Show that \( \left(1-\sin ^{2} \theta\right) \sec ^{2} \theta=1 \) Show that \( \tan x+\cot x=\sec x \operatorname{cosec} x \) Given that \( \sin x=\frac{5}{13} \) for \( 0^{\circ}<\theta<90^{\circ} \). Find \( \frac{\cos x-2 \sin x}{2 \tan x} \) Show that \( \left(\frac{1+\cos \theta}{1}\right)\left(\frac{\sec \theta-1}{}\right)=1 \)

Sure, let's address each of your problems step by step. --- ### **1. Show that \( \left(1 - \sin^2 \theta\right) \sec^2 \theta = 1 \)** **Proof:** 1. **Recall the Pythagorean Identity:** \[ \sin^2 \theta + \cos^2 \theta = 1 \] Therefore, \[ 1 - \sin^2 \theta = \cos^2 \theta \] 2. **Substitute into the Original Expression:** \[ \left(1 - \sin^2 \theta\right) \sec^2 \theta = \cos^2 \theta \cdot \sec^2 \theta \] 3. **Simplify Using the Definition of Secant:** \[ \sec \theta = \frac{1}{\cos \theta} \Rightarrow \sec^2 \theta = \frac{1}{\cos^2 \theta} \] So, \[ \cos^2 \theta \cdot \sec^2 \theta = \cos^2 \theta \cdot \frac{1}{\cos^2 \theta} = 1 \] **Conclusion:** \[ \left(1 - \sin^2 \theta\right) \sec^2 \theta = 1 \] --- ### **2. Show that \( \tan x + \cot x = \sec x \csc x \)** **Proof:** 1. **Express Tangent and Cotangent in Terms of Sine and Cosine:** \[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \] 2. **Add the Two Expressions:** \[ \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \] 3. **Find a Common Denominator:** \[ \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \] 4. **Use the Pythagorean Identity:** \[ \sin^2 x + \cos^2 x = 1 \] So, \[ \tan x + \cot x = \frac{1}{\sin x \cos x} \] 5. **Express in Terms of Secant and Cosecant:** \[ \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x} \] Therefore, \[ \sec x \csc x = \frac{1}{\cos x} \cdot \frac{1}{\sin x} = \frac{1}{\sin x \cos x} \] **Conclusion:** \[ \tan x + \cot x = \sec x \csc x \] --- ### **3. Given that \( \sin x = \frac{5}{13} \) for \( 0^\circ < x < 90^\circ \). Find \( \frac{\cos x - 2 \sin x}{2 \tan x} \)** **Solution:** 1. **Find \( \cos x \):** Using the Pythagorean identity: \[ \sin^2 x + \cos^2 x = 1 \Rightarrow \left(\frac{5}{13}\right)^2 + \cos^2 x = 1 \] \[ \frac{25}{169} + \cos^2 x = 1 \Rightarrow \cos^2 x = 1 - \frac{25}{169} = \frac{144}{169} \] Since \( 0^\circ < x < 90^\circ \), \( \cos x \) is positive: \[ \cos x = \frac{12}{13} \] 2. **Find \( \tan x \):** \[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \] 3. **Compute the Numerator (\( \cos x - 2 \sin x \)):** \[ \cos x - 2 \sin x = \frac{12}{13} - 2 \cdot \frac{5}{13} = \frac{12}{13} - \frac{10}{13} = \frac{2}{13} \] 4. **Compute the Denominator (\( 2 \tan x \)):** \[ 2 \tan x = 2 \cdot \frac{5}{12} = \frac{10}{12} = \frac{5}{6} \] 5. **Form the Fraction and Simplify:** \[ \frac{\cos x - 2 \sin x}{2 \tan x} = \frac{\frac{2}{13}}{\frac{5}{6}} = \frac{2}{13} \cdot \frac{6}{5} = \frac{12}{65} \] **Conclusion:** \[ \frac{\cos x - 2 \sin x}{2 \tan x} = \frac{12}{65} \] --- ### **4. Show that \( \left(\frac{1 + \cos \theta}{1}\right) \left(\frac{\sec \theta - 1}{\text{?}}\right) = 1 \)** **Clarification Needed:** The expression provided seems incomplete because the second fraction \( \frac{\sec \theta - 1}{\text{?}} \) is missing a denominator. To provide a meaningful proof, the complete expression is necessary. However, assuming that the intended expression is: \[ \left(1 + \cos \theta\right) \left( \sec \theta - 1 \right) = 1 \] Let's explore this assumption. **Attempted Proof:** 1. **Express Secant in Terms of Cosine:** \[ \sec \theta = \frac{1}{\cos \theta} \] So, \[ \sec \theta - 1 = \frac{1}{\cos \theta} - 1 = \frac{1 - \cos \theta}{\cos \theta} \] 2. **Multiply with \( 1 + \cos \theta \):** \[ (1 + \cos \theta) \left( \sec \theta - 1 \right) = (1 + \cos \theta) \cdot \frac{1 - \cos \theta}{\cos \theta} \] \[ = \frac{(1 + \cos \theta)(1 - \cos \theta)}{\cos \theta} \] 3. **Use the Difference of Squares:** \[ (1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta \] So, \[ \frac{(1 + \cos \theta)(1 - \cos \theta)}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \] 4. **Simplify the Expression:** \[ \frac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \cdot \tan \theta \] 5. **Assessing Equality to 1:** \[ \sin \theta \cdot \tan \theta = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \] This expression is **not** generally equal to 1 for all \( \theta \). It only equals 1 for specific angles where \( \frac{\sin^2 \theta}{\cos \theta} = 1 \), which does not hold true universally. **Conclusion:** Given the current form of the expression, \( \left(1 + \cos \theta\right) \left( \sec \theta - 1 \right) \) does **not** simplify to 1 for all \( \theta \). If there's a different intended expression or additional context, please provide the complete details so that a proper proof can be constructed. --- If you have any further questions or need clarification on the fourth problem, feel free to ask!