Given : \( \frac{\sin 2 x-\cos x}{1-\cos 2 x-\sin x}=\frac{\cos x}{\sin x} \) 6.2.1 Prove the above identity.

To prove the identity \[ \frac{\sin 2x - \cos x}{1 - \cos 2x - \sin x} = \frac{\cos x}{\sin x}, \] we'll start by using trigonometric identities to simplify both sides. First, recall the double angle identities: - \(\sin 2x = 2 \sin x \cos x\) - \(\cos 2x = 1 - 2 \sin^2 x\) Now, substitute these into the left-hand side: \[ \frac{\sin 2x - \cos x}{1 - \cos 2x - \sin x} = \frac{2 \sin x \cos x - \cos x}{1 - (1 - 2 \sin^2 x) - \sin x}. \] This simplifies to: \[ = \frac{(2 \sin x - 1)\cos x}{2 \sin^2 x - \sin x}. \] Now, let's rewrite the denominator: \[ 2 \sin^2 x - \sin x = \sin x(2 \sin x - 1). \] Thus, the left-hand side becomes: \[ \frac{(2 \sin x - 1) \cos x}{\sin x (2 \sin x - 1)}. \] For \(\sin x \neq 0\) and \(2 \sin x - 1 \neq 0\), we can cancel \(2 \sin x - 1\): \[ = \frac{\cos x}{\sin x}. \] Now, we observe that this is precisely the right-hand side of the original equation: \[ \frac{\cos x}{\sin x}. \] Since both sides are equal, we conclude that \[ \frac{\sin 2x - \cos x}{1 - \cos 2x - \sin x} = \frac{\cos x}{\sin x} \] is indeed an identity for all values of \(x\) where the expressions are defined.