Question

If $\theta = 30^\circ$, then the value of \[ \frac{\sin^2\theta - \cos 2\theta}{\sin 2\theta + \cos^2\theta} \] is:

• A. $\frac{3 - 2\sqrt{3}}{9}$
• B. $\frac{3 - 2\sqrt{3}}{3}$
• C. $\sqrt{3} + \frac{2\sqrt{3}}{5}$
• D. $\frac{3}{5}$

Hint:

Memorize standard values for $30^\circ$, $45^\circ$, $60^\circ$ to solve quickly.

Answer 1

The Correct Option is B

Concept: Substitute standard trigonometric values.
Step 1: Values at $30^\circ$.
\[ \sin 30^\circ = \frac{1}{2},\quad \cos 30^\circ = \frac{\sqrt{3}}{2} \] \[ \sin 2\theta = \sin 60^\circ = \frac{\sqrt{3}}{2},\quad \cos 2\theta = \cos 60^\circ = \frac{1}{2} \]
Step 2: Substitute.
\[ \frac{\left(\frac{1}{2}\right)^2 - \frac{1}{2}}{\frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right)^2} \] \[ = \frac{\frac{1}{4} - \frac{1}{2}}{\frac{\sqrt{3}}{2} + \frac{3}{4}} \]
Step 3: Simplify numerator.
\[ \frac{-1}{4} \]
Step 4: Simplify denominator.
\[ \frac{2\sqrt{3} + 3}{4} \]
Step 5: Final value.
\[ \frac{-1/4}{(2\sqrt{3}+3)/4} = \frac{-1}{2\sqrt{3}+3} \] Rationalizing: \[ = \frac{-1(2\sqrt{3}-3)}{(2\sqrt{3}+3)(2\sqrt{3}-3)} = \frac{3 - 2\sqrt{3}}{3} \]
Hence, the answer is $\frac{3 - 2\sqrt{3}{3}$.