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 : If $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{1}{7}, \theta$ lies in first quadrant, then the value of $\frac{\operatorname{cosec} \theta+\cot ^2 \theta}{\operatorname{cosec} \theta-\cot ^2 \theta}$ is:

Option 1: $\frac{19}{5}$

Option 2: $\frac{22}{3}$

Option 3: $\frac{37}{12}$

Option 4: $\frac{37}{19}$

Question : If $\tan ^2 \alpha=3+Q^2$, then $\sec \alpha+\tan ^3 \alpha \operatorname{cosec} \alpha=?$

Option 1: $\left(3+Q^2\right)^{\frac{3}{2}}$

Option 2: $\left(7+Q^2\right)^{\frac{3}{2}}$

Option 3: $\left(5-Q^2\right)^{\frac{3}{2}}$

Option 4: $\left(4+Q^2\right)^{\frac{3}{2}}$

Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:

Option 1: $2 \sec \theta$

Option 2: $2 \operatorname{cosec} \theta$

Option 3: $\operatorname{cosec} \theta$

Option 4: $\sec \theta$

Question : Which of the following is equal to $\frac{1}{\tan \theta}+\tan \theta$?

Option 1: $\sec \theta \operatorname{cosec} \theta$

Option 2: $1$

Option 3: $\frac{\operatorname{cosec} \theta}{\sec \theta}$

Option 4: $\tan ^2 \theta$