Question : If $\tan \frac{A}{2}=x$, then find $x$.

Option 1: $\frac{\sqrt{1+\cos A}}{\sqrt{1-\cos A}}$

Option 2: $\frac{\sqrt{1-\sin A}}{\sqrt{1+\cos A}}$

Option 3: $\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$

Option 4: $\frac{\sqrt{\cos A-1}}{\sqrt{1+\cos A}}$

Question : If $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1, \theta$ lies in the first quadrant, then the value of $\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}$ is:

Option 1: $\frac{2 \sqrt{3}}{27}$

Option 2: $\frac{5 \sqrt{3}}{27}$

Option 3: $\frac{2 \sqrt{3}}{9}$

Option 4: $\frac{7 \sqrt{3}}{54}$

Question : If $\cos \theta=\frac{\sqrt{3}}{2}$, then $\tan ^2 \theta \cos ^2 \theta=?$

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

Option 2: $\frac{1}{4}$

Option 3: $\frac{1}{2}$

Option 4: $\sqrt{3}$

Question : If $\tan \frac{\pi}{6}+\sec\frac{\pi}{6}=x$, then find $x$.

Option 1: $\sqrt{3}$

Option 2: $\frac{1}{\sqrt{3}}$

Option 3: $\frac{-1}{\sqrt{3}}$

Option 4: $\frac{2}{\sqrt{3}}$

Question : If $2 \frac{\cos ^2 x-\sec ^2 x}{\tan ^2 x}=a+b \cos 2 x$, then $a, b=$?

Option 1: $-\frac{3}{2}, -\frac{1}{2}$

Option 2: $\frac{3}{2}, \frac{1}{2}$

Option 3: $-3, -1$

Option 4: $3,1$