1. Given $\tan A = \frac{a}{a+1}$ and $\tan B = \frac{1}{2a+1}$, show that $A+B = \frac{\pi}{4}$. Use the tangent addition formula: $$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}.$$ Substitute values: $$\frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1} \cdot \frac{1}{2a+1}} = \frac{\frac{a(2a+1) + (a+1)}{(a+1)(2a+1)}}{1 - \frac{a}{(a+1)(2a+1)}}.$$ Simplify numerator: $$a(2a+1) + (a+1) = 2a^2 + a + a + 1 = 2a^2 + 2a + 1.$$ Denominator inside big fraction: $$1 - \frac{a}{(a+1)(2a+1)} = \frac{(a+1)(2a+1) - a}{(a+1)(2a+1)} = \frac{2a^2 + 3a + 1 - a}{(a+1)(2a+1)} = \frac{2a^2 + 2a + 1}{(a+1)(2a+1)}.$$ So, $$\tan(A+B) = \frac{\frac{2a^2 + 2a + 1}{(a+1)(2a+1)}}{\frac{2a^2 + 2a + 1}{(a+1)(2a+1)}} = 1.$$ Since $\tan(\frac{\pi}{4}) = 1$, we conclude $A+B=\frac{\pi}{4}$. 2. Evaluate without calculator: (1) $\tan 75^\circ$. Use angle addition: $75^\circ = 45^\circ + 30^\circ$, so $$\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} +1}{\sqrt{3} -1}.$$ Rationalize denominator: $$\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2 -1^2} = \frac{3 + 2\sqrt{3} +1}{3 -1}= \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}.$$ So $\tan 75^\circ = 2 + \sqrt{3}$. (2) $\sin 470^\circ \cos 130^\circ + \cos 470^\circ \sin 130^\circ$. Recall $\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) = \sin(\alpha + \beta)$. Calculate sum: $470^\circ + 130^\circ = 600^\circ = 600 - 360 = 240^\circ$. Thus, $$\sin 470^\circ \cos 130^\circ + \cos 470^\circ \sin 130^\circ = \sin 240^\circ = -\frac{\sqrt{3}}{2}.$$ 3. Prove: (a) $$\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ.$$ Rewrite numerator and denominator dividing by $\cos 11^\circ$: $$\frac{1 + \tan 11^\circ}{1 - \tan 11^\circ} = \tan\left(45^\circ + 11^\circ\right) = \tan 56^\circ,$$ using formula $\frac{1+\tan x}{1-\tan x} = \tan(45^\circ + x)$. (b) $$\sin(n+1)A \sin(n-1)A \cos(n+1)A \cos(n-1)A = \cos 2A.$$ This seems incomplete or incorrect as stated; clarification would be needed. 4. Simplify $$2 \sin \theta \sin 4\theta + 2 \cos \theta \cos 4\theta.$$ Use product-to-sum formulas: $$2 \sin \theta \sin 4\theta + 2 \cos \theta \cos 4\theta = 2(\cos(\theta - 4\theta)) = 2 \cos (-3\theta) = 2 \cos 3\theta.$$ Thus, simplified form is $$2 \cos 3\theta.$$ 5. Sequence $\{u_n\}$ is given by $u_0 = 1$, $u_n = \frac{(2n)^2}{n^{2+1}} = \frac{4n^2}{n^3} = \frac{4}{n}$ for $n \geq 1$. Calculate: $$u_1 = \frac{4}{1} = 4,$$ $$u_2 = \frac{4}{2} = 2,$$ $$u_3 = \frac{4}{3}.$$ 6. Determine monotonicity: (a) $1,2,3,...,n,...$ is increasing. (b) $1,-1,1,-1,...$ is not monotonic (alternates). (c) $\frac{n}{n+1}$ increases with $n$ because denominator grows slower, so sequence is increasing and bounded by 1 (non-decreasing). 7. Arithmetic sequence with $u_2=4$, $u_6=16$. Formula: $u_n = u_1 + (n-1)d$. So, $$u_2 = u_1 + d = 4,$$ $$u_6 = u_1 + 5d = 16.$$ Subtract equations: $$16 - 4 = (u_1 + 5d) - (u_1 + d) = 4d,$$ $$12 = 4d \Rightarrow d = 3.$$ 8. Find number of terms in sequence $1,4,7,10,...,25$. General term: $u_n = 1 + (n-1)3 = 3n - 2.$ Solve: $3n - 2 = 25 \Rightarrow 3n = 27 \Rightarrow n = 9.$ Number of terms is 9. 9. Sum of first 8 terms of geometric sequence $32, -16, 8, ...$. First term $a = 32$, common ratio $r = \frac{-16}{32} = -\frac{1}{2}$. Sum formula: $$S_n = a \frac{1 - r^n}{1 - r} = 32 \frac{1 - (-\frac{1}{2})^8}{1 + \frac{1}{2}} = 32 \frac{1 - \frac{1}{256}}{\frac{3}{2}} = 32 \frac{\frac{255}{256}}{\frac{3}{2}} = 32 \cdot \frac{255}{256} \cdot \frac{2}{3} = \frac{32 \times 255 \times 2}{256 \times 3} = \frac{16320}{768} = 21.25.$$ 10. Given sequence (details missing), so unable to answer without data. 11. Displacement $x = 10 \cos t$; find at $t=0, \frac{\pi}{3}, \frac{3\pi}{4}$. Calculate: $$x(0) = 10 \cos 0 = 10 \times 1 = 10,$$ $$x(\frac{\pi}{3}) = 10 \cos \frac{\pi}{3} = 10 \times \frac{1}{2} = 5,$$ $$x(\frac{3\pi}{4}) = 10 \cos \frac{3\pi}{4} = 10 \times \left(-\frac{\sqrt{2}}{2}\right) = -5 \sqrt{2}.$$ 12. Solve inequality $\cos 3x \leq \frac{\sqrt{3}}{2}$. Recall $\cos \theta \leq \frac{\sqrt{3}}{2}$ means $\theta \in [0,360]^\circ$ except intervals where $\cos \theta > \frac{\sqrt{3}}{2}$, namely angles less than $30^\circ$ and greater than $330^\circ$. Set $y = 3x$. Solve for $y$: $$\cos y \leq \frac{\sqrt{3}}{2}.$$ Solutions for $y$ in $[0,1080]$ (since $x$ in $[0,360]$): $$y \in [30^\circ, 330^\circ] \cup [390^\circ, 690^\circ] \cup [750^\circ, 1070^\circ].$$ Divide by 3: $$x \in [10^\circ, 110^\circ] \cup [130^\circ, 230^\circ] \cup [250^\circ, 356.67^\circ].$$ 13. Solve $\sin 2x + 5 \cos 2x = 3$ for $0^\circ \leq x \leq 360^\circ$. Max value of $|\sin 2x + 5 \cos 2x|$ is $\sqrt{1^2 + 5^2} = \sqrt{26} \approx 5.10$. Since 3 < 5.10, solutions exist. Rewrite left side: $$R \sin(2x + \alpha) = 3,$$ where $$R = \sqrt{1^2 + 5^2} = \sqrt{26},$$ $$\tan \alpha = \frac{5}{1} = 5,$$ $$\alpha = \tan^{-1}(5) \approx 78.69^\circ.$$ So, $$\sin(2x + 78.69^\circ) = \frac{3}{\sqrt{26}} \approx 0.588.$$ General solutions: $$2x + 78.69^\circ = \sin^{-1}(0.588) = 36^\circ \text{ or } 180^\circ - 36^\circ = 144^\circ.$$ Plus multiples of $360^\circ$: $$2x + 78.69^\circ = 36^\circ + 360^\circ k,$$ $$2x + 78.69^\circ = 144^\circ + 360^\circ k,$$ for integers $k$. Solve for $x$: $$x = \frac{36^\circ - 78.69^\circ + 360^\circ k}{2} = \frac{-42.69^\circ + 360^\circ k}{2},$$ $$x = \frac{144^\circ - 78.69^\circ + 360^\circ k}{2} = \frac{65.31^\circ + 360^\circ k}{2}.$$ For $x$ in $0^\circ$ to $360^\circ$, find $k$ such that $0 \leq x \leq 360$: Calculate possible $x$ values: - For $k=1$, first equation: $x = \frac{317.31}{2} = 158.66^\circ$ - For $k=0$, second equation: $x = \frac{65.31}{2} = 32.65^\circ$ - For $k=0$, first equation: $x = \frac{-42.69}{2} = -21.345^\circ$ (discarded) - For $k=1$, second eq: $x = \frac{425.31}{2} = 212.655^\circ$ - For $k=2$, first eq: $x = \frac{677.31}{2} = 338.655^\circ$ - For $k=2$, second eq: $x = \frac{785.31}{2} = 392.655^\circ$ (discarded) Final solutions within range: $$x \approx 32.65^\circ, 158.66^\circ, 212.655^\circ, 338.655^\circ.$$ 14. Geometric sequence with positive terms, sixth term $= \sqrt{3}$ and eighth term $= \sqrt{7}$. Let first term $a$, ratio $r$. Use: $$a r^{5} = \sqrt{3},$$ $$a r^{7} = \sqrt{7}.$$ Divide second equation by first: $$\frac{a r^{7}}{a r^{5}} = r^{2} = \frac{\sqrt{7}}{\sqrt{3}} = \sqrt{\frac{7}{3}}.$$ Thus: $$r = \left(\sqrt{\frac{7}{3}}\right)^{\frac{1}{2}} = \left(\frac{7}{3}\right)^{\frac{1}{4}}.$$ Find $a$: $$a = \frac{\sqrt{3}}{r^{5}} = \sqrt{3} \times r^{-5} = \sqrt{3} \times \left(\frac{7}{3}\right)^{-\frac{5}{4}} = \sqrt{3} \times \left(\frac{3}{7}\right)^{\frac{5}{4}}.$$ Simplify if desired. Final Answers: 1. $A + B = \frac{\pi}{4}$. 2. $\tan 75^\circ = 2 + \sqrt{3}$; $\sin 470^\circ \cos 130^\circ + \cos 470^\circ \sin 130^\circ = - \frac{\sqrt{3}}{2}$. 3. (a) proved equal; (b) needs clarification. 4. $2 \cos 3\theta$. 5. $u_1=4$, $u_2=2$, $u_3=\frac{4}{3}$. 6. (a) Increasing; (b) Not monotonic; (c) Increasing. 7. Common difference $d=3$. 8. Number of terms = 9. 9. Sum of first 8 terms = 21.25. 11. $x(0) = 10$, $x(\pi/3) = 5$, $x(3\pi/4) = -5\sqrt{2}$. 12. Solutions for $x$ in $[10^\circ, 110^\circ] \cup [130^\circ, 230^\circ] \cup [250^\circ, 356.67^\circ]$. 13. Solutions approximately $32.65^\circ$, $158.66^\circ$, $212.66^\circ$, $338.66^\circ$. 14. $r = (\frac{7}{3})^{1/4}$, $a = \sqrt{3} \left(\frac{3}{7}\right)^{5/4}$.