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 $\tan(\alpha -\beta)=1,\sec(\alpha +\beta)=\frac{2}{\sqrt{3}}$ 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 : Using $\operatorname{cosec}(\alpha+\beta)=\frac{\sec \alpha \times \sec \beta \times \operatorname{cosec} \alpha \times \operatorname{cosec} \beta}{\sec \alpha \times \operatorname{cosec} \beta+\operatorname{cosec} \alpha \times \sec \beta}$, find the value of $\operatorname{cosec} 75°$.

Option 1: $\frac{\sqrt{6}+\sqrt{2}}{4}$

Option 2: $\frac{\sqrt{6}-\sqrt{2}}{4}$

Option 3: $\sqrt{6}-\sqrt{2}$

Option 4: $\sqrt{6}+\sqrt{2}$

Question : In a triangle $XYZ$ right-angled at $Y$, if $XY=2\sqrt{6}$ and $XZ-YZ=2$, then $\sec X+\tan X$ is:

Option 1: $\frac{1}{\sqrt{6}}$

Option 2: $\sqrt{6}$

Option 3: $2\sqrt{6}$

Option 4: $\frac{\sqrt{6}}{2}$

Question : If $\tan\alpha=2$, then the value of $\frac{\operatorname{cosec}^{2}\alpha-\sec^{2}\alpha}{\operatorname{cosec}^{2}\alpha+\sec^{2}\alpha}$ is:

Option 1: $-\frac{15}{9}$

Option 2: $-\frac{3}{5}$

Option 3: $\frac{3}{5}$

Option 4: $\frac{17}{5}$