frove that \( \frac{1-\cos x}{\sin x}=\sin x \)

We start with the expression \[ \frac{1-\cos x}{\sin x}=\sin x. \] A common method to "prove" an identity is to manipulate one side to see if it can be transformed into the other. In this case, multiply both sides of the equation by \(\sin x\) (assuming \(\sin x \neq 0\)): \[ 1-\cos x = \sin^2 x. \] Recall the Pythagorean identity: \[ \sin^2 x=1-\cos^2 x. \] If we substitute the identity into the equation we get: \[ 1-\cos x = 1-\cos^2 x. \] Subtract \(1\) from both sides: \[ -\cos x = -\cos^2 x. \] Multiply both sides by \(-1\): \[ \cos x = \cos^2 x. \] This equation can be rewritten as: \[ \cos^2 x - \cos x = 0. \] Factor out \(\cos x\): \[ \cos x\,( \cos x-1)=0. \] This factorization tells us that the equation holds if and only if \[ \cos x=0\quad \text{or} \quad \cos x=1. \] Thus, the original equation \[ \frac{1-\cos x}{\sin x}=\sin x \] holds only for those values of \(x\) for which \(\cos x=0\) or \(\cos x=1\). That is, it is true when - \(\cos x=1\), which happens for \(x=2\pi k\) (with \(k\) an integer), or - \(\cos x=0\), which happens for \(x=\frac{\pi}{2}+ \pi k\) (with \(k\) an integer). For any other value of \(x\) (with \(\sin x\neq 0\) to avoid division by zero), the equality does not hold in general. Therefore, the identity \(\frac{1-\cos x}{\sin x}=\sin x\) is valid only for those \(x\) satisfying \[ \cos x\,( \cos x-1)=0. \]