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 $\cos x=-\frac{1}{2}, x$ lies in third quadrant, then $\tan x=?$

Option 1: $\sqrt{3}$

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

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

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

Question : If $\cos27^{\circ}$ = $x$, then the value of $\tan 63°$ is:

Option 1: $\frac{x}{\sqrt{1–x^2}}$

Option 2: $\frac{x}{\sqrt{1+x^2}}$

Option 3: $\frac{\sqrt{1–x^2}}{x}$

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

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 $(x+\frac{1}{x})=6$ and $x>1$, find the value of $(x^2–\frac{1}{x^2})$.

Option 1: $18 \sqrt{2}$

Option 2: $30 \sqrt{2}$

Option 3: $24 \sqrt{2}$

Option 4: $12 \sqrt{10}$