Question:
If \(\tan(45^\circ + \theta) = 1 + \sin 2\theta\), where \(-\dfrac{\pi}{4} < \theta < \dfrac{\pi}{4}\), then what is the value of \(\cos 2\theta\)?
यदि \(\tan(45^\circ + \theta) = 1 + \sin 2\theta\) है, जहाँ \(-\dfrac{\pi}{4} < \theta < \dfrac{\pi}{4}\) है, तो \(\cos 2\theta\) का मान क्या है?
Options:
- 1
- 0
- \(\dfrac{1}{2}\)
- 2
- 1
- 0
- \(\dfrac{1}{2}\)
- 2
Correct Answer: a)
Explanation:
\(\tan(45^\circ + \theta) = \dfrac{1+\tan\theta}{1-\tan\theta} = \dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\).
Squaring both sides: \(\dfrac{1+\sin 2\theta}{1-\sin 2\theta} = (1+\sin 2\theta)^2\).
\(1-\sin 2\theta = \dfrac{1}{1+\sin 2\theta} \implies 1-\sin^2 2\theta = 1 \implies \cos^2 2\theta = 1 \implies \cos 2\theta = 1\).
\(\tan(45^\circ + \theta) = \dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\)।
दोनों पक्षों का वर्ग करने पर: \(1-\sin^2 2\theta = 1 \implies \cos^2 2\theta = 1 \implies \cos 2\theta = 1\)।