Question

Evaluate: $$ \sin^2\left(\frac{2\pi}{3}\right) + \cos^2\left(\frac{5\pi}{6}\right) - \tan^2\left(\frac{3\pi}{4}\right) $$

• A. 0
• B. \( \frac{1}{2} \)
• C. 1
• D. \( \frac{1}{3} \)

Hint:

Know the standard values for trigonometric functions at multiples of \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} \).

Answer 1

The Correct Option is B

We evaluate each term one by one.
1. \( \sin\left(\frac{2\pi}{3}\right) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \Rightarrow \sin^2 = \frac{3}{4} \)
2. \( \cos\left(\frac{5\pi}{6}\right) = \cos(150^\circ) = -\frac{\sqrt{3}}{2} \Rightarrow \cos^2 = \frac{3}{4} \)
3. \( \tan\left(\frac{3\pi}{4}\right) = \tan(135^\circ) = -1 \Rightarrow \tan^2 = 1 \) So, \[ \sin^2\left(\frac{2\pi}{3}\right) + \cos^2\left(\frac{5\pi}{6}\right) - \tan^2\left(\frac{3\pi}{4}\right) = \frac{3}{4} + \frac{3}{4} - 1 = \frac{6}{4} - 1 = \frac{2}{4} = \boxed{ \frac{1}{2} } \]