ESTION 9 Simplify as far as possible: \( \frac{\sin ^{2} \theta}{\sin \left(180^{\circ}-\theta\right) \cdot \cos \left(90^{\circ}+\theta\right)+\tan 45^{\circ}} \) Simplify without the use of a calculator: \( \frac{\sin 104^{\circ}\left(2 \cos ^{2} 15^{\circ}-1\right)}{\tan 38^{\circ} \cdot \sin ^{2} 412^{\circ}} \)

We start by simplifying each expression step by step. ### Problem 1 Simplify \[ \frac{\sin ^{2} \theta}{\sin \left(180^{\circ}-\theta\right) \cdot \cos \left(90^{\circ}+\theta\right)+\tan 45^{\circ}}. \] 1. Use the identities: - \(\sin (180^{\circ}-\theta)=\sin\theta\), - \(\cos (90^{\circ}+\theta)=-\sin\theta\), - \(\tan 45^{\circ}=1\). 2. Substitute these into the expression: \[ \text{Denominator} = \sin\theta \cdot (-\sin\theta) + 1 = -\sin^2\theta+1. \] 3. Recognize that: \[ 1-\sin^2\theta = \cos^2\theta. \] 4. Thus, the entire expression becomes: \[ \frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta. \] ### Problem 2 Simplify \[ \frac{\sin 104^{\circ}\left(2 \cos ^{2} 15^{\circ}-1\right)}{\tan 38^{\circ} \cdot \sin ^{2} 412^{\circ}}. \] 1. **Simplify the numerator:** - Notice the double-angle identity: \[ \cos 2x=2\cos^2 x-1. \] Here, with \(x=15^\circ\), we have: \[ 2\cos^2 15^{\circ}-1=\cos30^{\circ}. \] - Also, use the identity: \[ \sin104^{\circ}=\sin(180^{\circ}-104^{\circ})=\sin76^{\circ}. \] - So the numerator becomes: \[ \sin76^{\circ}\cos30^{\circ}. \] 2. **Simplify the denominator:** - Write \(\tan38^{\circ}\) as: \[ \tan 38^{\circ}=\frac{\sin 38^{\circ}}{\cos 38^{\circ}}. \] - For the sine squared term, consider: \[ 412^{\circ}=412^{\circ}-360^{\circ}=52^{\circ}, \] so: \[ \sin 412^{\circ}=\sin52^{\circ} \quad \text{and} \quad \sin^2 412^{\circ}=\sin^2 52^{\circ}. \] - Notice that \(38^{\circ}+52^{\circ}=90^{\circ}\), hence: \[ \sin52^{\circ}=\cos38^{\circ}. \] Therefore, \[ \sin^2 412^{\circ}=\cos^2 38^{\circ}. \] - The denominator becomes: \[ \tan 38^{\circ}\cdot \cos^2 38^{\circ}=\frac{\sin 38^{\circ}}{\cos 38^{\circ}} \cdot \cos^2 38^{\circ}=\sin38^{\circ}\cos38^{\circ}. \] 3. **Combine numerator and denominator:** - The expression now is: \[ \frac{\sin76^{\circ}\cos30^{\circ}}{\sin38^{\circ}\cos38^{\circ}}. \] - Use the double-angle formula for sine: \[ \sin76^{\circ}=\sin(2\cdot38^{\circ})=2\sin38^{\circ}\cos38^{\circ}. \] - Substitute this into the expression: \[ \frac{2\sin38^{\circ}\cos38^{\circ}\cos30^{\circ}}{\sin38^{\circ}\cos38^{\circ}}. \] - Cancel the common factors \(\sin38^{\circ}\cos38^{\circ}\): \[ 2\cos30^{\circ}. \] - Finally, recall that: \[ \cos30^{\circ}=\frac{\sqrt{3}}{2}. \] - Therefore, the expression simplifies to: \[ 2\cdot\frac{\sqrt{3}}{2}=\sqrt{3}. \] ### Final Answers 1. \(\tan^2 \theta\) 2. \(\sqrt{3}\)