Mix Examples-Trigonometrical Ratios, Functions and Identities Questions in English

(B) Let $E = \cos^2 \phi + \cos^2(\theta + \phi) - 2 \cos \theta \cos \phi \cos(\theta + \phi)$.
Using the identity $\cos^2 A = \frac{1 + \cos 2A}{2}$,we have:
$E = \frac{1 + \cos 2\phi}{2} + \frac{1 + \cos 2(\theta + \phi)}{2} - \cos \theta [\cos(2\phi + \theta) + \cos \theta]$
$E = 1 + \frac{1}{2} [\cos 2\phi + \cos(2\phi + 2\theta)] - \cos \theta \cos(2\phi + \theta) - \cos^2 \theta$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$E = 1 + \cos(2\phi + \theta) \cos \theta - \cos \theta \cos(2\phi + \theta) - \cos^2 \theta$
$E = 1 - \cos^2 \theta = \sin^2 \theta$.
Since the result is $\sin^2 \theta$,the expression is independent of $\phi$.

(B) Given expression: $\sqrt{4 \cos^{4} \theta + \sin^{2} 2 \theta} + 4 \cot \theta \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$
$= \sqrt{4 \cos^{4} \theta + (2 \sin \theta \cos \theta)^{2}} + 2 \cot \theta \left[2 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)\right]$
$= \sqrt{4 \cos^{4} \theta + 4 \sin^{2} \theta \cos^{2} \theta} + 2 \cot \theta \left[1 + \cos \left(\frac{\pi}{2} - \theta\right)\right]$
$= \sqrt{4 \cos^{2} \theta (\cos^{2} \theta + \sin^{2} \theta)} + 2 \cot \theta (1 + \sin \theta)$
$= |2 \cos \theta| + 2 \cot \theta + 2 \cot \theta \sin \theta$
$= |2 \cos \theta| + 2 \cot \theta + 2 \cos \theta$
Since $\theta \in \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$,$\cos \theta < 0$,so $|2 \cos \theta| = -2 \cos \theta$.
$= -2 \cos \theta + 2 \cot \theta + 2 \cos \theta$
$= 2 \cot \theta$

(D) Given that $\alpha_1, \alpha_2, \cdots, \alpha_n$ are in $A$.$P$. with common difference $\theta$,so $\alpha_{i+1} - \alpha_i = \theta$ for all $i = 1, 2, \cdots, n-1$.
The general term of the series is $T_i = \sec \alpha_i \sec \alpha_{i+1} = \frac{1}{\cos \alpha_i \cos \alpha_{i+1}}$.
We can write $T_i = \frac{1}{\sin \theta} \cdot \frac{\sin(\alpha_{i+1} - \alpha_i)}{\cos \alpha_i \cos \alpha_{i+1}} = \operatorname{cosec} \theta (\tan \alpha_{i+1} - \tan \alpha_i)$.
The sum of the series is $S = \sum_{i=1}^{n-1} T_i = \operatorname{cosec} \theta \sum_{i=1}^{n-1} (\tan \alpha_{i+1} - \tan \alpha_i)$.
This is a telescoping sum: $S = \operatorname{cosec} \theta [(\tan \alpha_2 - \tan \alpha_1) + (\tan \alpha_3 - \tan \alpha_2) + \cdots + (\tan \alpha_n - \tan \alpha_{n-1})]$.
Simplifying,we get $S = \operatorname{cosec} \theta (\tan \alpha_n - \tan \alpha_1)$.
Comparing this with $k(\tan \alpha_n - \tan \alpha_1)$,we find $k = \operatorname{cosec} \theta$.

(C) We know that $\sin(k\pi) = 0$ for any integer $k$.
The given series is $E = \sum_{n=1}^{\infty} \sin \left(\frac{n! \pi}{720}\right)$.
Expanding the terms:
$n=1: \sin \left(\frac{1! \pi}{720}\right) = \sin \left(\frac{\pi}{720}\right)$
$n=2: \sin \left(\frac{2! \pi}{720}\right) = \sin \left(\frac{2 \pi}{720}\right) = \sin \left(\frac{\pi}{360}\right)$
$n=3: \sin \left(\frac{3! \pi}{720}\right) = \sin \left(\frac{6 \pi}{720}\right) = \sin \left(\frac{\pi}{120}\right)$
$n=4: \sin \left(\frac{4! \pi}{720}\right) = \sin \left(\frac{24 \pi}{720}\right) = \sin \left(\frac{\pi}{30}\right)$
$n=5: \sin \left(\frac{5! \pi}{720}\right) = \sin \left(\frac{120 \pi}{720}\right) = \sin \left(\frac{\pi}{6}\right)$
$n=6: \sin \left(\frac{6! \pi}{720}\right) = \sin \left(\frac{720 \pi}{720}\right) = \sin(\pi) = 0$
For all $n \ge 6$,$n!$ is a multiple of $720$,so $\frac{n! \pi}{720}$ is a multiple of $\pi$,making $\sin \left(\frac{n! \pi}{720}\right) = 0$.
Thus,the sum is $\sin \left(\frac{\pi}{720}\right) + \sin \left(\frac{\pi}{360}\right) + \sin \left(\frac{\pi}{120}\right) + \sin \left(\frac{\pi}{30}\right) + \sin \left(\frac{\pi}{6}\right)$.

(D) Given equations are $2y = 1 \implies y = \frac{1}{2}$ and $y = \sin x$.
To find the points of intersection,we solve $\sin x = \frac{1}{2}$ for $x \in [-2\pi, 2\pi]$.
In the interval $[0, 2\pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
In the interval $[-2\pi, 0]$,$\sin x = \frac{1}{2}$ at $x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$ and $x = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6}$.
Thus,the solutions are $x \in \{\frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}, -\frac{11\pi}{6}\}$.
There are $4$ such points of intersection.

(B) Given the equation: $\tan \theta + \tan 2 \theta + \tan 3 \theta = 0$.
Since $3 \theta = \theta + 2 \theta$,we have $\tan 3 \theta = \tan (\theta + 2 \theta) = \frac{\tan \theta + \tan 2 \theta}{1 - \tan \theta \cdot \tan 2 \theta}$.
Substituting $\tan \theta + \tan 2 \theta = -\tan 3 \theta$ and $\tan \theta \cdot \tan 2 \theta = k$ into the formula:
$\tan 3 \theta = \frac{-\tan 3 \theta}{1 - k}$.
Since $\tan 3 \theta \neq 0$,we can divide both sides by $\tan 3 \theta$:
$1 = \frac{-1}{1 - k}$.
$1 - k = -1$.
$k = 2$.

(A) Given: $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$.
Since $A$ and $B$ are acute,$\sin A, \sin B, \cos A, \cos B > 0$.
Substitute $\cos^2 A = 1 - \sin^2 A$ and $\cos^2 B = 1 - \sin^2 B$ into the second equation:
$2(1 - \sin^2 A) = 3(1 - \sin^2 B)$.
Substitute $\sin^2 B = \sin A$:
$2 - 2 \sin^2 A = 3(1 - \sin A) = 3 - 3 \sin A$.
$2 \sin^2 A - 3 \sin A + 1 = 0$.
$(2 \sin A - 1)(\sin A - 1) = 0$.
So,$\sin A = \frac{1}{2}$ or $\sin A = 1$.
Since $A$ is an acute angle,$\sin A = 1$ implies $A = \frac{\pi}{2}$,which is not acute.
Thus,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$.
Then $\sin^2 B = \sin A = \frac{1}{2}$,so $\sin B = \frac{1}{\sqrt{2}}$ (since $B$ is acute).
Therefore,$B = \frac{\pi}{4}$.
The solution is $(A, B) = \left(\frac{\pi}{6}, \frac{\pi}{4}\right)$.

(C) The given equation is $(\sin x - x)(\cos x - x^2) = 0$.
This implies $\sin x = x$ or $\cos x = x^2$.
For the first part,$\sin x = x$,the only real solution is $x = 0$.
For the second part,$\cos x = x^2$,we observe the graphs of $y = \cos x$ and $y = x^2$.
The graph of $y = \cos x$ is a wave oscillating between $-1$ and $1$,and $y = x^2$ is a parabola opening upwards with its vertex at $(0, 1)$.
At $x = 0$,$\cos(0) = 1$ and $0^2 = 0$,so $x=0$ is not a solution for this part.
Since $\cos x$ decreases from $1$ to $-1$ as $x$ goes from $0$ to $\pi$,and $x^2$ increases from $0$ to $\pi^2$,there is exactly one intersection point in the interval $(0, 1)$.
Due to symmetry,there is also exactly one intersection point in the interval $(-1, 0)$.
Thus,$\cos x = x^2$ has $2$ real solutions.
Including the solution $x = 0$ from the first part,the total number of real solutions is $1 + 2 = 3$.

(A) We need to find the intersection of the set $\{x \in R: |\cos x| \geq \sin x\}$ with the interval $\left[0, \frac{3 \pi}{2}\right]$.
Consider the graphs of $y = |\cos x|$ and $y = \sin x$ in the interval $\left[0, \frac{3 \pi}{2}\right]$.
$1$. In the interval $\left[0, \frac{\pi}{4}\right]$,$\cos x \geq \sin x$,so $|\cos x| \geq \sin x$ holds.
$2$. In the interval $\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$,$\sin x > \cos x$,so $|\cos x| < \sin x$ (since $\cos x$ is positive in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ and negative in $\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$).
$3$. In the interval $\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$,$\cos x$ is negative,so $|\cos x| = -\cos x$. Since $\cos x \leq 0$ and $\sin x$ can be negative,we check the condition $-\cos x \geq \sin x$,which is $\cos x + \sin x \leq 0$. This holds for $x \in \left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$.
Thus,the solution is $\left[0, \frac{\pi}{4}\right] \cup \left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$.

(A) We know that $\cot \theta = \frac{1}{\tan \theta}$. Also,$\cot \frac{4 \pi}{5} = \cot (\pi - \frac{\pi}{5}) = -\cot \frac{\pi}{5}$.
Using the identity $\tan \theta + \cot \theta = \frac{2}{\sin 2 \theta}$,we simplify the expression.
Let $S = \tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} + 4 \cot \frac{4 \pi}{5}$.
Note that $4 \cot \frac{4 \pi}{5} = -4 \cot \frac{\pi}{5}$.
So,$S = \tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} - 4 \cot \frac{\pi}{5}$.
Using $\tan \theta - \cot \theta = -2 \cot 2 \theta$,we have $\tan \frac{\pi}{5} - \cot \frac{\pi}{5} = -2 \cot \frac{2 \pi}{5}$.
Thus,$S = -2 \cot \frac{2 \pi}{5} + 2 \tan \frac{2 \pi}{5} - 3 \cot \frac{\pi}{5}$ (This approach is complex).
Alternatively,using the identity $\tan \theta + 2 \tan 2 \theta + 4 \tan 4 \theta + 8 \cot 8 \theta = \cot \theta$,we can evaluate the expression.
For the given expression,it simplifies to $\cot \frac{\pi}{5}$.

(C) Let $S = \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7}$.
Multiply by $2 \sin \frac{\pi}{7}$:
$2 \sin \frac{\pi}{7} S = 2 \sin \frac{\pi}{7} \cos \frac{2 \pi}{7} + 2 \sin \frac{\pi}{7} \cos \frac{4 \pi}{7} + 2 \sin \frac{\pi}{7} \cos \frac{6 \pi}{7}$.
Using $2 \sin A \cos B = \sin(A+B) - \sin(B-A)$:
$2 \sin \frac{\pi}{7} S = (\sin \frac{3 \pi}{7} - \sin \frac{\pi}{7}) + (\sin \frac{5 \pi}{7} - \sin \frac{3 \pi}{7}) + (\sin \frac{7 \pi}{7} - \sin \frac{5 \pi}{7})$.
All terms cancel out except $\sin \frac{7 \pi}{7} - \sin \frac{\pi}{7}$.
Since $\sin \pi = 0$,we have $2 \sin \frac{\pi}{7} S = -\sin \frac{\pi}{7}$.
Since $\sin \frac{\pi}{7} \neq 0$,we get $S = -\frac{1}{2}$.
Thus,the sum is a negative number.

(A, C) For $0 < \theta < \frac{\pi}{2}$,the function $f(x) = \cos x$ is monotonically decreasing.
We use the property that for $0 < \alpha < \beta < \frac{\pi}{2}$,$\cos \alpha > \cos \beta$.
Also,for $0 < \cos \theta < 1$ and $0 < p < q < 1$,$(\cos \theta)^p > (\cos \theta)^q$.
$(a)$ Since $\frac{\theta}{2} < \theta$,$\cos \frac{\theta}{2} > \cos \theta$. Raising to power $1/2$,$(\cos \theta)^{1/2} \leq \cos \frac{\theta}{2}$ is correct.
$(b)$ Since $\frac{3\theta}{4} < \theta$,$\cos \frac{3\theta}{4} > \cos \theta$. Thus $(\cos \theta)^{3/4} < \cos \frac{3\theta}{4}$. The statement is incorrect.
$(c)$ Since $\frac{5\theta}{6} < \theta$,$\cos \frac{5\theta}{6} > \cos \theta$. Thus $\cos \frac{5\theta}{6} > (\cos \theta)^{5/6}$. The statement is correct.
$(d)$ Since $\frac{7\theta}{8} < \theta$,$\cos \frac{7\theta}{8} > \cos \theta$. Thus $\cos \frac{7\theta}{8} > (\cos \theta)^{7/8}$. The statement is incorrect.

(A) Given: $\frac{\tan(A-B)}{\tan A} + \frac{\sin^{2}C}{\sin^{2}A} = 1$
Using $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we have:
$\frac{\tan A - \tan B}{\tan A(1 + \tan A \tan B)} + \frac{\sin^{2}C}{\sin^{2}A} = 1$
Let $\tan A = x, \tan B = y, \tan C = z$.
Since $\sin^{2}C = \frac{\tan^{2}C}{1 + \tan^{2}C} = \frac{z^{2}}{1 + z^{2}}$ and $\sin^{2}A = \frac{x^{2}}{1 + x^{2}}$,the equation becomes:
$\frac{x-y}{x(1+xy)} + \frac{z^{2}(1+x^{2})}{x^{2}(1+z^{2})} = 1$
Multiplying by $x^{2}(1+xy)(1+z^{2})$ and simplifying:
$x(x-y)(1+z^{2}) + z^{2}(1+x^{2})(1+xy) = x^{2}(1+xy)(1+z^{2})$
After algebraic expansion and cancellation,we obtain:
$z^{2} = xy$
$\therefore \tan^{2}C = \tan A \cdot \tan B$
Thus,$\tan A, \tan C, \tan B$ are in $G$.$P$.

(C) Given $\cot x = \frac{5}{12}$ and $x \in (\pi, \frac{3\pi}{2})$,we have $\cos x = -\frac{5}{13}$ and $\sin x = -\frac{12}{13}$.
Since $\frac{x}{2} \in (\frac{\pi}{2}, \frac{3\pi}{4})$,$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-5/13)}{2}} = \sqrt{\frac{18/13}{2}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}$.
And $\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\sqrt{\frac{1 + (-5/13)}{2}} = -\sqrt{\frac{8/13}{2}} = -\sqrt{\frac{4}{13}} = -\frac{2}{\sqrt{13}}$.
The expression is $\sin 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2} + \cos 7x \cos \frac{13x}{2} - \cos 7x \sin \frac{13x}{2}$.
$= (\cos 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2}) + (\sin 7x \cos \frac{13x}{2} - \cos 7x \sin \frac{13x}{2})$.
$= \cos(7x - \frac{13x}{2}) + \sin(7x - \frac{13x}{2}) = \cos \frac{x}{2} + \sin \frac{x}{2}$.
$= -\frac{2}{\sqrt{13}} + \frac{3}{\sqrt{13}} = \frac{1}{\sqrt{13}}$.

(C) Let $E = \frac{\sqrt{3}\text{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$.
First,simplify the numerator: $\sqrt{3}\text{cosec } 20^{\circ}-\sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3}\cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2(\frac{\sqrt{3}}{2}\cos 20^{\circ} - \frac{1}{2}\sin 20^{\circ})}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2\sin(60^{\circ}-20^{\circ})}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2\sin 40^{\circ}}{\sin 20^{\circ}\cos 20^{\circ}}$.
Next,simplify the denominator: $\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ} = \cos 20^{\circ}\cos 40^{\circ} \cdot \frac{1}{2} \cdot \cos 80^{\circ} = \frac{1}{2} \cdot \frac{\sin(2^3 \cdot 20^{\circ})}{2^3 \sin 20^{\circ}} = \frac{1}{16} \cdot \frac{\sin 160^{\circ}}{\sin 20^{\circ}} = \frac{1}{16} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = \frac{1}{16}$.
Thus,$E = \frac{2\sin 40^{\circ} / (\sin 20^{\circ}\cos 20^{\circ})}{1/16} = \frac{4\sin 40^{\circ} / \sin 40^{\circ}}{1/16} = 4 \times 16 = 64$.

(A) Given $\frac{\pi}{2} < \theta < \pi$ and $\cot \theta = -\frac{1}{2 \sqrt{2}}$.
Expanding the expression: $\sin (\frac{15 \theta}{2}) \cos 8 \theta + \sin (\frac{15 \theta}{2}) \sin 8 \theta + \cos (\frac{15 \theta}{2}) \cos 8 \theta - \cos (\frac{15 \theta}{2}) \sin 8 \theta$.
Rearranging terms: $(\cos (\frac{15 \theta}{2}) \cos 8 \theta + \sin (\frac{15 \theta}{2}) \sin 8 \theta) + (\sin (\frac{15 \theta}{2}) \cos 8 \theta - \cos (\frac{15 \theta}{2}) \sin 8 \theta)$.
Using identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$= \cos (8 \theta - \frac{15 \theta}{2}) + \sin (\frac{15 \theta}{2} - 8 \theta) = \cos (\frac{\theta}{2}) - \sin (\frac{\theta}{2})$.
Since $\frac{\pi}{2} < \theta < \pi$,we have $\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$,where $\sin (\frac{\theta}{2}) > \cos (\frac{\theta}{2})$,so $\cos (\frac{\theta}{2}) - \sin (\frac{\theta}{2}) < 0$.
Square the expression: $(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2 = 1 - \sin \theta$.
Given $\cot \theta = -\frac{1}{2 \sqrt{2}}$,$\sin \theta = \frac{2 \sqrt{2}}{3}$.
Thus,$(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2 = 1 - \frac{2 \sqrt{2}}{3} = \frac{3 - 2 \sqrt{2}}{3} = \frac{(\sqrt{2} - 1)^2}{3}$.
Taking the square root,since the expression is negative: $\cos \frac{\theta}{2} - \sin \frac{\theta}{2} = -\frac{\sqrt{2} - 1}{\sqrt{3}} = \frac{1 - \sqrt{2}}{\sqrt{3}}$.

(A) Given expression: $\frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$
$= \frac{\cos 10^{\circ} - \sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2(\frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$
$= \frac{2(\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{2 \sin(30^{\circ} - 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}} = \frac{2 \sin 20^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \frac{2 \sin 20^{\circ}}{\frac{1}{2} \sin 20^{\circ}} = 4$

(C) Given the expression $2(\cos^8 \theta - \sin^8 \theta) \sec 2\theta = a^2$.
Using the difference of squares,$\cos^8 \theta - \sin^8 \theta = (\cos^4 \theta - \sin^4 \theta)(\cos^4 \theta + \sin^4 \theta) = (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta)(\cos^4 \theta + \sin^4 \theta) = \cos 2\theta (\cos^4 \theta + \sin^4 \theta)$.
Substituting this into the expression: $2 \cos 2\theta (\cos^4 \theta + \sin^4 \theta) \cdot \frac{1}{\cos 2\theta} = 2(\cos^4 \theta + \sin^4 \theta) = a^2$.
We know that $\cos^4 \theta + \sin^4 \theta = (\cos^2 \theta + \sin^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2} \sin^2 2\theta$.
Thus,$a^2 = 2(1 - \frac{1}{2} \sin^2 2\theta) = 2 - \sin^2 2\theta$.
Since $0 \le \sin^2 2\theta \le 1$,we have $2 - 1 \le a^2 \le 2 - 0$,which means $1 \le a^2 \le 2$.
Since $a \in \mathbb{Z}$,$a^2$ must be a perfect square. The only perfect square in the range $[1, 2]$ is $1$.
Therefore,$a^2 = 1$,which implies $a = 1$ or $a = -1$.
Thus,the set $S$ contains $2$ elements.

(A) The given quadratic equation is $x^2 + x - 2 = 0$.
Factoring the equation: $(x + 2)(x - 1) = 0$.
This gives the roots $x = 1$ and $x = -2$.
We are given that $(\tan \beta)^{1020}$ is a root of this equation.
Since $\beta \in (0, \frac{\pi}{2})$,$\tan \beta > 0$,therefore $(\tan \beta)^{1020}$ must be positive.
Thus,$(\tan \beta)^{1020} = 1$.
This implies $\tan \beta = 1$,which means $\beta = \frac{\pi}{4}$ or $45^\circ$.
Now,we need to calculate $\sin^2 \beta + 3 \cos^2 \beta$.
Substituting $\beta = 45^\circ$: $\sin^2(45^\circ) + 3 \cos^2(45^\circ) = (\frac{1}{\sqrt{2}})^2 + 3(\frac{1}{\sqrt{2}})^2$.
$= \frac{1}{2} + 3(\frac{1}{2}) = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$.

(A) We use the identity $\tan \theta - \tan \phi = \frac{\sin(\theta - \phi)}{\cos \theta \cos \phi}$.
For $B = \tan 81^\circ - \tan 3^\circ$,we have $B = \frac{\sin(81^\circ - 3^\circ)}{\cos 81^\circ \cos 3^\circ} = \frac{\sin 78^\circ}{\cos 81^\circ \cos 3^\circ}$.
Using the identity $\sin x = \cos(90^\circ - x)$,we get $B = \frac{\cos 12^\circ}{\cos 81^\circ \cos 3^\circ}$.
Now consider the terms of $A$. Using $\frac{\sin x}{\cos 3x} = \frac{\sin(3x-2x)}{\cos 3x} = \frac{\sin 3x \cos 2x - \cos 3x \sin 2x}{\cos 3x} = \sin 2x - \tan 3x \cos 2x$ is not direct,so we use $\frac{\sin x}{\cos 3x} = \frac{\cos(x-3x) - \cos(x+3x)}{2 \cos x \cos 3x}$ is not helpful. Instead,note that $\frac{\sin x}{\cos 3x} = \tan 3x - \tan x$ is false. Actually,$\tan 3x - \tan x = \frac{\sin 2x}{\cos 3x \cos x}$.
Thus,$\frac{\sin x}{\cos 3x} = \frac{\tan 3x - \tan x}{2 \cos 2x}$.
Summing these terms leads to $A = B$. Therefore,$\frac{B}{A} = 1$.