EXERCISE 1 1.1 Use the diagram alongside to find the length OP. 1.2 Now write down the value of: 1.2.1 \( \sin \theta \) 1.2.2 \( 61 \cos ^{2} \theta+1 \) \( 1.2 .3 \tan \left(180^{\circ}-\theta\right) \) 2 If \( 7 \sin \theta+4=0 \) and \( \cos \theta>0 \) find, without a calculator, the value of \( 2.1 \quad 1-49 \cos ^{2} \theta \) \( 2.2 \cos \theta \cdot \tan \theta \) 3 If \( \cos \mathrm{A}=\frac{5}{6} \) and \( \mathrm{A}+\mathrm{B}=90^{\circ} \), use a sketch to find the value of \( \frac{\tan \mathrm{A} \cdot \tan \mathrm{B} \cdot \sin \mathrm{A}}{\cos \mathrm{B}} \) 4 Prove the following identities: \( 4.1 \quad \cos ^{2} x+\sin x \cdot \cos x \cdot \tan x=1 \) \( 4.2 \quad \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=2 \cos ^{2} \theta-1 \)

Let's solve the exercise step by step. ### Exercise 1 #### 1.1 Find the length OP. To find the length \( OP \), we need more information about the diagram or the context in which \( OP \) is defined. Please provide the necessary details or the diagram. #### 1.2 Write down the values: 1.2.1 \( \sin \theta \) 1.2.2 \( 61 \cos^{2} \theta + 1 \) 1.2.3 \( \tan(180^{\circ} - \theta) \) To solve these, we need to express them in terms of \( \sin \theta \) and \( \cos \theta \). ### 2. Given \( 7 \sin \theta + 4 = 0 \) and \( \cos \theta > 0 \): From \( 7 \sin \theta + 4 = 0 \): \[ \sin \theta = -\frac{4}{7} \] Since \( \cos \theta > 0 \), \( \theta \) is in the fourth quadrant. #### 2.1 Find \( 1 - 49 \cos^{2} \theta \): Using the Pythagorean identity: \[ \sin^{2} \theta + \cos^{2} \theta = 1 \] We can find \( \cos^{2} \theta \): \[ \cos^{2} \theta = 1 - \sin^{2} \theta = 1 - \left(-\frac{4}{7}\right)^{2} = 1 - \frac{16}{49} = \frac{33}{49} \] Now, substituting into \( 1 - 49 \cos^{2} \theta \): \[ 1 - 49 \cos^{2} \theta = 1 - 49 \cdot \frac{33}{49} = 1 - 33 = -32 \] #### 2.2 Find \( \cos \theta \cdot \tan \theta \): First, we find \( \cos \theta \): \[ \cos \theta = \sqrt{1 - \sin^{2} \theta} = \sqrt{1 - \left(-\frac{4}{7}\right)^{2}} = \sqrt{\frac{33}{49}} = \frac{\sqrt{33}}{7} \] Now, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{4}{7}}{\frac{\sqrt{33}}{7}} = -\frac{4}{\sqrt{33}} \). Thus, \[ \cos \theta \cdot \tan \theta = \frac{\sqrt{33}}{7} \cdot \left(-\frac{4}{\sqrt{33}}\right) = -\frac{4}{7} \] ### 3. Given \( \cos A = \frac{5}{6} \) and \( A + B = 90^{\circ} \): Since \( A + B = 90^{\circ} \), we have \( \sin B = \cos A = \frac{5}{6} \) and \( \cos B = \sin A \). Using the identity \( \tan A = \frac{\sin A}{\cos A} \) and \( \tan B = \frac{\sin B}{\cos B} \): \[ \tan A = \frac{\sqrt{1 - \cos^{2} A}}{\cos A} = \frac{\sqrt{1 - \left(\frac{5}{6}\right)^{2}}}{\frac{5}{6}} = \frac{\sqrt{\frac{11}{36}}}{\frac{5}{6}} = \frac{\sqrt{11}}{5} \] \[ \tan B = \frac{\sin B}{\cos B} = \frac{\frac{5}{6}}{\sqrt{1 - \left(\frac{5}{6}\right)^{2}}} = \frac{\frac{5}{6}}{\frac{\sqrt{11}}{6}} = \frac{5}{\sqrt{11}} \] Now, we can find: \[ \frac{\tan A \cdot \tan B \cdot \sin A}{\cos B} = \frac{\left(\frac{\sqrt{11}}{5}\right) \cdot \left(\frac{5}{\sqrt{11}}\right) \cdot \sin A}{\cos B} \] Since \( \sin A = \sqrt{1 - \cos^{2} A} = \frac{\sqrt{11}}{6} \) and \( \cos B = \sin A = \frac{\sqrt{11}}{6} \): \[ = \frac{\left(\frac{\sqrt{11}}{5}\right) \cdot \left(\frac{5}{\sqrt{11}}\right) \cdot \frac{\sqrt{11}}{6}}{\frac{\sqrt{11}}{6}} = 1 \] ### 4. Prove the following identities: #### 4.1 Prove \( \cos^{2} x + \sin x \cdot \cos x \cdot \tan x = 1 \): Using \( \tan x = \frac{\sin x}{\cos x} \): \[ \sin x \cdot \cos x \cdot \tan x = \sin x \cdot \cos x \cdot \frac{\sin x}{\cos x} = \sin^{2} x \] Thus, \[ \cos^{2} x + \sin^{2} x = 1 \] #### 4.2 Prove \( \frac{1 - \tan^{2} \theta}{1 + \tan^{2} \theta} = 2 \cos^{2} \theta - 1 \): Using \( \tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta} \): \[ \frac{1 - \frac{\sin^{2} \theta}{\cos^{2} \theta}}{1 + \frac{\sin^{2} \theta}{\cos^{2} \theta}} = \frac{\cos^{2} \theta - \sin^{2} \theta}{\cos^{2} \theta + \sin^{2} \theta} = \cos^{2} \theta - \sin^{2} \theta \] Using \( \cos^{2} \theta - \sin^{2} \theta = 2 \cos^{2} \theta - 1 \). Thus, both identities are proven. If you have any specific values or additional information for part 1.1, please provide them for further assistance!