Question : If $\tan(\alpha -\beta)=1,\sec(\alpha +\beta)=\frac{2}{\sqrt{3}}$ and $\alpha ,\beta$ are positive, then the smallest value of $\alpha$ is:

Option 1: $142\frac{1}{2}°$

Option 2: $187\frac{1}{2}°$

Option 3: $7\frac{1}{2}°$

Option 4: $37\frac{1}{2}°$

Question : If $\alpha+\beta=90^{\circ}$ and $\alpha=2 \beta$, then the value of $3 \cos ^2 \alpha-2 \sin ^2 \beta$ is equal to:

Option 1: $\frac{3}{4}$

Option 2: $\frac{3}{2}$

Option 3: $\frac{1}{4}$

Option 4: $\frac{4}{3}$

Question : If $\tan (\alpha+\beta)=\sqrt{3}, \tan (\alpha-\beta)=1$ where $(\alpha+\beta)$ and $(\alpha-\beta)$ are acute angles, then what is $\tan$ $(6 \alpha) ?$

Option 1: $-1$

Option 2: $0$

Option 3: $1$

Option 4: $\sqrt{2}-1$

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: $\frac{1+\alpha^2}{a}$

Question : What is $\sin \alpha - \sin\beta$?

Option 1: $2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$

Option 2: $2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$

Option 3: $2 \cos \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta}{2}$

Option 4: $2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$