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

Sure! Let's prove each identity step by step. --- ### **1. Prove that** \[ \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x} \] **Proof:** Start with the left-hand side (LHS): \[ \frac{\cos x}{1 - \sin x} \] To simplify, multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 + \sin x\): \[ \frac{\cos x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x} = \frac{\cos x (1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \] Simplify the denominator using the identity \(a^2 - b^2 = (a - b)(a + b)\): \[ (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x \] Now, the expression becomes: \[ \frac{\cos x (1 + \sin x)}{\cos^2 x} = \frac{1 + \sin x}{\cos x} \] Thus, \[ \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x} \] Q.E.D. --- ### **2. Prove that** \[ \frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x} = \frac{1}{\tan x} - \frac{1}{\sin x} \] **Proof:** Let's start by simplifying the left-hand side (LHS): \[ \frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x} \] **Step 1:** Simplify the numerator. Recall the Pythagorean identity: \[ \cos^{2} x + \sin^{2} x = 1 \quad \Rightarrow \quad \sin^{2} x = 1 - \cos^{2} x \] Substitute \(\sin^{2} x\) in the numerator: \[ \cos^{2} x - \cos x - (1 - \cos^{2} x) = \cos^{2} x - \cos x - 1 + \cos^{2} x = 2\cos^{2} x - \cos x - 1 \] **Step 2:** Factor the numerator. We need to factor \(2\cos^{2} x - \cos x - 1\). Let’s find two numbers that multiply to \(-2\) (from \(2 \times -1\)) and add to \(-1\): \[ 2\cos^{2} x - 2\cos x + \cos x - 1 = (2\cos^{2} x - 2\cos x) + (\cos x - 1) = 2\cos x(\cos x - 1) + 1(\cos x - 1) = (2\cos x + 1)(\cos x - 1) \] So, the numerator factors as: \[ (2\cos x + 1)(\cos x - 1) \] **Step 3:** Simplify the denominator. Factor out \(\sin x\): \[ 2 \sin x \cos x + \sin x = \sin x (2\cos x + 1) \] **Step 4:** Combine numerator and denominator. \[ \frac{(2\cos x + 1)(\cos x - 1)}{\sin x (2\cos x + 1)} = \frac{\cos x - 1}{\sin x} \] Here, \(2\cos x + 1\) cancels out from the numerator and denominator. **Step 5:** Simplify the right-hand side (RHS). Compute RHS: \[ \frac{1}{\tan x} - \frac{1}{\sin x} = \frac{\cos x}{\sin x} - \frac{1}{\sin x} = \frac{\cos x - 1}{\sin x} \] Thus, both LHS and RHS simplify to \(\frac{\cos x - 1}{\sin x}\). Therefore, \[ \frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x} = \frac{1}{\tan x} - \frac{1}{\sin x} \] Q.E.D. --- Both identities have been successfully proven!