ \cot \left(\frac{\mathrm{\pi }}{2}+\mathrm{arcsec} \mathrm{x}\right)=-\tan \left(\mathrm{arcsec} \mathrm{x}\right) \\ =-\tan \left(\arcsin \frac{1}{\sqrt{17}}-2\arccos \frac{2}{\sqrt{5 }}\right)\to \boxed{2} \end{gathered}$$
$$\begin{gathered} \mathrm{กำหนดให้} \arcsin \frac{1}{\sqrt{17}}=\mathrm{A} \arccos \frac{2}{\sqrt{15}}=\mathrm{B} \\ \sin \mathrm{A}=\frac{1}{\sqrt{17}} \cos \mathrm{B}=\frac{2}{\sqrt{15}} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \tan \mathrm{A}=\frac{1}{4} \tan \mathrm{B}=\frac{1}{2} \\ \tan 2\mathrm{B}=\frac{2\tan \mathrm{B}}{1-{\tan }^2 \mathrm{B}} \\ =\frac{2\left({\displaystyle \frac{1}{2}}\right)}{1-{\left(\frac{1}{2}\right)}^2}=\frac{4}{3} \end{gathered}$$
$$\begin{gathered} \mathrm{จาก} \boxed{2} \cot \left(\frac{\mathrm{\pi }}{2}+\mathrm{arcsec} \mathrm{x}\right)=-\tan \left(\mathrm{A}-2\mathrm{B}\right) \\ =-\left(\frac{\tan \mathrm{A}-\tan 2\mathrm{B}}{1+\tan \mathrm{A} \tan 2\mathrm{B}}\right) \\ =-\left(\frac{{\displaystyle \frac{1}{4}-\frac{4}{3}}}{1+\frac{1}{4}\left(\frac{4}{3}\right)}\right) \\ =-\left(\frac{{\displaystyle -\frac{13}{12}}}{{\displaystyle \frac{16}{12}}}\right) \\ =\frac{13}{16} \end{gathered}$$