Q.30. If \tan\theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}, 0\leq\alpha\leq\frac{\pi}{2}, then the value of \cos2\theta is
A. \cos2\alpha
B. \sin\alpha
C. \cos\alpha
D. \sin2\alpha
Answer: D
Explanation :
\tan\theta = \frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \frac{\sin\theta}{\cos \theta} = \frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \cos\alpha\cos\theta+\sin\alpha\sin\theta=\sin\alpha\cos\theta-\cos\alpha\sin\theta \sin \alpha \sin \theta + \cos \alpha \sin \theta = \sin \alpha \cos \theta - \cos \alpha \sin \theta \cos(\alpha - \theta) = \sin(\alpha - \theta)\alpha - \theta = \frac{\pi}{4} … 0 \leq \alpha \leq \frac{\pi}{2}
\theta = \alpha - \frac{\pi}{4} 2\theta = 2\alpha - \frac{\pi}{2} \cos 2\theta = \cos(2\alpha - \frac{\pi}{2}) = \cos[-(\frac{\pi}{2} - 2\alpha)]= \cos(\frac{\pi}{2} - 2\alpha) … -\cos(-\theta) = \cos\theta
\cos 2\theta = \sin 2\alpha