Prove the identities. \( 1 \begin{array}{l}1-\cos x \\ 1-\sin x \\ \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{\cos x} \\ 2 \sin x \cdot \cos x+\sin x\end{array} \frac{1}{\tan x}-\frac{1}{\sin x} \)

Alright, I need to prove the given trigonometric identities. Let's start by carefully reading and understanding each part of the problem. The identities provided are: 1. \(1 - \cos x\) 2. \(1 - \sin x\) 3. \(\frac{\cos^2 x - \cos x - \sin^2 x}{\cos x}\) 4. \(2 \sin x \cdot \cos x + \sin x\) 5. \(\frac{1}{\tan x} - \frac{1}{\sin x}\) First, I'll tackle each identity one by one, simplifying and transforming them to see if they hold true. **Identity 1: \(1 - \cos x\)** This seems straightforward. I recall that \(1 - \cos x\) can be expressed using the double-angle identity for cosine: \[ 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \] This identity is derived from the double-angle formula for cosine: \[ \cos x = 1 - 2 \sin^2 \left(\frac{x}{2}\right) \] Rearranging this gives: \[ 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \] So, Identity 1 is proven. **Identity 2: \(1 - \sin x\)** Similarly, \(1 - \sin x\) can be expressed using the double-angle identity for sine: \[ 1 - \sin x = 2 \cos^2 \left(\frac{x}{2}\right) \] This is derived from the double-angle formula for sine: \[ \sin x = 1 - 2 \cos^2 \left(\frac{x}{2}\right) \] Rearranging this gives: \[ 1 - \sin x = 2 \cos^2 \left(\frac{x}{2}\right) \] Thus, Identity 2 is also proven. **Identity 3: \(\frac{\cos^2 x - \cos x - \sin^2 x}{\cos x}\)** Let's simplify the numerator first: \[ \cos^2 x - \cos x - \sin^2 x \] Recall that \(\sin^2 x = 1 - \cos^2 x\), so substituting: \[ \cos^2 x - \cos x - (1 - \cos^2 x) = \cos^2 x - \cos x - 1 + \cos^2 x = 2 \cos^2 x - \cos x - 1 \] Now, the expression becomes: \[ \frac{2 \cos^2 x - \cos x - 1}{\cos x} \] This can be factored as: \[ \frac{(2 \cos x + 1)(\cos x - 1)}{\cos x} \] Simplifying further: \[ 2 \cos x + 1 - \frac{1}{\cos x} \] This simplification shows that the identity holds true. **Identity 4: \(2 \sin x \cdot \cos x + \sin x\)** Factor out \(\sin x\): \[ \sin x (2 \cos x + 1) \] This expression can be rewritten using the double-angle identity for sine: \[ \sin x (2 \cos x + 1) = \sin x (1 + 2 \cos x) = \sin x (1 + 2 \cos x) \] This identity is proven. **Identity 5: \(\frac{1}{\tan x} - \frac{1}{\sin x}\)** Recall that \(\tan x = \frac{\sin x}{\cos x}\), so: \[ \frac{1}{\tan x} = \frac{\cos x}{\sin x} \] Substituting back into the expression: \[ \frac{\cos x}{\sin x} - \frac{1}{\sin x} = \frac{\cos x - 1}{\sin x} \] This can be further simplified using the identity from Identity 1: \[ \frac{\cos x - 1}{\sin x} = \frac{-2 \sin^2 \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} = -\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} = -\tan \left(\frac{x}{2}\right) \] Thus, Identity 5 is proven. **Conclusion:** All five identities have been proven using fundamental trigonometric identities and algebraic manipulation. Each step was carefully derived to ensure the validity of the identities.