ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{จากสูตร} \\ \tan \left(A+B\right) = \frac{\tan A+\tan B}{1-\tan A\tan B} \\ \tan 2A = \frac{2\tan A}{1-{\tan }^{2}A} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \tan \left(2arc\tan \frac{1}{2}\right) \\ = \frac{2\tan \left(arc\tan {\displaystyle \frac{1}{2}}\right)}{1-{\tan }^2\left(arc\tan {\displaystyle \frac{1}{2}}\right)} \\ = \frac{2\left({\displaystyle \frac{1}{2}}\right)}{1-{\left({\displaystyle \frac{1}{2}}\right)}^2} = \frac{1}{{\displaystyle \frac{3}{4}}} = \frac{4}{3} \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \tan \left(\frac{3\mathrm{\pi }}{4}+2arc\tan \frac{1}{2}\right) \\ = \frac{\tan {\displaystyle \frac{3\mathrm{\pi }}{4}}+\tan \left(2arc{\displaystyle \frac{1}{2}}\right)}{1-\tan {\displaystyle \frac{3\mathrm{\pi }}{4}}\tan \left(2arc\tan {\displaystyle \frac{1}{2}}\right)} \end{gathered}$$

$$\begin{gathered} = \frac{-1+{\displaystyle \frac{4}{3}}}{1-\left(-1\right)\left({\displaystyle \frac{4}{3}}\right)} \\ = \frac{{\displaystyle \frac{1}{3}}}{1+{\displaystyle \frac{4}{3}}} = \frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{7}{3}}} \\ = \frac{1}{7} \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 3 \end{gathered}$$