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: $2+\sqrt{3}$
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Correct Answer: $2-\sqrt{3}$
Solution : Given: $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$ ⇒ $\tan (45°-30°)=\frac{\tan 45°-\tan 30°}{1+\tan 45° \tan 30°}$ ⇒ $\tan15° = \frac{(1-\frac{1}{\sqrt3})}{(1+\frac{1}{\sqrt3})}$ ⇒ $\tan15° = \frac{(\sqrt3-1)}{(\sqrt3+1)}$ After rationalizing, ⇒ $\tan15°=2-\sqrt{3}$ Hence, the correct answer is $2-\sqrt{3}$.
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Question : What is the value of $\frac{\tan 45^{\circ}-\tan 15^{\circ}}{1+\tan 45^{\circ} \tan 15^{\circ}}$?
Option 1: $\sqrt{3}$
Option 2: $\frac{1}{\sqrt{2}}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: ${\sqrt{2}}$
Question : What is the value of $\cos ^2 15°?$
Option 1: $(2+\sqrt{3})$
Option 2: $\frac{(2+\sqrt{3})}{4}$
Option 3: $\frac{(2+\sqrt{3})}{2}$
Option 4: $\frac{(1+\sqrt{3})}{2}$
Question : What is the value of $\cos45° \sin15°$?
Option 1: $\frac{(\sqrt{3}-1)}{2}$
Option 2: $\frac{(\sqrt{3}-1)}{4}$
Option 3: $(\sqrt{3}+1)$
Option 4: $2 \sqrt{3}-1$
Question : Find the value of $\frac{\cos^2 15^{\circ}-\sin^2 15^{\circ}}{\cos^2 145^{\circ}+\sin^2 145^{\circ}}$.
Option 1: $\frac{1}{\sqrt{3}}$
Option 2: $\frac{1}{1-\sqrt{3}}$
Option 3: $\frac{\sqrt{3}}{2}$
Option 4: $\frac{2}{\sqrt{3}}$
Question : If $2 \cot \theta = 3$, find the value of $\frac{\sqrt{13} \sin \theta – 3 \tan \theta}{3 \tan \theta + \sqrt{13} \cos \theta}$
Option 1: $\frac{1}{\sqrt{13}}$
Option 2: $\frac{2}{\sqrt{13}}$
Option 3: 0
Option 4: $\frac{2}{3}$
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