Question

The value of \( \frac{\sqrt{3}}{\sin 15^\circ} - \frac{1}{\cos 15^\circ} \) is equal to:

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$
• E. $\frac{\sqrt{3}}{2}$

Hint:

Whenever you see $\sqrt{3}$ and $1$ paired with $\sin$ and $\cos$, look to convert them into $\sin 60^\circ$ and $\cos 60^\circ$ (or $30^\circ$). It is a standard trick to trigger the compound angle formulas.

Answer 1

The Correct Option is A

Concept: This expression can be simplified by finding a common denominator and using the compound angle identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and the double angle identity $\sin 2\theta = 2\sin\theta\cos\theta$.

Step 1: Find a common denominator.
\[ \frac{\sqrt{3} \cos 15^\circ - \sin 15^\circ}{\sin 15^\circ \cos 15^\circ} \]

Step 2: Simplify the numerator using the $R\sin(\theta \pm \alpha)$ method.
Multiply and divide the numerator by $2$: \[ 2 \left( \frac{\sqrt{3}}{2} \cos 15^\circ - \frac{1}{2} \sin 15^\circ \right) = 2 \left( \sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ \right) \] Using the identity $\sin(A - B)$: \[ 2 \sin(60^\circ - 15^\circ) = 2 \sin 45^\circ \]

Step 3: Simplify the denominator.
Using $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$: \[ \sin 15^\circ \cos 15^\circ = \frac{1}{2} \sin(2 \times 15^\circ) = \frac{1}{2} \sin 30^\circ \]

Step 4: Calculate the final value.
\[ \frac{2 \sin 45^\circ}{\frac{1}{2} \sin 30^\circ} = \frac{2 \times \frac{1}{\sqrt{2}}}{\frac{1}{2} \times \frac{1}{2}} = \frac{\sqrt{2}}{\frac{1}{4}} = 4\sqrt{2} \]