Question : If $\tan 40^{\circ}=\alpha$, then find $\frac{\tan 320^{\circ}-\tan 310^{\circ}}{1+\tan 320^{\circ} \cdot \tan 310^{\circ}}$.
Option 1: $\frac{1-\alpha^2}{\alpha}$
Option 2: $\frac{1+\alpha^2}{2 \alpha}$
Option 3: $\frac{1-\alpha^2}{2 \alpha}$
Option 4: $\frac{1+\alpha^2}{a}$
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Correct Answer: $\frac{1-\alpha^2}{2 \alpha}$
Solution :
$\tan 40^{\circ}=\alpha$
$\frac{\tan 320^{\circ}-\tan 310^{\circ}}{1+\tan 320^{\circ} \cdot \tan 310^{\circ}}$
$= \frac{\tan(360°-40°) - \tan(270°+40°)}{1+\tan(360°-40°)\cdot\tan(270°+40°)}$
$= \frac{-\tan 40° + \cot 40°}{1-\tan40° \cdot\cot 40°}$
$= \frac{-\alpha + \frac{1}{\alpha}}{1+\alpha×\frac{1}{\alpha}}$
$= \frac{-\alpha^2 + 1}{2\alpha}$
$=\frac{1-\alpha^2}{2\alpha}$
Hence, the correct answer is $\frac{1-\alpha^2}{2\alpha}$.
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