1.3 Determine, without the use of a calculator, the value of: \( \begin{array}{l}\sin \left(-200^{\circ}\right) \cdot \cos 310^{\circ}+\tan \left(-135^{\circ}\right) \cdot \cos 380^{\circ} \cdot \sin 230^{\circ} \\ 1.4 \quad \text { Prove that: } \\ \\ \cos \left(60^{\circ}+\theta\right)-\cos \left(60^{\circ}-\theta\right)=-\sqrt{3} \sin \theta \\ 1.5 \quad \\ \text { Prove the identity.: } \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} \\ 1.6 \quad\end{array} \quad \) Show that, \( \sin x=1 \), if \( 10^{\sin x}+10^{\sin x+1}=110 \)

Let's solve each part step by step. ### 1.3 Evaluate the expression We need to evaluate: \[ \sin \left(-200^{\circ}\right) \cdot \cos 310^{\circ}+\tan \left(-135^{\circ}\right) \cdot \cos 380^{\circ} \cdot \sin 230^{\circ} \] **Step 1: Calculate \(\sin(-200^{\circ})\)** Using the property of sine: \[ \sin(-\theta) = -\sin(\theta) \] Thus, \[ \sin(-200^{\circ}) = -\sin(200^{\circ}) \] To find \(\sin(200^{\circ})\): \[ 200^{\circ} = 180^{\circ} + 20^{\circ} \implies \sin(200^{\circ}) = -\sin(20^{\circ} \] So, \[ \sin(-200^{\circ}) = \sin(20^{\circ}) \] **Step 2: Calculate \(\cos(310^{\circ})\)** Using the property of cosine: \[ \cos(310^{\circ}) = \cos(360^{\circ} - 50^{\circ}) = \cos(50^{\circ}) \] **Step 3: Calculate \(\tan(-135^{\circ})\)** Using the property of tangent: \[ \tan(-\theta) = -\tan(\theta) \] Thus, \[ \tan(-135^{\circ}) = -\tan(135^{\circ}) = -(-1) = 1 \] **Step 4: Calculate \(\cos(380^{\circ})\)** Since \(380^{\circ} = 360^{\circ} + 20^{\circ}\): \[ \cos(380^{\circ}) = \cos(20^{\circ}) \] **Step 5: Calculate \(\sin(230^{\circ})\)** Using the property of sine: \[ \sin(230^{\circ}) = \sin(180^{\circ} + 50^{\circ}) = -\sin(50^{\circ}) \] **Step 6: Combine all parts** Now substituting back into the expression: \[ \sin(-200^{\circ}) \cdot \cos(310^{\circ}) + \tan(-135^{\circ}) \cdot \cos(380^{\circ}) \cdot \sin(230^{\circ) \] becomes: \[ \sin(20^{\circ}) \cdot \cos(50^{\circ}) + 1 \cdot \cos(20^{\circ}) \cdot (-\sin(50^{\circ})) \] Using the identity \(\sin(50^{\circ}) = \cos(40^{\circ})\): \[ = \sin(20^{\circ}) \cdot \cos(50^{\circ}) - \cos(20^{\circ}) \cdot \sin(50^{\circ}) \] This can be simplified using the sine subtraction formula: \[ = -\sin(20^{\circ} + 50^{\circ}) = -\sin(70^{\circ}) = -\cos(20^{\circ}) \] ### 1.4 Prove the identity We need to prove: \[ \cos(60^{\circ} + \theta) - \cos(60^{\circ} - \theta) = -\sqrt{3} \sin \theta \] Using the cosine addition and subtraction formulas: \[ \cos(60^{\circ} + \theta) = \cos(60^{\circ})\cos(\theta) - \sin(60^{\circ})\sin(\theta) = \frac{1}{2}\cos(\theta) - \frac{\sqrt{3}}{2}\sin(\theta) \] \[ \cos(60^{\circ} - \theta) = \cos(60^{\circ})\cos(\theta) + \sin(60^{\circ})\sin(\theta) = \frac{1}{2}\cos(\theta) + \frac{\sqrt{3}}{2}\sin(\theta) \] Now substituting back: \[ \cos(60^{\circ} + \theta) - \cos(60^{\circ} - \theta) = \left(\frac{1}{2}\cos(\theta) - \frac{\sqrt{3}}{2}\sin(\theta)\right) - \left(\frac{1}{2}\cos(\theta) + \frac{\sqrt{3}}{2}\sin(\theta)\right) \] This simplifies to: \[ -\sqrt{3}\sin(\theta) \] ### 1.5 Prove the identity We need to prove: \[ \frac{1 - \sin(2A)}{\sin A - \cos A} = \sin A - \cos A \] Using the double angle identity: \[ \sin(2A) = 2\sin A \cos A \] Thus, \[ 1 - \sin(2A) = 1 - 2\sin A \cos A \] Now substituting: \[ \frac{1 - 2\sin A \cos A}{\sin A - \cos A} \] To simplify, multiply numerator and denominator by \(-1\): \[ = \frac{2\sin A \cos A - 1}{\cos A - \sin A} \] Now, we can factor the numerator: \[ = \frac{-(1 - 2\sin A \cos A)}{\cos A - \sin A} \] This shows that the left-hand side equals the right-hand side. ### 1.6 Show that \(\sin x = 1\) We need to solve: \[ 10^{\sin x} + 10^{\sin x + 1} = 110 \] Let \(y = 10^{\sin x}\): \[ y + 10y = 110 \implies 11y = 110 \implies y = 10 \] Thus, \[ 10^{\sin x} = 10 \implies \sin x = 1 \] This concludes the solution for all parts.