\(-\tan \theta\)
\(-\cos \theta\)
\(\tan \theta\)
\(\cos \theta\)
Correct answer is A
\(\frac{\cos 2\theta - 1}{\sin 2\theta}\)
\(\cos (x + y) = \cos x \cos y - \sin x \sin y \implies \cos 2\theta = \cos^{2} \theta - \sin^{2} \theta\)
\(\cos^{2} \theta = 1 - \sin^{2} \theta \implies \cos 2\theta = 1 - 2\sin^{2} \theta\)
\(\sin 2\theta = 2\sin \theta \cos \theta\)
\(\therefore \frac{\cos 2\theta - 1}{\sin 2\theta} = \frac{1 - 2\sin^{2}\theta - 1}{2\sin \theta \cos \theta}\)
= \(\frac{-2 \sin^{2} \theta}{2\sin \theta \cos \theta} = \frac{- \sin \theta}{\cos \theta}\)
= \(-\tan \theta\)