1. **Problem statement:** Given $\sin a = \cos b$ where $a,b$ are acute angles, find $\tan(a+b)$. Since $a,b$ are acute and $\sin a = \cos b$, then $a = b$ because sine and cosine are cofunctions. Using the formula for tangent of sum: $$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} = \frac{2 \tan a}{1 - \tan^2 a}$$ Substitute with $\tan a = t$ and solve accordingly; possible values include 1, -1, $1 - \sqrt{3}$, or undefined. 2. **Problem statement:** Same as 1, duplicate question; answer is the same. 3. **Problem statement:** Given $\tan 5x = \cot 4x$ with $0^\circ < x < 90^\circ$, find $\sin 3x$. Since $\cot 4x = \tan (90^\circ - 4x)$, equality implies $5x = 90^\circ - 4x$, so $9x = 90^\circ$, so $x = 10^\circ$. Then $\sin 3x = \sin 30^\circ = \frac{1}{2}$. 4. **Problem statement:** If $\sin 2\theta \sec \theta = 1$ where $\theta \in [0^\circ, 90^\circ]$, find $\theta$. Use $\sec \theta = \frac{1}{\cos \theta}$, then: $$\sin 2\theta \times \frac{1}{\cos \theta} = 1 \implies \frac{2 \sin \theta \cos \theta}{\cos \theta} = 1 \implies 2 \sin \theta = 1 \implies \sin \theta = \frac{1}{2}$$ Thus $\theta = 30^\circ$. 5. **Problem statement:** If $5 \cos(270^\circ + \theta) = 4$ and $0 < \theta < 90^\circ$, find $\sin \theta$. Recall $\cos(270^\circ + \theta) = -\sin \theta$, so: $$5 \times (-\sin \theta) = 4 \implies -5 \sin \theta = 4 \implies \sin \theta = -\frac{4}{5}$$ Since $\theta$ is in $0^\circ < \theta < 90^\circ$ and sine cannot be negative there, no solution within the interval. 6. **Problem statement:** Find $\sin(-30^\circ)$. We know $\sin(-\alpha) = -\sin \alpha$, so: $$\sin(-30^\circ) = -\sin 30^\circ = -\frac{1}{2}$$. 7. **Problem statement:** Find $\cos 240^\circ + \cos 420^\circ$. Using periodicity $\cos \theta = \cos(\theta \pm 360^\circ)$, $$\cos 420^\circ = \cos (420^\circ - 360^\circ) = \cos 60^\circ = \frac{1}{2}$$ Also, $\cos 240^\circ = -\frac{1}{2}$. Sum: $$-\frac{1}{2} + \frac{1}{2} = 0$$. 8. **Problem statement:** If $2 \sin x = -1$ for positive acute $x$, find $x$. Since $x$ is acute (between $0^\circ$ and $90^\circ$), sine must be positive. $2 \sin x = -1 \implies \sin x = -\frac{1}{2}$, which is impossible in $0^\circ < x < 90^\circ$. No solution. 9. **Problem statement:** Find the range of $f(x) = \sin x$. Since sine function ranges from -1 to 1, $$\text{Range} = [-1, 1]$$. Ex 5: 1) Evaluate $\cos 330^\circ \cdot \cos 240^\circ + \sin 750^\circ \cdot \sin 120^\circ$. Using cosine addition formula, $$\cos A \cos B + \sin A \sin B = \cos(A-B)$$. Here $A = 330^\circ$, $B=240^\circ$. So expression equals: $$\cos(330^\circ - 240^\circ) = \cos 90^\circ = 0$$. 2) Evaluate $\sin 120^\circ \cos 330^\circ - \cos 420^\circ \sin 330^\circ$. Using sine subtraction formula, $$\sin A \cos B - \cos A \sin B = \sin (A - B)$$. $A = 120^\circ$, $B=330^\circ$, so: $$\sin(120^\circ - 330^\circ) = \sin (-210^\circ) = -\sin 210^\circ = -(-\frac{1}{2}) = \frac{1}{2}$$. 3) Calculate $\cos 120^\circ + \tan 225^\circ + \csc 135^\circ + \sec 25^\circ + \sec 25^\circ + \csc 65^\circ$. $\cos 120^\circ = -\frac{1}{2}$, $\tan 225^\circ = 1$, $\csc 135^\circ = \frac{1}{\sin 135^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$, $\sec 25^\circ = \frac{1}{\cos 25^\circ}$ (approximate twice), $\csc 65^\circ = \frac{1}{\sin 65^\circ}$. Exact sum depends on numerical values, leave in symbolic form. 4) Evaluate $\sin 510^\circ + \cos (-60^\circ) + \tan ...$ (incomplete). $\sin 510^\circ = \sin (510^\circ - 360^\circ) = \sin 150^\circ = \frac{1}{2}$, $\cos (-60^\circ) = \cos 60^\circ = \frac{1}{2}$. Sum so far is $1 + \tan ...$ (incomplete). Ex 6: Given terminal side cuts unit circle at $P(,)$ The value of $\tan(270^\circ + \theta) = \tan(270^\circ + \theta)$. Since $\tan(270^\circ + \theta) = -\cot \theta = - \frac{1}{\tan \theta}$. Ex 7: Solve these equations: (i) $\frac{\sqrt{3}}{2} x - \frac{\sqrt{3}}{2} = 0 \implies \frac{\sqrt{3}}{2} (x - 1) = 0 \implies x=1$. (ii) $2 \sqrt{3} x - \sqrt{3} = 0 \implies \sqrt{3} (2x - 1) = 0 \implies x=\frac{1}{2}$. (iii) $\sqrt{3} \tan + 1 = 0$ (incomplete equation, likely $\sqrt{3} \tan x + 1=0$), solve for $\tan x = -\frac{1}{\sqrt{3}}$. (iv) $4 \cos^2 x = 1 \implies \cos^2 x = \frac{1}{4} \implies \cos x = \pm \frac{1}{2}$. Ex 8: Choose: The point $( - \frac{\sqrt{3}}{2}, \frac{1}{2})$ lies on unit circle corresponding to angles $150^\circ$ and $210^\circ$ in standard position. Final answers summarized in content.