If \tan(\pi \cos \theta)=\cot(\pi \sin \theta), 0 \lt \theta \lt \frac{\pi}{2} then what is the value of 8\sin^{2}\left(\theta+\frac{\pi}{4}\right)?
- A. 16
- B. 2
- C. 1 ✓
- D. \frac{1}{2}
Correct Answer: C. 1
Explanation
Given \tan(\pi\cos\theta) = \tan(\frac{\pi}{2} - \pi\sin\theta). Thus, \pi\cos\theta = \frac{\pi}{2} - \pi\sin\theta \implies \cos\theta + \sin\theta = \frac{1}{2}. Squaring gives 1 + 2\sin\theta\cos\theta = \frac{1}{4} \implies \sin2\theta = -\frac{3}{4}. The required expression is 8\sin^2(\theta+\frac{\pi}{4}) = 8\left(\frac{1 - \cos(2\theta + \pi/2)}{2}\right) = 4(1 + \sin2\theta) = 4(1 - \frac{3}{4}) = 1.