Question : Using the identity, $\tan 2 \alpha = \frac{2 \tan \alpha}{1 – \tan ^2 \alpha},$ find the value of $\tan15°,$ correct to three decimal places. [Use $\sqrt{3} = 1.732$]

Option 1: 0.268

Option 2: 0.27

Option 3: 0.267

Option 4: 0.269

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 $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, find the value of $\tan 15°$.

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

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

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

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

Question : If $x=\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}$, then the value of $\frac{\sqrt{3}-x}{\sqrt{3}+x}$ (corrected to two decimal places) is:

Option 1: 0.25

Option 2: 0.17

Option 3: 0.19

Option 4: 0.27

Question : The value of tan 25° tan 35° tan 45° tan 55° tan 65° is:

Option 1: $2$

Option 2: $1$

Option 3: $\sqrt{3}$

Option 4: $0$

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