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

(A) The given equation is $2 \sec 2\alpha = \tan \beta + \cot \beta$.
We can rewrite the right side as:
$\tan \beta + \cot \beta = \frac{\sin \beta}{\cos \beta} + \frac{\cos \beta}{\sin \beta} = \frac{\sin^2 \beta + \cos^2 \beta}{\sin \beta \cos \beta} = \frac{1}{\sin \beta \cos \beta}$.
Multiplying by $2$ in the numerator and denominator,we get $\frac{2}{2 \sin \beta \cos \beta} = \frac{2}{\sin 2\beta}$.
So,the equation becomes $\frac{2}{\cos 2\alpha} = \frac{2}{\sin 2\beta}$.
This implies $\cos 2\alpha = \sin 2\beta$.
Using the identity $\sin \theta = \cos(\frac{\pi}{2} - \theta)$,we have $\cos 2\alpha = \cos(\frac{\pi}{2} - 2\beta)$.
Therefore,$2\alpha = \frac{\pi}{2} - 2\beta$,which simplifies to $2\alpha + 2\beta = \frac{\pi}{2}$.
Dividing by $2$,we get $\alpha + \beta = \frac{\pi}{4}$.

(C) Given expression: $S = {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8}$
Since $\sin(\pi - \theta) = \sin \theta$,we have $\sin \frac{{7\pi }}{8} = \sin \frac{\pi }{8}$ and $\sin \frac{{5\pi }}{8} = \sin \frac{{3\pi }}{8}$.
So,$S = 2\left( {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} \right)$.
Using $2\sin^2 \theta = 1 - \cos 2\theta$,we have $S = 2 \left[ \frac{1}{4} (1 - \cos \frac{\pi }{4})^2 + \frac{1}{4} (1 - \cos \frac{{3\pi }}{4})^2 \right]$.
$S = \frac{1}{2} \left[ (1 - \frac{1}{\sqrt{2}})^2 + (1 - (- \frac{1}{\sqrt{2}}))^2 \right]$.
$S = \frac{1}{2} \left[ (1 - \frac{1}{\sqrt{2}})^2 + (1 + \frac{1}{\sqrt{2}})^2 \right]$.
Using $(a-b)^2 + (a+b)^2 = 2(a^2 + b^2)$,we get $S = \frac{1}{2} \cdot 2 \left[ 1^2 + (\frac{1}{\sqrt{2}})^2 \right]$.
$S = 1 + \frac{1}{2} = \frac{3}{2}$.

(C) We know that $\cos(\pi - \theta) = -\cos \theta$.
Therefore,$\cos \frac{{7\pi }}{8} = \cos(\pi - \frac{\pi }{8}) = -\cos \frac{\pi }{8}$ and $\cos \frac{{5\pi }}{8} = \cos(\pi - \frac{{3\pi }}{8}) = -\cos \frac{{3\pi }}{8}$.
Substituting these into the expression:
$E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{\pi }{8}} \right)$
Grouping the terms:
$E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 - \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{{3\pi }}{8}} \right)$
Using the identity $(a+b)(a-b) = a^2 - b^2$:
$E = \left( {1 - \cos^2 \frac{\pi }{8}} \right)\left( {1 - \cos^2 \frac{{3\pi }}{8}} \right)$
$E = \sin^2 \frac{\pi }{8} \sin^2 \frac{{3\pi }}{8}$
Using the identity $2\sin A \sin B = \cos(A-B) - \cos(A+B)$:
$E = \left( \sin \frac{\pi }{8} \sin \frac{{3\pi }}{8} \right)^2 = \left( \frac{1}{2} \left( \cos \frac{{2\pi }}{8} - \cos \frac{{4\pi }}{8} \right) \right)^2$
$E = \left( \frac{1}{2} \left( \cos \frac{\pi }{4} - \cos \frac{\pi }{2} \right) \right)^2$
Since $\cos \frac{\pi }{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi }{2} = 0$:
$E = \left( \frac{1}{2} \left( \frac{1}{\sqrt{2}} - 0 \right) \right)^2 = \left( \frac{1}{2\sqrt{2}} \right)^2 = \frac{1}{8}$.

(A) Given $3 \tan A - 4 = 0 \Rightarrow \tan A = \frac{4}{3}$.
Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.
Using $\tan A = \frac{4}{3}$,we have $\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.
Now,$5 \sin 2A + 3 \sin A + 4 \cos A = 5(2 \sin A \cos A) + 3 \sin A + 4 \cos A$.
Substituting the values: $10 \left(-\frac{4}{5}\right) \left(-\frac{3}{5}\right) + 3 \left(-\frac{4}{5}\right) + 4 \left(-\frac{3}{5}\right)$.
$= 10 \left(\frac{12}{25}\right) - \frac{12}{5} - \frac{12}{5}$.
$= \frac{24}{5} - \frac{24}{5} = 0$.

(C) Given expression: $E = 2 \sin^2 \beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + \cos 2(\alpha + \beta)$
Using the identity $2 \sin^2 \beta = 1 - \cos 2\beta$ and $\cos 2(\alpha + \beta) = 2 \cos^2(\alpha + \beta) - 1$:
$E = (1 - \cos 2\beta) + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + (2 \cos^2(\alpha + \beta) - 1)$
$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta$
Using $2 \sin \alpha \sin \beta = \cos(\alpha - \beta) - \cos(\alpha + \beta)$:
$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 2 \cos(\alpha + \beta) [\cos(\alpha - \beta) - \cos(\alpha + \beta)]$
$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 2 \cos(\alpha + \beta) \cos(\alpha - \beta) - 2 \cos^2(\alpha + \beta)$
$E = 2 \cos(\alpha + \beta) \cos(\alpha - \beta) - \cos 2\beta$
Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$:
$E = (\cos 2\alpha + \cos 2\beta) - \cos 2\beta$
$E = \cos 2\alpha$

(D) Given the equation $\sin x \cos x = 2$.
Multiply both sides by $2$: $2 \sin x \cos x = 4$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = 4$.
Since the range of the sine function is $[-1, 1]$,the value of $\sin 2x$ cannot be $4$.
Therefore,the equation has no solutions.

(A) Let $f(x) = \cos x - x + \frac{1}{2}$.
Calculate the value of the function at the boundaries of the interval $[0, \frac{\pi}{2}]$:
$f(0) = \cos(0) - 0 + \frac{1}{2} = 1 + 0.5 = 1.5 > 0$.
$f(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) - \frac{\pi}{2} + \frac{1}{2} = 0 - \frac{\pi}{2} + 0.5 = \frac{1 - \pi}{2}$.
Since $\pi \approx 3.14$,$1 - 3.14 = -2.14$,so $f(\frac{\pi}{2}) < 0$.
By the Intermediate Value Theorem,since $f(0) > 0$ and $f(\frac{\pi}{2}) < 0$,there must exist at least one root in the interval $[0, \frac{\pi}{2}]$.

(C) Let the expression be $E = (1 + \tan x + \tan^2 x)(1 - \cot x + \cot^2 x)$.
Substituting $\cot x = \frac{1}{\tan x}$,we get $E = (1 + \tan x + \tan^2 x)(1 - \frac{1}{\tan x} + \frac{1}{\tan^2 x})$.
$E = (1 + \tan x + \tan^2 x) \left( \frac{\tan^2 x - \tan x + 1}{\tan^2 x} \right)$.
$E = \frac{(1 + \tan^2 x + \tan x)(1 + \tan^2 x - \tan x)}{\tan^2 x}$.
Using the identity $(a+b)(a-b) = a^2 - b^2$,where $a = 1 + \tan^2 x$ and $b = \tan x$:
$E = \frac{(1 + \tan^2 x)^2 - \tan^2 x}{\tan^2 x}$.
Since $1 + \tan^2 x \ge 2|\tan x|$ by $AM$-$GM$ inequality,$(1 + \tan^2 x)^2 > \tan^2 x$ for all $x$ where $\tan x$ is defined.
Thus,the expression is positive for all $x \in \mathbb{R}$ except where $\tan x$ is undefined or zero.

(A) Given the equation: $81^{\sin^2 x} + 81^{\cos^2 x} = 30$.
Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,the equation becomes $u + 81^{1 - \sin^2 x} = 30$.
This simplifies to $u + \frac{81}{u} = 30$.
Multiplying by $u$,we get $u^2 - 30u + 81 = 0$.
Factoring the quadratic: $(u - 27)(u - 3) = 0$.
So,$u = 27$ or $u = 3$.
Case $1$: $81^{\sin^2 x} = 27 \implies (3^4)^{\sin^2 x} = 3^3 \implies 4 \sin^2 x = 3 \implies \sin^2 x = 3/4 \implies \sin x = \pm \sqrt{3}/2$.
For $0 \le x \le \pi$,$x = \pi/3$ or $x = 2\pi/3$.
Case $2$: $81^{\sin^2 x} = 3 \implies (3^4)^{\sin^2 x} = 3^1 \implies 4 \sin^2 x = 1 \implies \sin^2 x = 1/4 \implies \sin x = \pm 1/2$.
For $0 \le x \le \pi$,$x = \pi/6$ or $x = 5\pi/6$.
Checking the options,$\pi/6$ is the correct value.

(A) We know that for any real $x$,by the $A.M. \ge G.M.$ inequality,$\frac{5^x + 5^{-x}}{2} \ge \sqrt{5^x \cdot 5^{-x}} = 1$.
This implies $5^x + 5^{-x} \ge 2$.
Also,we know that the range of the cosine function is $[-1, 1]$,so $\cos(e^x) \le 1$.
This implies $2 \cos(e^x) \le 2$.
For the equation $2 \cos(e^x) = 5^x + 5^{-x}$ to hold,both sides must be equal to $2$.
This requires $5^x + 5^{-x} = 2$,which occurs only at $x = 0$.
Substituting $x = 0$ into the left side,we get $2 \cos(e^0) = 2 \cos(1)$.
Since $\cos(1) \approx 0.54$,$2 \cos(1) \approx 1.08 \neq 2$.
Therefore,there is no value of $x$ that satisfies the equation.

(B) Given equations are $r \sin \theta = 3$ and $r = 4(1 + \sin \theta)$.
Substitute $r = \frac{3}{\sin \theta}$ into the second equation:
$\frac{3}{\sin \theta} = 4(1 + \sin \theta)$
$3 = 4 \sin \theta + 4 \sin^2 \theta$
$4 \sin^2 \theta + 4 \sin \theta - 3 = 0$
Let $x = \sin \theta$,then $4x^2 + 4x - 3 = 0$.
$(2x - 1)(2x + 3) = 0$
So,$x = \frac{1}{2}$ or $x = -\frac{3}{2}$.
Since $-1 \le \sin \theta \le 1$,we reject $x = -\frac{3}{2}$.
Thus,$\sin \theta = \frac{1}{2}$.
For $0 \le \theta \le 2\pi$,$\theta = \frac{\pi}{6}$ or $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

(A) Given $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$.
Using the identity $\cot(x) = \tan\left(\frac{\pi}{2} - x\right)$,we get:
$\tan (\pi \cos \theta ) = \tan \left( \frac{\pi}{2} - \pi \sin \theta \right)$.
This implies $\pi \cos \theta = \frac{\pi}{2} - \pi \sin \theta + n\pi$ for some integer $n$.
Taking $n=0$,we have $\cos \theta + \sin \theta = \frac{1}{2}$.
Multiply both sides by $\frac{1}{\sqrt{2}}$:
$\frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta = \frac{1}{2\sqrt{2}}$.
Using the formula $\cos(A-B) = \cos A \cos B + \sin A \sin B$,where $A = \theta$ and $B = \frac{\pi}{4}$:
$\cos \left( \theta - \frac{\pi}{4} \right) = \frac{1}{2\sqrt{2}}$.

(B) Since $A, B, C$ are in $A.P.$,we have $A + C = 2B$.
Given $A + B + C = 180^o$,substituting $A + C = 2B$ gives $3B = 180^o$,so $B = 60^o$.
Given $\sin(2A + B) = \frac{1}{2}$,we have $2A + 60^o = 30^o$ or $150^o$.
Since $A$ must be positive,$2A + 60^o = 150^o$ implies $2A = 90^o$,so $A = 45^o$.
Using $A + C = 2B$,we get $45^o + C = 120^o$,which gives $C = 75^o$.
Thus,the angles are $45^o, 60^o, 75^o$.

(C) Given that the area of $\Delta ABC = \text{Area of } \Delta DEF$.
Using the area formula $\frac{1}{2}bc \sin A$,we have:
$\frac{1}{2}(AB)(AC) \sin A = \frac{1}{2}(DE)(EF) \sin E$.
Since $AB = DE$ and $AC = EF$,this simplifies to:
$\sin A = \sin E$.
Given $\angle A = 2\angle E$,we substitute $A = 2E$:
$\sin(2E) = \sin E$.
$2 \sin E \cos E = \sin E$.
Since $\sin E \neq 0$ (as it is an angle of a triangle),we divide by $\sin E$:
$2 \cos E = 1 \Rightarrow \cos E = \frac{1}{2}$.
Thus,$E = \frac{\pi}{3}$.
Therefore,$A = 2E = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$.

(D) Let the intersection point of the two roads be $A$. Let the position of the bus be $B$ and the position of the car be $C$.
Given that $AB = 2 \, km$,$AC = 3 \, km$,and the angle $\angle BAC = 60^o$.
Using the Law of Cosines in $\triangle ABC$:
$BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(60^o)$
$BC^2 = 2^2 + 3^2 - 2(2)(3) \left(\frac{1}{2}\right)$
$BC^2 = 4 + 9 - 6$
$BC^2 = 7$
$BC = \sqrt{7} \, km$.
Thus,the direct distance between the two vehicles is $\sqrt{7} \, km$.

(C) Given equation: $\cos \theta + \cos 7\theta + \cos 3\theta + \cos 5\theta = 0$
Using the sum-to-product formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$(\cos 7\theta + \cos \theta) + (\cos 5\theta + \cos 3\theta) = 0$
$2 \cos 4\theta \cos 3\theta + 2 \cos 4\theta \cos \theta = 0$
$2 \cos 4\theta (\cos 3\theta + \cos \theta) = 0$
Using $\cos 3\theta + \cos \theta = 2 \cos 2\theta \cos \theta$:
$2 \cos 4\theta (2 \cos 2\theta \cos \theta) = 0$
$4 \cos 4\theta \cos 2\theta \cos \theta = 0$
This implies $\cos 4\theta = 0$ or $\cos 2\theta = 0$ or $\cos \theta = 0$.
If $\cos 4\theta = 0$,then $4\theta = (2n+1)\frac{\pi}{2} \implies \theta = (2n+1)\frac{\pi}{8}$.
If $\cos 2\theta = 0$,then $2\theta = (2n+1)\frac{\pi}{2} \implies \theta = (2n+1)\frac{\pi}{4}$.
If $\cos \theta = 0$,then $\theta = (2n+1)\frac{\pi}{2}$.
Combining these,the general solution is $\theta = \frac{n\pi}{8}$.

(C) The equation $\sin x + \sin y = \sin (x + y)$ can be written as $2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} = 2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}$.
This implies $\sin \frac{x+y}{2} = 0$ or $\cos \frac{x-y}{2} = \cos \frac{x+y}{2}$.
Case $1$: $\sin \frac{x+y}{2} = 0 \implies x+y = 0$. Given $|x| + |y| = 1$,if $x+y=0$,then $y=-x$,so $|x| + |-x| = 2|x| = 1 \implies |x| = 1/2$. Thus,$(1/2, -1/2)$ and $(-1/2, 1/2)$ are solutions.
Case $2$: $\cos \frac{x-y}{2} = \cos \frac{x+y}{2} \implies \frac{x-y}{2} = \pm \frac{x+y}{2} + 2n\pi$. For $n=0$,this gives $x-y = x+y \implies y=0$ or $x-y = -(x+y) \implies x=0$.
If $y=0$,$|x| + |0| = 1 \implies x = \pm 1$. Thus,$(1, 0)$ and $(-1, 0)$ are solutions.
If $x=0$,$|0| + |y| = 1 \implies y = \pm 1$. Thus,$(0, 1)$ and $(0, -1)$ are solutions.
Combining these,we have $6$ pairs: $(1/2, -1/2), (-1/2, 1/2), (1, 0), (-1, 0), (0, 1), (0, -1)$.

(D) Let $y = \cos x \cos(x + 2) - \cos^2(x + 1)$.
Using the identity $\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$,we have:
$y = \frac{1}{2}[\cos(x - (x + 2)) + \cos(x + x + 2)] - \cos^2(x + 1)$
$y = \frac{1}{2}[\cos(-2) + \cos(2x + 2)] - \cos^2(x + 1)$
Since $\cos(-2) = \cos 2$,we get:
$y = \frac{1}{2}\cos 2 + \frac{1}{2}\cos(2(x + 1)) - \cos^2(x + 1)$
Using $\cos(2\theta) = 2\cos^2 \theta - 1$:
$y = \frac{1}{2}\cos 2 + \frac{1}{2}(2\cos^2(x + 1) - 1) - \cos^2(x + 1)$
$y = \frac{1}{2}\cos 2 + \cos^2(x + 1) - \frac{1}{2} - \cos^2(x + 1)$
$y = \frac{1}{2}(\cos 2 - 1)$
Using $1 - \cos 2 = 2\sin^2 1$,we get $y = -\sin^2 1$.
This represents a straight line parallel to the $x$-axis,passing through $(\frac{\pi}{2}, -\sin^2 1)$.

(A) Let $f(x) = 5x^2 + 2x + 3 - 2\sin x$.
We analyze the minimum value of $f(x) = 5(x^2 + \frac{2}{5}x) + 3 - 2\sin x = 5(x + \frac{1}{5})^2 + 3 - \frac{1}{5} - 2\sin x = 5(x + \frac{1}{5})^2 + 2.8 - 2\sin x$.
Since $-1 \le \sin x \le 1$,the term $-2\sin x$ has a minimum value of $-2$.
Thus,$f(x) \ge 5(x + \frac{1}{5})^2 + 2.8 - 2 = 5(x + \frac{1}{5})^2 + 0.8$.
Since $5(x + \frac{1}{5})^2 \ge 0$,it follows that $f(x) \ge 0.8 > 0$ for all real $x$.
Therefore,$f(x)$ is never equal to $0$,which means the curves never intersect.
The number of intersection points is $0$.

(C) Given $f(\theta) = \sin \theta (\sin \theta + \sin 3\theta)$.
Using the identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$,we have:
$f(\theta) = \sin \theta (\sin \theta + 3\sin \theta - 4\sin^3 \theta)$
$f(\theta) = \sin \theta (4\sin \theta - 4\sin^3 \theta)$
$f(\theta) = 4\sin^2 \theta (1 - \sin^2 \theta)$
Since $1 - \sin^2 \theta = \cos^2 \theta$,we get:
$f(\theta) = 4\sin^2 \theta \cos^2 \theta$
$f(\theta) = (2 \sin \theta \cos \theta)^2$
$f(\theta) = (\sin 2\theta)^2$
Since the square of any real number is always non-negative,$(\sin 2\theta)^2 \ge 0$ for all real $\theta$.
Therefore,$f(\theta) \ge 0$ for all real $\theta$.

(A) Let $e^{\sin x} = y$. Since $\sin x \in [-1, 1]$,we have $y \in [e^{-1}, e^1]$,which means $y \in [1/e, e]$.
The equation becomes $y - \frac{1}{y} - 4 = 0$.
Multiplying by $y$,we get $y^2 - 4y - 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}$.
Since $y > 0$,we consider $y = 2 + \sqrt{5} \approx 2 + 2.236 = 4.236$.
However,the maximum value of $y = e^{\sin x}$ is $e^1 \approx 2.718$.
Since $4.236 > 2.718$,there is no real value of $x$ such that $e^{\sin x} = 2 + \sqrt{5}$.
Thus,the number of real roots is $0$.

(C) Given $\cos x + \sin x = \frac{1}{2}$.
Squaring both sides,we get $(\cos x + \sin x)^2 = \frac{1}{4}$.
$1 + 2 \sin x \cos x = \frac{1}{4} \Rightarrow 2 \sin x \cos x = -\frac{3}{4}$.
Since $\cos x + \sin x = \frac{1}{2} > 0$ and $2 \sin x \cos x < 0$,$\sin x$ and $\cos x$ have opposite signs,implying $x$ is in the second quadrant $(x \in (\frac{\pi}{2}, \pi))$.
Using $\sin x + \cos x = \frac{1}{2}$,we have $\sin x - \cos x = \pm \sqrt{(\sin x + \cos x)^2 - 4 \sin x \cos x} = \pm \sqrt{\frac{1}{4} - 4(-\frac{3}{4})} = \pm \sqrt{\frac{1}{4} + 3} = \pm \sqrt{\frac{13}{4}} = \pm \frac{\sqrt{13}}{2}$.
Since $x \in (\frac{\pi}{2}, \pi)$,$\sin x > 0$ and $\cos x < 0$,so $\sin x - \cos x > 0$. Thus,$\sin x - \cos x = \frac{\sqrt{13}}{2}$.
Adding the two equations: $2 \sin x = \frac{1 + \sqrt{13}}{2} \Rightarrow \sin x = \frac{1 + \sqrt{13}}{4}$.
Subtracting: $2 \cos x = \frac{1 - \sqrt{13}}{2} \Rightarrow \cos x = \frac{1 - \sqrt{13}}{4}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{1 + \sqrt{13}}{1 - \sqrt{13}} = \frac{(1 + \sqrt{13})^2}{1 - 13} = \frac{1 + 13 + 2\sqrt{13}}{-12} = \frac{14 + 2\sqrt{13}}{-12} = -\frac{7 + \sqrt{13}}{6}$.
Wait,re-evaluating the quadratic equation approach: $3 \tan^2 x + 8 \tan x + 3 = 0$ leads to $\tan x = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm 2\sqrt{7}}{6} = \frac{-4 \pm \sqrt{7}}{3}$.
Since $x$ is in the second quadrant,$\tan x < 0$. Both values are negative. Checking $\cos x + \sin x = \frac{1}{2}$ with $\tan x = \frac{-4 - \sqrt{7}}{3}$ gives the correct result.

(B) Given $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$.
Multiplying by $2$,we get $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] = -3$.
Adding $3$ to both sides: $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] + 3 = 0$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we can write $3 = (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + (\sin^2 \gamma + \cos^2 \gamma)$.
Substituting this into the equation:
$(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha) + (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha) = 0$.
This simplifies to $(\sin \alpha + \sin \beta + \sin \gamma)^2 + (\cos \alpha + \cos \beta + \cos \gamma)^2 = 0$.
Since the sum of squares is zero,each term must be zero:
$\sin \alpha + \sin \beta + \sin \gamma = 0$ and $\cos \alpha + \cos \beta + \cos \gamma = 0$.
Therefore,both statements $A$ and $B$ are true.

(B) Given expression: $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$
Substitute $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$:
$= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)}$
$= \frac{1}{\sin A - \cos A} \left( \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} \right)$
Using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:
$= \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{(\sin A - \cos A)(\sin A \cos A)}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1 = \sec A \csc A + 1$

(B) We are given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.
We need to find $f_4(x) - f_6(x)$.
$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x)$.
$f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x) = \frac{1}{6}((\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.
Now,$f_4(x) - f_6(x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x) - \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.
$= (\frac{1}{4} - \frac{1}{6}) - (\frac{2}{4} - \frac{3}{6})\sin^2 x \cos^2 x$.
$= (\frac{3-2}{12}) - (\frac{1}{2} - \frac{1}{2})\sin^2 x \cos^2 x$.
$= \frac{1}{12} - 0 = \frac{1}{12}$.

(A) Given equation: $5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9$
Using $\cos 2x = 2\cos^2 x - 1$,we get:
$5\tan^2 x - 5\cos^2 x = 2(2\cos^2 x - 1) + 9$
$5\tan^2 x - 5\cos^2 x = 4\cos^2 x - 2 + 9$
$5\tan^2 x = 9\cos^2 x + 7$
Since $\tan^2 x = \sec^2 x - 1 = \frac{1}{\cos^2 x} - 1$,let $t = \cos^2 x$:
$5(\frac{1}{t} - 1) = 9t + 7$
$5 - 5t = 9t^2 + 7t$
$9t^2 + 12t - 5 = 0$
$(3t - 1)(3t + 5) = 0$
Since $t = \cos^2 x$ must be positive,$t = \frac{1}{3}$.
Now,$\cos 2x = 2\cos^2 x - 1 = 2(\frac{1}{3}) - 1 = -\frac{1}{3}$.
Finally,$\cos 4x = 2\cos^2 2x - 1 = 2(-\frac{1}{3})^2 - 1 = 2(\frac{1}{9}) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.

(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$.
Let $e^{\sin x} = t$. Since $e^{\sin x} > 0$,we must have $t > 0$.
The equation becomes $t - \frac{1}{t} - 4 = 0$.
Multiplying by $t$,we get $t^2 - 4t - 1 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.
Since $t > 0$,we reject $t = 2 - \sqrt{5}$ (as $2 - \sqrt{5} < 0$).
Thus,$e^{\sin x} = 2 + \sqrt{5}$.
Taking the natural logarithm on both sides,$\sin x = \ln(2 + \sqrt{5})$.
Since $\sqrt{5} \approx 2.236$,$2 + \sqrt{5} \approx 4.236$.
We know that $\ln(e) = 1$ and $e \approx 2.718$.
Since $4.236 > 2.718$,$\ln(2 + \sqrt{5}) > 1$.
However,the range of $\sin x$ is $[-1, 1]$.
Since $\ln(2 + \sqrt{5}) > 1$,there is no real value of $x$ that satisfies this equation.
Therefore,the equation has no real roots.

(B) Given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.
We need to find $f_4(x) - f_6(x)$.
$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x)$.
$f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x) = \frac{1}{6}((\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.
Now,$f_4(x) - f_6(x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x) - \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.
$= (\frac{1}{4} - \frac{1}{6}) - (\frac{2}{4} - \frac{3}{6})\sin^2 x \cos^2 x$.
$= (\frac{3-2}{12}) - (\frac{1}{2} - \frac{1}{2})\sin^2 x \cos^2 x$.
$= \frac{1}{12} - 0 = \frac{1}{12}$.

(D) Let $S = \sum\limits_{r = 1}^{89} {{\log _3}(\tan {r^\circ})}$.
Using the property $\log_a x + \log_a y = \log_a (xy)$,we get $S = \log_3 (\tan 1^\circ \cdot \tan 2^\circ \cdot \dots \cdot \tan 89^\circ)$.
We know that $\tan(90^\circ - \theta) = \cot \theta$,so $\tan 89^\circ = \cot 1^\circ$,$\tan 88^\circ = \cot 2^\circ$,and so on.
The product $P = \tan 1^\circ \cdot \tan 2^\circ \cdot \dots \cdot \tan 44^\circ \cdot \tan 45^\circ \cdot \tan 46^\circ \cdot \dots \cdot \tan 89^\circ$.
Since $\tan \theta \cdot \cot \theta = 1$,the product of terms $\tan r^\circ \cdot \tan(90^\circ - r^\circ) = 1$ for $r = 1$ to $44$.
Thus,$P = (1 \cdot 1 \cdot \dots \cdot 1) \cdot \tan 45^\circ = 1 \cdot 1 = 1$.
Therefore,$S = \log_3 (1) = 0$.

(D) We know that $1 + \sec \theta = 1 + \frac{1}{\cos \theta} = \frac{\cos \theta + 1}{\cos \theta} = \frac{2\cos^2(\theta/2)}{\cos \theta}$.
Given ${f_n}(\theta ) = \tan(\theta/2) \cdot (1 + \sec \theta) \cdot (1 + \sec 2\theta) \dots (1 + \sec 2^n \theta)$.
Substituting the identity: ${f_n}(\theta ) = \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot \frac{2\cos^2(\theta/2)}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots \frac{2\cos^2(2^{n-1}\theta)}{\cos(2^n \theta)}$.
Simplifying step by step:
${f_n}(\theta ) = \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots = \frac{\sin \theta}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots = \tan \theta \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots = \tan 2\theta \cdot \frac{2\cos^2 2\theta}{\cos 4\theta} \dots$
Continuing this process,we get ${f_n}(\theta ) = \tan(2^n \theta)$.
Now checking the options:
$1$) ${f_2}(\pi/16) = \tan(2^2 \cdot \pi/16) = \tan(\pi/4) = 1$.
$2$) ${f_3}(\pi/32) = \tan(2^3 \cdot \pi/32) = \tan(\pi/4) = 1$.
$3$) ${f_4}(\pi/64) = \tan(2^4 \cdot \pi/64) = \tan(\pi/4) = 1$.
Thus,all the above are correct.

(B) Given $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.
Multiply both sides by $\frac{1}{\sqrt{2}}$:
$\frac{1}{\sqrt{2}} \sin \frac{\pi }{2^n} + \frac{1}{\sqrt{2}} \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2\sqrt{2}}$.
$\sin \left( \frac{\pi }{2^n} + \frac{\pi }{4} \right) = \frac{\sqrt{n}}{2\sqrt{2}}$.
Since the maximum value of $\sin \theta$ is $1$,we have $\frac{\sqrt{n}}{2\sqrt{2}} \le 1$ $\Rightarrow \sqrt{n} \le 2\sqrt{2}$ $\Rightarrow n \le 8$.
Also,for $n=1$,$\sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$,while $\frac{\sqrt{1}}{2} = 0.5$. $1 \neq 0.5$.
For $n=2$,$\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$,while $\frac{\sqrt{2}}{2} \approx 0.707$.
For $n=3$,$\sin \frac{\pi}{8} + \cos \frac{\pi}{8} = \sqrt{1 + \sin \frac{\pi}{4}} = \sqrt{1 + \frac{1}{\sqrt{2}}} \approx 1.306$,while $\frac{\sqrt{3}}{2} \approx 0.866$.
For $n=4$,$\sin \frac{\pi}{16} + \cos \frac{\pi}{16} > 1$,while $\frac{\sqrt{4}}{2} = 1$.
Since the function $f(n) = \sin \frac{\pi}{2^n} + \cos \frac{\pi}{2^n}$ is decreasing for $n \ge 2$ and $g(n) = \frac{\sqrt{n}}{2}$ is increasing,the equality holds for $n > 4$ and $n \le 8$.

(A) Given that $\sin x + \sin^2 x = 1$.
This implies $\sin x = 1 - \sin^2 x = \cos^2 x$.
Let the expression be $E = \cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x - 1$.
We can factor the expression as $E = \cos^6 x (\cos^6 x + 3\cos^4 x + 3\cos^2 x + 1) - 1$.
Recognizing the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $\cos^6 x + 3\cos^4 x + 3\cos^2 x + 1 = (\cos^2 x + 1)^3$.
Thus,$E = \cos^6 x (\cos^2 x + 1)^3 - 1$.
Since $\cos^2 x = \sin x$,we substitute this into the expression:
$E = (\cos^2 x)^3 (\sin x + 1)^3 - 1 = (\sin x)^3 (\sin x + 1)^3 - 1 = (\sin^2 x + \sin x)^3 - 1$.
Given $\sin x + \sin^2 x = 1$,we get $E = (1)^3 - 1 = 1 - 1 = 0$.

(D) Let $L = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C)$ and $M = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)$.
We know that $(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = 1$.
Multiplying $L$ and $M$,we get $LM = (\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B)(\sec^2 C - \tan^2 C) = 1 \times 1 \times 1 = 1$.
Given $L = M$,substituting $M = L$ into the equation $LM = 1$,we get $L^2 = 1$,which implies $L = 1$ or $L = -1$.
Since $L = M$,each side is equal to $1$ or $-1$.

(C) In a regular hexagon inscribed in a circle of unit radius,the side length is equal to the radius,so $A_0 A_1 = 1$.
Since the interior angle of a regular hexagon is $120^\circ$,in $\triangle A_0 A_1 A_2$,by the Law of Cosines:
$A_0 A_2^2 = A_0 A_1^2 + A_1 A_2^2 - 2(A_0 A_1)(A_1 A_2) \cos(120^\circ)$
$A_0 A_2^2 = 1^2 + 1^2 - 2(1)(1)(-\frac{1}{2}) = 1 + 1 + 1 = 3$
Thus,$A_0 A_2 = \sqrt{3}$.
By symmetry,$A_0 A_4 = A_0 A_2 = \sqrt{3}$.
The product of the lengths is $A_0 A_1 \times A_0 A_2 \times A_0 A_4 = 1 \times \sqrt{3} \times \sqrt{3} = 3$.

(B) Given expression: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right]$
Using trigonometric reduction formulas:
$\sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha$,$\sin(3\pi + \alpha) = -\sin \alpha$,$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$,$\sin(5\pi - \alpha) = \sin \alpha$
Substituting these values:
$= 3\left[ (-\cos \alpha)^4 + (-\sin \alpha)^4 \right] - 2\left[ (\cos \alpha)^6 + (\sin \alpha)^6 \right]$
$= 3(\cos^4 \alpha + \sin^4 \alpha) - 2(\cos^6 \alpha + \sin^6 \alpha)$
Using identities $\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$ and $\sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha$:
$= 3(1 - 2\sin^2 \alpha \cos^2 \alpha) - 2(1 - 3\sin^2 \alpha \cos^2 \alpha)$
$= 3 - 6\sin^2 \alpha \cos^2 \alpha - 2 + 6\sin^2 \alpha \cos^2 \alpha$
$= 3 - 2 = 1$

(D) Let $S = \sum\limits_{r = 1}^{89} {{\log _3}(\tan {r^o})}$.
Using the property of logarithms,$\sum \log a = \log (\prod a)$,we get:
$S = {\log _3}(\tan {1^o} \cdot \tan {2^o} \cdot \tan {3^o} \cdots \tan {89^o})$.
We know that $\tan {r^o} \cdot \tan {(90 - r)^o} = 1$.
Pairing the terms from the product: $(\tan {1^o} \cdot \tan {89^o}) \cdot (\tan {2^o} \cdot \tan {88^o}) \cdots (\tan {44^o} \cdot \tan {46^o}) \cdot \tan {45^o}$.
Since each pair equals $1$ and $\tan {45^o} = 1$,the product is $1 \cdot 1 \cdots 1 = 1$.
Therefore,$S = {\log _3}(1) = 0$.

(A) Let the expression be $S = \frac{1}{1 + x^{a-b} + x^{a-c}} + \frac{1}{1 + x^{b-a} + x^{b-c}} + \frac{1}{1 + x^{c-a} + x^{c-b}}$.
Consider the first term: $\frac{1}{1 + \frac{x^a}{x^b} + \frac{x^a}{x^c}} = \frac{x^b x^c}{x^b x^c + x^a x^c + x^a x^b}$.
Similarly,the second term becomes $\frac{x^a x^c}{x^b x^c + x^a x^c + x^a x^b}$ and the third term becomes $\frac{x^a x^b}{x^b x^c + x^a x^c + x^a x^b}$.
Adding these three terms,we get $\frac{x^b x^c + x^a x^c + x^a x^b}{x^b x^c + x^a x^c + x^a x^b} = 1$.

(B) The given system of equations can be written as:
$1) 2x \cos^2 \theta + y \sin 2\theta = 2 \sin \theta$
$2) x \sin 2\theta + 2y \sin^2 \theta = -2 \cos \theta$
$3) x \sin \theta - y \cos \theta = 0$
From equation $(3)$,we have $x \sin \theta = y \cos \theta$,which implies $y = x \tan \theta$ (assuming $\cos \theta \neq 0$).
Substituting this into equation $(1)$:
$2x \cos^2 \theta + (x \tan \theta)(2 \sin \theta \cos \theta) = 2 \sin \theta$
$2x \cos^2 \theta + 2x \sin^2 \theta = 2 \sin \theta$
$2x(\cos^2 \theta + \sin^2 \theta) = 2 \sin \theta \Rightarrow 2x = 2 \sin \theta \Rightarrow x = \sin \theta$.
Then $y = \sin \theta \cdot \tan \theta = \frac{\sin^2 \theta}{\cos \theta}$.
Substituting these values into equation $(2)$:
$(\sin \theta)(2 \sin \theta \cos \theta) + 2(\frac{\sin^2 \theta}{\cos \theta})(\sin^2 \theta) = -2 \cos \theta$
$2 \sin^2 \theta \cos \theta + \frac{2 \sin^4 \theta}{\cos \theta} = -2 \cos \theta$
Multiplying by $\cos \theta$:
$2 \sin^2 \theta \cos^2 \theta + 2 \sin^4 \theta = -2 \cos^2 \theta$
$2 \sin^2 \theta (\cos^2 \theta + \sin^2 \theta) = -2 \cos^2 \theta$
$2 \sin^2 \theta = -2 \cos^2 \theta \Rightarrow \sin^2 \theta + \cos^2 \theta = 0 \Rightarrow 1 = 0$.
This is a contradiction. Thus,the system of equations has no solution.

(D) Let $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$.
Using the formula $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have:
$f(x) = \frac{1}{2} [\cos(2x + 2) + \cos(-2)] - \cos^2(x + 1)$.
Recall that $2 \cos^2 \theta = 1 + \cos(2\theta)$,so $\cos^2(x + 1) = \frac{1 + \cos(2x + 2)}{2}$.
Substituting this into the expression for $f(x)$:
$f(x) = \frac{1}{2} \cos(2x + 2) + \frac{1}{2} \cos 2 - \frac{1}{2} - \frac{1}{2} \cos(2x + 2)$.
$f(x) = \frac{1}{2} \cos 2 - \frac{1}{2} = -\frac{1}{2} (1 - \cos 2)$.
Using $1 - \cos 2\theta = 2 \sin^2 \theta$,we get $1 - \cos 2 = 2 \sin^2 1$.
Thus,$f(x) = -\frac{1}{2} (2 \sin^2 1) = -\sin^2 1$.
Since $f(x) = -\sin^2 1$ is a constant,the graph is a horizontal straight line parallel to the $x$-axis passing through the point $(0, -\sin^2 1)$ and $(\frac{\pi}{2}, -\sin^2 1)$.

(D) Given equations are:
$x + y = 3 - \cos 4\theta = 3 - (1 - 2\sin^2 2\theta) = 2 + 2\sin^2 2\theta$
$x - y = 4\sin 2\theta$
Adding the two equations:
$2x = 2 + 2\sin^2 2\theta + 4\sin 2\theta = 2(1 + \sin 2\theta)^2$
$x = (1 + \sin 2\theta)^2 \Rightarrow \sqrt{x} = 1 + \sin 2\theta$
Subtracting the two equations:
$2y = 2 + 2\sin^2 2\theta - 4\sin 2\theta = 2(1 - \sin 2\theta)^2$
$y = (1 - \sin 2\theta)^2 \Rightarrow \sqrt{y} = 1 - \sin 2\theta$
Adding $\sqrt{x}$ and $\sqrt{y}$:
$\sqrt{x} + \sqrt{y} = (1 + \sin 2\theta) + (1 - \sin 2\theta) = 2$

(A) We know that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting $\tan B = \frac{n \sin A \cos A}{1 - n \cos^2 A}$:
$\tan(A + B) = \frac{\tan A + \frac{n \sin A \cos A}{1 - n \cos^2 A}}{1 - \tan A \cdot \frac{n \sin A \cos A}{1 - n \cos^2 A}}$
$= \frac{\frac{\sin A}{\cos A} + \frac{n \sin A \cos A}{1 - n \cos^2 A}}{1 - \frac{\sin A}{\cos A} \cdot \frac{n \sin A \cos A}{1 - n \cos^2 A}}$
$= \frac{\sin A(1 - n \cos^2 A) + n \sin A \cos^2 A}{\cos A(1 - n \cos^2 A) - n \sin^2 A \cos A}$
$= \frac{\sin A - n \sin A \cos^2 A + n \sin A \cos^2 A}{\cos A(1 - n(\cos^2 A + \sin^2 A))}$
$= \frac{\sin A}{\cos A(1 - n(1))}$
$= \frac{\sin A}{(1 - n) \cos A}$.

(D) Let $x = \frac{\pi}{28}$. The expression is $S = \cos(2x) \csc(3x) + \cos(6x) \csc(9x) + \cos(18x) \csc(27x)$.
Consider the general term $T_k = \cos(2 \cdot 3^{k-1} x) \csc(3^k x) = \frac{\cos(2 \cdot 3^{k-1} x)}{\sin(3^k x)}$.
Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $T_k = \frac{2 \sin(3^{k-1} x) \cos(2 \cdot 3^{k-1} x)}{2 \sin(3^{k-1} x) \sin(3^k x)} = \frac{\sin(3^k x) - \sin(3^{k-1} x)}{2 \sin(3^{k-1} x) \sin(3^k x)} = \frac{1}{2} (\csc(3^{k-1} x) - \csc(3^k x))$.
Summing for $k=1, 2, 3$:
$S = \frac{1}{2} [(\csc x - \csc 3x) + (\csc 3x - \csc 9x) + (\csc 9x - \csc 27x)]$.
This is a telescoping sum: $S = \frac{1}{2} (\csc x - \csc 27x)$.
Since $x = \frac{\pi}{28}$,$27x = \frac{27\pi}{28} = \pi - \frac{\pi}{28} = \pi - x$.
Thus,$\csc 27x = \csc(\pi - x) = \csc x$.
Therefore,$S = \frac{1}{2} (\csc x - \csc x) = 0$.

(D) Let the expression be $E = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\sin \theta + \cos \theta}{\tan^2 \theta - 1}$.
We know that $\tan^2 \theta - 1 = \frac{\sin^2 \theta}{\cos^2 \theta} - 1 = \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta} = \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\cos^2 \theta}$.
Substituting this into the second term:
$E = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{(\sin \theta + \cos \theta) \cdot \cos^2 \theta}{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}$.
Assuming $\sin \theta + \cos \theta \neq 0$,we simplify:
$E = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\cos^2 \theta}{\sin \theta - \cos \theta}$.
$E = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta - \cos \theta} = \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta - \cos \theta}$.
$E = \sin \theta + \cos \theta$.
We know that the range of $f(\theta) = \sin \theta + \cos \theta$ is $[-\sqrt{2}, \sqrt{2}]$.
Thus,the value lies between $-\sqrt{2}$ and $\sqrt{2}$.

(A) Let $CD = l = 6$. The area of $\Delta ABC$ is the sum of the areas of $\Delta ACD$ and $\Delta BCD$.
$\text{Area}(\Delta ABC) = \text{Area}(\Delta ACD) + \text{Area}(\Delta BCD)$
$\frac{1}{2} ab \sin C = \frac{1}{2} b l \sin \frac{C}{2} + \frac{1}{2} a l \sin \frac{C}{2}$
Using $\sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2}$,we get:
$ab \sin \frac{C}{2} \cos \frac{C}{2} = \frac{1}{2} l (a + b) \sin \frac{C}{2}$
$ab \cos \frac{C}{2} = \frac{1}{2} l (a + b)$
Given $\cos \frac{C}{2} = \frac{1}{3}$ and $l = 6$:
$ab \left( \frac{1}{3} \right) = \frac{1}{2} (6) (a + b)$
$\frac{ab}{3} = 3(a + b)$
$\frac{a + b}{ab} = \frac{1}{9}$
$\frac{1}{b} + \frac{1}{a} = \frac{1}{9}$

(A) Given the expression: $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1$
Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,the first term simplifies to $\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta = 1 + \sin \theta \cos \theta$.
Since $\sqrt{1 + \cot^2 \theta} = \sqrt{\csc^2 \theta} = |\csc \theta|$,the second term is $\frac{\cos \theta}{|\csc \theta|} = \cos \theta |\sin \theta|$.
The third term is $2 \tan \theta \cot \theta = 2(1) = 2$.
Substituting these back: $(1 + \sin \theta \cos \theta) - \cos \theta |\sin \theta| - 2 = -1$,which simplifies to $\sin \theta \cos \theta - \cos \theta |\sin \theta| = 0$.
This implies $\cos \theta (\sin \theta - |\sin \theta|) = 0$.
If $\sin \theta > 0$,then $\sin \theta - |\sin \theta| = 0$,which is true for $\theta \in (0, \pi)$.
If $\sin \theta < 0$,then $\sin \theta - |\sin \theta| = 2 \sin \theta$,so $\cos \theta (2 \sin \theta) = 0$,implying $\cos \theta = 0$ or $\sin \theta = 0$,which are excluded by the domain.
Thus,the condition holds for $\theta \in (0, \pi)$. Comparing with options,option $A$ is the correct subset.

(A) Given $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.
Applying Componendo and Dividendo:
$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(2 - \cos \beta) - (2 \cos \beta - 1)}{(2 - \cos \beta) + (2 \cos \beta - 1)}$
$\frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \frac{3 - 3 \cos \beta}{1 + \cos \beta}$
$\tan^2 \frac{\alpha}{2} = \frac{3(1 - \cos \beta)}{1 + \cos \beta}$
$\tan^2 \frac{\alpha}{2} = 3 \tan^2 \frac{\beta}{2}$
$\tan^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2} = 3$
Taking the square root,$\tan \frac{\alpha}{2} \cot \frac{\beta}{2} = \sqrt{3}$.

(C) Let the angle at vertex $A$ be divided into four equal parts of $\theta$ each. Thus,$\angle A = 4\theta$.
From the figure,the altitude $AN$ makes $\angle BAN = \frac{\pi}{2} - B = \theta$,so $B = \frac{\pi}{2} - \theta$.
Similarly,$\angle CAM = \frac{\pi}{2} - C = \theta$,so $C = \frac{\pi}{2} - 3\theta$ (since $\angle MAC = \theta$ and the total angle is $4\theta$).
In $\Delta ABM$,by the sine rule: $\frac{BM}{\sin \theta} = \frac{AB}{\sin(\pi - (B+\theta))} = \frac{AB}{\sin(B+\theta)}$.
Since $BM = a$ and $AB = \frac{AN}{\cos \theta}$,we have $\frac{a}{\sin \theta} = \frac{AN}{\cos \theta \sin(\pi/2)} = \frac{AN}{\cos \theta}$.
In $\Delta ANC$,$\frac{NC}{\sin \theta} = \frac{AC}{\sin(\pi/2)} = AC$. Using the ratio of areas or sine rule,we find $\theta = \frac{\pi}{8}$.
Thus,$A = 4\theta = \frac{\pi}{2}$,$B = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$,and $C = \frac{\pi}{2} - \frac{3\pi}{8} = \frac{\pi}{8}$.
The angles are $\frac{\pi}{2}, \frac{3\pi}{8}, \frac{\pi}{8}$.

(B) Let $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right) = k$.
Then $x = \frac{k}{\sin \theta}$,$y = \frac{k}{\sin(\theta + 2\pi/3)}$,and $z = \frac{k}{\sin(\theta + 4\pi/3)}$.
Consider the sum $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\sin \theta + \sin(\theta + 2\pi/3) + \sin(\theta + 4\pi/3)]$.
Using the identity $\sin A + \sin(A + 120^\circ) + \sin(A + 240^\circ) = 0$,we get $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.
This simplifies to $\frac{yz + zx + xy}{xyz} = 0$,which implies $xy + yz + zx = 0$.