3 If \( \cos A=\frac{S}{6} \) and \( A+B-90^{\circ} \), use a sketch to find the value of \( \frac{\tan A \cdot \tan B \cdot \sin A}{\cos B} \). Prove the following identities: \( 4.1 \quad \cos ^{2} x+\sin x \cdot \cos x \cdot \tan x=1 \) \( 4.2 \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=2 \cos ^{2} \theta-1 \) sinf 5 Simplify each of the following. osi \( 5.1 \frac{2 \sin \left(90^{\circ}-x\right)-\cos \left(360^{\circ}-x\right)}{2 \cos \left(90^{\circ}+x\right)-\sin \left(180^{\circ}+x\right)} \) e \( 5.2 \frac{\tan \left(180^{\circ}-x\right) \tan 45^{\circ} \cdot \cos (-x)}{\sin \left(180^{\circ}-x\right)} \)

Let's break down the problem step by step. ### Part 1: Finding \( \frac{\tan A \cdot \tan B \cdot \sin A}{\cos B} \) Given: - \( \cos A = \frac{S}{6} \) - \( A + B = 90^\circ \) (which implies \( B = 90^\circ - A \)) From the identity \( \tan B = \frac{\sin B}{\cos B} \) and using the co-function identity, we have: \[ \tan B = \cot A = \frac{1}{\tan A} \] Now, we can express \( \sin A \) and \( \cos B \): - \( \sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{S}{6}\right)^2} \) - \( \cos B = \sin A \) Now substituting these into the expression: \[ \frac{\tan A \cdot \tan B \cdot \sin A}{\cos B} = \frac{\tan A \cdot \frac{1}{\tan A} \cdot \sin A}{\sin A} \] This simplifies to: \[ \frac{\sin A}{\sin A} = 1 \] ### Part 2: Proving the identities #### Identity 4.1: \( \cos^2 x + \sin x \cdot \cos x \cdot \tan x = 1 \) Starting with the left-hand side: \[ \cos^2 x + \sin x \cdot \cos x \cdot \tan x \] Substituting \( \tan x = \frac{\sin x}{\cos x} \): \[ = \cos^2 x + \sin x \cdot \cos x \cdot \frac{\sin x}{\cos x} \] This simplifies to: \[ = \cos^2 x + \sin^2 x \] Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \): \[ = 1 \] Thus, the identity is proven. #### Identity 4.2: \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = 2 \cos^2 \theta - 1 \) Starting with the left-hand side: \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \] Substituting \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \): \[ = \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} \] Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \): \[ = \cos^2 \theta - \sin^2 \theta \] Using the double angle identity \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \): \[ = \cos(2\theta) \] Now, using the identity \( \cos(2\theta) = 2\cos^2 \theta - 1 \): \[ = 2\cos^2 \theta - 1 \] Thus, the identity is proven. ### Part 3: Simplifying the expressions #### Expression 5.1: \[ \frac{2 \sin(90^\circ - x) - \cos(360^\circ - x)}{2 \cos(90^\circ + x) - \sin(180^\circ + x)} \] Using the co-function identities: - \( \sin(90^\circ - x) = \cos x \) - \( \cos(360^\circ - x) = \cos x \) - \( \cos(90^\circ + x) = -\sin x \) - \( \sin(180^\circ + x) = -\sin x \) Substituting these into the expression: \[ = \frac{2 \cos x - \cos x}{2(-\sin x) - (-\sin x)} = \frac{\cos x}{-2\sin x + \sin x} = \frac{\cos x}{-\sin x} = -\cot x \] #### Expression 5.2: \[ \frac{\tan(180^\circ - x) \tan 45^\circ \cdot \cos(-x)}{\sin(180^\circ - x)} \] Using the identities: - \( \tan(180^\circ - x) = -\tan x \) - \( \tan 45^\circ = 1 \) - \( \cos(-x) = \cos x \) - \( \sin(180^\circ - x) = \sin x \) Substituting these into the expression: \[ = \frac{-\tan x \cdot 1 \cdot \cos x}{\sin x} = -\frac{\tan x \cdot \cos x}{\sin x} = -\frac{\sin x}{\sin x} = -1 \] ### Final Answers: 1. \( \frac{\tan A \cdot \tan B \cdot \sin A}{\cos B} = 1 \) 2. Identity 4.1 is proven. 3. Identity 4.2 is proven. 4. Expression 5.1 simplifies to \( -\cot x \). 5. Expression 5.2 simplifies to \( -1 \).