ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{\theta }=\arctan \left(\frac{\sqrt{\mathrm{x}}+1}{1-\sqrt{\mathrm{x}}}\right)-\arctan \left(\sqrt{\mathrm{x}}\right)\to \boxed{1} \\ \mathrm{กำหนดให้} \mathrm{A}=\arctan \left(\frac{\sqrt{\mathrm{x}}+1}{1-\sqrt{\mathrm{x}}}\right) , \mathrm{B}=\arctan \left(\sqrt{\mathrm{x}}\right) \\ \tan \mathrm{A}=\frac{\sqrt{\mathrm{x}}+1}{1-\sqrt{\mathrm{x}}} \tan \mathrm{B}=\sqrt{\mathrm{x}} \\ \mathrm{แทน} \mathrm{A} , \mathrm{B} \mathrm{ใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \mathrm{\theta }=\mathrm{A}-\mathrm{B} ; \mathrm{take} \tan \mathrm{ทั้งสองข้าง} \\ \mathrm{tan\theta }=\tan (\mathrm{A}-\mathrm{B}) \\ =\frac{\tan \mathrm{A}-\tan \hspace{0.17em}\mathrm{B}}{1+\tan \mathrm{A} \tan \mathrm{B}} ; \mathrm{แทนค่า} \tan \mathrm{A} , \tan \mathrm{B} \end{gathered}$$

$$\begin{gathered} =\frac{\left(\frac{\sqrt{\mathrm{x}}+1}{1-\sqrt{\mathrm{x}}}\right) -\left(\sqrt{\mathrm{x}}\right)}{1+\left(\frac{\sqrt{\mathrm{x}}+1}{1-\sqrt{\mathrm{x}}}\right)\left(\sqrt{\mathrm{x}}\right)} \\ =\frac{\left[\frac{\sqrt{\mathrm{x}}+1-\left(1-\sqrt{\mathrm{x}}\right)\left(\sqrt{\mathrm{x}}\right)}{\cancel{\left(1-\sqrt{\mathrm{x}}\right)}} \right]}{\left[\frac{\left(1\right)\left(1-\sqrt{\mathrm{x}}\right)+\left(\sqrt{\mathrm{x}}+1\right)\left(\sqrt{\mathrm{x}}\right)}{\cancel{\left(1-\sqrt{\mathrm{x}}\right)}} \right]} \end{gathered}$$

$$\begin{gathered} =\frac{\bcancel{\sqrt{\mathrm{x}}}+1\bcancel{-\sqrt{\mathrm{x}}}+\mathrm{x}}{1\cancel{-\sqrt{\mathrm{x}}}+\mathrm{x}\cancel{+\sqrt{\mathrm{x}}}} \\ =\frac{\cancel{\mathrm{x}+1}}{\cancel{\mathrm{x}+1}} \\ \tan \mathrm{\theta }=1 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} 0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2} \\ \mathrm{แสดงว่า} \mathrm{\theta }=45° \\ \mathrm{ดังนั้น} \tan \mathrm{\theta }+\cot \mathrm{\theta }=1+1 \\ =2 \end{gathered}$$