ข้อสอบ PAT 1 - เมษายน 2557

$\mathrm{จากโจทย์}\hspace{0.17em} \mathrm{y}=\arctan \left(\frac{3\mathrm{cos\theta }}{1-3\mathrm{sin\theta }}\right)-\mathrm{arccot}\left(\frac{\mathrm{cos\theta }}{3-\mathrm{sin\theta }}\right)\to \boxed{1}$

$$\begin{gathered} \mathrm{กำหนดให้} \mathrm{A}=\arctan \left(\frac{3\mathrm{cos\theta }}{1-3\mathrm{sin\theta }}\right) \\ \tan \mathrm{A}=\frac{3\mathrm{cos\theta }}{1-3\mathrm{sin\theta }} \\ -\mathrm{นำมาวาดรูปสามเหลี่ยมมุม} \mathrm{A} \end{gathered}$$​

$$\begin{gathered} \mathrm{กำหนดให้} \mathrm{B}=\mathrm{arccot}\left(\frac{\mathrm{cos\theta }}{3-\mathrm{sin\theta }}\right) \\ \cot \mathrm{B}=\frac{\mathrm{cos\theta }}{3-\mathrm{sin\theta }} \\ \tan \mathrm{B}=\frac{3-\sin \left(\theta \right)}{\cos \left(\theta \right)} \\ -\mathrm{นำมาวาดรูปสามเหลี่ยมมุม} \mathrm{B} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะ}\mathrm{ได้} \\ \mathrm{y}=\mathrm{A}-\mathrm{B} \\ \tan \mathrm{y}=\tan \left(\mathrm{A}-\mathrm{B}\right) \\ =\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \tan \mathrm{B}} ; \mathrm{แทนค่า} \mathrm{tanA} , \tan \mathrm{B} \\ =\frac{\left({\displaystyle \frac{3\mathrm{cos\theta }}{1-3\mathrm{sin\theta }}}\right)-\left({\displaystyle \frac{3-\mathrm{sin\theta }}{\mathrm{cos\theta }}}\right)}{1+\left({\displaystyle \frac{3\mathrm{cos\theta }}{1-3\mathrm{sin\theta }}}\right)\left({\displaystyle \frac{3-\mathrm{sin\theta }}{\mathrm{cos\theta }}}\right)} \\ =\frac{\left({\displaystyle \frac{3{\cos }^2\mathrm{\theta }-\left(3-\mathrm{sin\theta }\right)\left(1-3\mathrm{sin\theta }\right)}{\left(1-3\mathrm{sin\theta }\right)\left(\mathrm{cos\theta }\right)}}\right)}{\left({\displaystyle \frac{\left(1-3\mathrm{sin\theta }\right)\left(\mathrm{cos\theta }\right)+9\mathrm{cos\theta }-3\mathrm{sin\theta cos\theta }}{\left(1-3\mathrm{sin\theta }\right)\left(\mathrm{cos\theta }\right)}}\right)} \end{gathered}$$
$$\begin{gathered} =\frac{3{\cos }^2\mathrm{\theta }-3-10\mathrm{sin\theta }+3{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }-3\mathrm{sin\theta cos\theta }+9\mathrm{cos\theta }-3\mathrm{sin\theta cos\theta }} \\ =\frac{3-3{\sin }^2\mathrm{\theta }-3+10\mathrm{sin\theta }-3{\sin }^2\mathrm{\theta }}{10\mathrm{cos\theta }-6\mathrm{sin\theta cos\theta }} \\ =\frac{10\mathrm{sin\theta }-6{\sin }^2\mathrm{\theta }}{10\mathrm{cos\theta }-6\mathrm{sin\theta cos\theta }} \\ =\frac{\mathrm{sin\theta }\left(10-6\mathrm{sin\theta }\right)}{\mathrm{cos\theta }\left(10-6\mathrm{sin\theta }\right)} \\ =\tan \mathrm{\theta } \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \hspace{0.17em}\hspace{0.17em}\tan \mathrm{y}=\tan \mathrm{\theta } \\ \hspace{0.17em}\hspace{0.17em} \mathrm{y}=\arctan \left(\mathrm{tan\theta }\right) \end{gathered}$$​