Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities Questions in English

(A) Let $f(A) = \cos A \sin \left( A - \frac{\pi}{6} \right)$.
Using the identity $2 \sin x \cos y = \sin(x+y) + \sin(x-y)$,we have:
$f(A) = \frac{1}{2} \left[ \sin \left( A + A - \frac{\pi}{6} \right) + \sin \left( A - (A - \frac{\pi}{6}) \right) \right]$
$f(A) = \frac{1}{2} \left[ \sin \left( 2A - \frac{\pi}{6} \right) + \sin \left( \frac{\pi}{6} \right) \right]$
$f(A) = \frac{1}{2} \sin \left( 2A - \frac{\pi}{6} \right) + \frac{1}{4}$.
For $f(A)$ to be maximum,$\sin \left( 2A - \frac{\pi}{6} \right)$ must be maximum,which is $1$.
So,$2A - \frac{\pi}{6} = \frac{\pi}{2}$.
$2A = \frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.
$A = \frac{\pi}{3}$.

(A) By the $A.M.-G.M.$ inequality,we have $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$.
This simplifies to $2^{\sin x} + 2^{\cos x} \ge 2 \cdot 2^{\frac{\sin x + \cos x}{2}} = 2^{1 + \frac{\sin x + \cos x}{2}}$.
We know that $\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$,which has a minimum value of $-\sqrt{2}$.
Thus,the minimum value of the expression $2^{\sin x} + 2^{\cos x}$ is $2^{1 - \frac{\sqrt{2}}{2}} = 2^{1 - (1/\sqrt{2})}$.
The inequality $2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}$ holds for all $x$ except where $\sin x + \cos x = -\sqrt{2}$.
$\sin x + \cos x = -\sqrt{2}$ occurs when $\sin(x + \frac{\pi}{4}) = -1$,which implies $x + \frac{\pi}{4} = \frac{3\pi}{2}$,so $x = \frac{5\pi}{4}$.

(C) Given $A + B + C = 180^{\circ}$,we have $A + B = 180^{\circ} - C$.
Taking tangent on both sides: $\tan(A + B) = \tan(180^{\circ} - C)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(180^{\circ} - C) = -\tan C$,we get:
$\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$.
Multiplying both sides by $(1 - \tan A \tan B)$,we get:
$\tan A + \tan B = -\tan C(1 - \tan A \tan B)$.
$\tan A + \tan B = -\tan C + \tan A \tan B \tan C$.
Rearranging the terms,we get:
$\tan A + \tan B + \tan C = \tan A \tan B \tan C$.

(D) Given the function $f(x) = a \cos(bx + c) + d$.
We know that the range of the cosine function is $[-1, 1]$,i.e.,$-1 \le \cos(bx + c) \le 1$.
Multiplying by $a$ (assuming $a > 0$),we get $-a \le a \cos(bx + c) \le a$.
Adding $d$ to all parts,we get $d - a \le a \cos(bx + c) + d \le d + a$.
Thus,the range of $f(x)$ is $[d - a, d + a]$.

(D) We know that any function of the form $f(x) = a \cos x + b \sin x$ can be written as $f(x) = \sqrt{a^2 + b^2} \cos(x + \alpha)$,where $\tan \alpha = \frac{b}{a}$.
Here,$a = 1$ and $b = -1$.
Thus,$f(x) = \sqrt{1^2 + (-1)^2} \cos(x + \alpha) = \sqrt{2} \cos(x + \alpha)$.
Since the range of $\cos(x + \alpha)$ is $[-1, 1]$,the range of $f(x)$ is $[-\sqrt{2}, \sqrt{2}]$.

(B) Given $f(x) = \cos 2x - \sin 2x$.
We can write this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 2x - \frac{1}{\sqrt{2}} \sin 2x \right)$.
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we have $f(x) = \sqrt{2} \cos\left( 2x + \frac{\pi}{4} \right)$.
The range of $\cos\theta$ is $[-1, 1]$,so the range of $f(x)$ is $[-\sqrt{2}, \sqrt{2}]$.
Since $\sqrt{2} \approx 1.414$,the interval $[-\sqrt{2}, \sqrt{2}]$ is approximately $[-1.414, 1.414]$.
Thus,the set $[-1, 1]$ is a subset of $[-\sqrt{2}, \sqrt{2}]$.
Therefore,the correct option is $B$.

(D) Given $f(\theta) = \sec^2 \theta + \cos^2 \theta$.
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that for positive real numbers $a$ and $b$,$\frac{a+b}{2} \ge \sqrt{ab}$.
Here,$f(\theta) = \sec^2 \theta + \cos^2 \theta \ge 2 \sqrt{\sec^2 \theta \cdot \cos^2 \theta} = 2 \sqrt{1} = 2$.
Since $\theta > \frac{\pi}{3}$,$\sec \theta > \sec(\frac{\pi}{3}) = 2$,so $\sec^2 \theta > 4$.
As $\theta$ increases,$\sec^2 \theta$ increases towards $\infty$ and $\cos^2 \theta$ remains in the interval $[0, 1)$.
Thus,the minimum value is $2$ and the function can take any value up to $\infty$.
Therefore,the interval is $[2, \infty)$.

(B) Given the function $f(x) = |\sin 4x + 3|$.
We know that for any real $x$,the range of $\sin 4x$ is $[-1, 1]$.
To find the minimum value of $f(x)$,we substitute the minimum value of $\sin 4x$,which is $-1$:
$f_{\text{min}} = |-1 + 3| = |2| = 2$.
To find the maximum value of $f(x)$,we substitute the maximum value of $\sin 4x$,which is $1$:
$f_{\text{max}} = |1 + 3| = |4| = 4$.
Thus,the maximum and minimum values are $4$ and $2$ respectively.

(D) Let $f(x) = 2\cos 2x - \cos 4x$.
Using the identity $\cos 4x = 2\cos^2 2x - 1$,we can rewrite the function as:
$f(x) = 2\cos 2x - (2\cos^2 2x - 1) = -2\cos^2 2x + 2\cos 2x + 1$.
Let $t = \cos 2x$. Since $0 \le x \le \pi$,the range of $2x$ is $0 \le 2x \le 2\pi$,so the range of $t = \cos 2x$ is $[-1, 1]$.
Now,we have a quadratic function $g(t) = -2t^2 + 2t + 1$ for $t \in [-1, 1]$.
To find the minimum value,we check the endpoints of the interval $[-1, 1]$ because the parabola opens downwards.
For $t = 1$: $g(1) = -2(1)^2 + 2(1) + 1 = -2 + 2 + 1 = 1$.
For $t = -1$: $g(-1) = -2(-1)^2 + 2(-1) + 1 = -2 - 2 + 1 = -3$.
Comparing these values,the minimum value is $-3$.

(A) Let $f(A) = \cos A \cos B = \cos A \cos \left( \frac{\pi}{2} - A \right) = \cos A \sin A = \frac{1}{2} \sin 2A.$
Since the maximum value of $\sin \theta$ is $1,$ the maximum value of $\frac{1}{2} \sin 2A$ occurs when $\sin 2A = 1.$
Thus,the maximum value is $\frac{1}{2} \times 1 = \frac{1}{2}.$
Alternatively,using derivatives: $f'(A) = \cos 2A.$
Setting $f'(A) = 0$ gives $2A = \frac{\pi}{2},$ so $A = \frac{\pi}{4}.$
Since $f''(A) = -2 \sin 2A,$ at $A = \frac{\pi}{4},$ $f''(\frac{\pi}{4}) = -2 \sin \frac{\pi}{2} = -2 < 0,$ confirming a maximum.
The maximum value is $\cos \frac{\pi}{4} \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2}.$

(C) Let $f(x) = 2 + \sqrt{3} \cos x + \sin x$.
To find the maximum value of $e^{f(x)}$,we need to find the maximum value of $f(x)$.
We know that the expression $a \cos x + b \sin x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = \sqrt{3}$ and $b = 1$.
So,$\sqrt{a^2 + b^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Thus,the maximum value of $\sqrt{3} \cos x + \sin x$ is $2$.
Therefore,the maximum value of $f(x) = 2 + (\sqrt{3} \cos x + \sin x)$ is $2 + 2 = 4$.
Hence,the maximum value of $e^{f(x)}$ is $e^{4}$.

(B) We are given the expression $P = \prod_{i=1}^{n} (1 + a_i + a_i^2)$ where $\prod_{i=1}^{n} a_i = 1$ and $a_i > 0$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for each term $(1 + a_i + a_i^2)$,we have:
$\frac{1 + a_i + a_i^2}{3} \ge \sqrt[3]{1 \cdot a_i \cdot a_i^2} = \sqrt[3]{a_i^3} = a_i$.
Therefore,$1 + a_i + a_i^2 \ge 3a_i$.
Multiplying these inequalities for $i = 1$ to $n$,we get:
$P = \prod_{i=1}^{n} (1 + a_i + a_i^2) \ge \prod_{i=1}^{n} (3a_i) = 3^n \prod_{i=1}^{n} a_i$.
Since $\prod_{i=1}^{n} a_i = 1$,we have $P \ge 3^n \cdot 1 = 3^n$.
The equality holds when $1 = a_i = a_i^2$,which implies $a_i = 1$ for all $i$.
Thus,the minimum value is $3^n$.

(D) Given $p^2 + q^2 = 1$ where $p, q > 0$.
Using the inequality between Arithmetic Mean and Quadratic Mean (or Cauchy-Schwarz inequality):
$\frac{p + q}{2} \leq \sqrt{\frac{p^2 + q^2}{2}}$
Substituting the given value:
$\frac{p + q}{2} \leq \sqrt{\frac{1}{2}}$
$\frac{p + q}{2} \leq \frac{1}{\sqrt{2}}$
$p + q \leq \frac{2}{\sqrt{2}}$
$p + q \leq \sqrt{2}$
Thus,the maximum value of $p + q$ is $\sqrt{2}$.

(C) Let $y = a \sec x + b \csc x$.
To find the minimum value,we differentiate with respect to $x$:
$\frac{dy}{dx} = a \sec x \tan x - b \csc x \cot x$.
Setting $\frac{dy}{dx} = 0$ for critical points:
$a \sec x \tan x = b \csc x \cot x$.
Converting to sine and cosine:
$\frac{a \sin x}{\cos^2 x} = \frac{b \cos x}{\sin^2 x} \implies a \sin^3 x = b \cos^3 x \implies \tan^3 x = \frac{b}{a}$.
Thus,$\tan x = (b/a)^{1/3}$.
Using the identity $\sec^2 x = 1 + \tan^2 x$ and $\csc^2 x = 1 + \cot^2 x$:
$\sec x = \sqrt{1 + (b/a)^{2/3}} = \frac{\sqrt{a^{2/3} + b^{2/3}}}{a^{1/3}}$ and $\csc x = \sqrt{1 + (a/b)^{2/3}} = \frac{\sqrt{a^{2/3} + b^{2/3}}}{b^{1/3}}$.
Substituting these into the expression for $y$:
$y = a \left( \frac{\sqrt{a^{2/3} + b^{2/3}}}{a^{1/3}} \right) + b \left( \frac{\sqrt{a^{2/3} + b^{2/3}}}{b^{1/3}} \right)$.
$y = a^{2/3} \sqrt{a^{2/3} + b^{2/3}} + b^{2/3} \sqrt{a^{2/3} + b^{2/3}}$.
$y = (a^{2/3} + b^{2/3}) \sqrt{a^{2/3} + b^{2/3}} = (a^{2/3} + b^{2/3})^{3/2}$.

(C) Let $f(x) = \sin x + \cos 2x$.
Using the identity $\cos 2x = 1 - 2\sin^2 x$,we get:
$f(x) = \sin x + 1 - 2\sin^2 x$.
Let $t = \sin x$,where $t \in [-1, 1]$.
Then $f(t) = -2t^2 + t + 1$.
This is a downward-opening parabola. The vertex occurs at $t = -b/(2a) = -1/(2 \times -2) = 1/4$.
Since $1/4 \in [-1, 1]$,the maximum value is $f(1/4) = -2(1/4)^2 + (1/4) + 1$.
$f(1/4) = -2(1/16) + 1/4 + 1 = -1/8 + 2/8 + 8/8 = 9/8$.

(A) Given $A + B = \frac{\pi}{2}$,we have $B = \frac{\pi}{2} - A$.
Let $f(A) = \cos A \cos B = \cos A \cos(\frac{\pi}{2} - A) = \cos A \sin A$.
We can rewrite this as $f(A) = \frac{1}{2} \sin(2A)$.
To find the maximum value,we know that the maximum value of $\sin(2A)$ is $1$.
Therefore,the maximum value of $f(A) = \frac{1}{2} \times 1 = \frac{1}{2}$.

(A) Given expression is $f(x) = 27^{\cos 2x} 81^{\sin 2x}$.
We can write this as $f(x) = (3^3)^{\cos 2x} (3^4)^{\sin 2x} = 3^{3 \cos 2x + 4 \sin 2x}$.
To find the minimum value of $f(x)$,we need to find the minimum value of the exponent $g(x) = 3 \cos 2x + 4 \sin 2x$.
The range of the function $a \cos \theta + b \sin \theta$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$,so the minimum value of $g(x) = -\sqrt{3^2 + 4^2} = -\sqrt{9 + 16} = -\sqrt{25} = -5$.
Therefore,the minimum value of $f(x) = 3^{-5} = \frac{1}{3^5} = \frac{1}{243}$.

(A) Given the function $f(x) = 3^{x+1} + 3^{-(x+1)}$.
We can rewrite this as $f(x) = 3^{x+1} + \frac{1}{3^{x+1}}$.
Since $3^{x+1} > 0$ for all real $x$,we can apply the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.
For any two positive numbers $a$ and $b$,$\frac{a+b}{2} \geq \sqrt{ab}$,which implies $a+b \geq 2\sqrt{ab}$.
Let $a = 3^{x+1}$ and $b = \frac{1}{3^{x+1}}$.
Then $f(x) = a + b \geq 2\sqrt{a \cdot b}$.
$f(x) \geq 2\sqrt{3^{x+1} \cdot \frac{1}{3^{x+1}}}$.
$f(x) \geq 2\sqrt{1}$.
$f(x) \geq 2$.
Thus,the minimum value of the function is $2$.

(C) Let $f(x) = a \sin x + b \cos x$. The maximum value of this expression is given by $\sqrt{a^2 + b^2}$.
Here,$a = 3$ and $b = 4$.
Maximum value $= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Alternatively,using derivatives: $f'(x) = 3 \cos x - 4 \sin x$.
Setting $f'(x) = 0$ gives $\tan x = 3/4$.
For $\tan x = 3/4$,we have $\sin x = 3/5$ and $\cos x = 4/5$ (in the first quadrant).
Substituting these values,$f(x) = 3(3/5) + 4(4/5) = 9/5 + 16/5 = 25/5 = 5$.

(D) Given $u = \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} + \sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}$.
Squaring both sides:
$u^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta) + 2\sqrt{(a^2 \cos^2 \theta + b^2 \sin^2 \theta)(a^2 \sin^2 \theta + b^2 \cos^2 \theta)}$.
$u^2 = a^2 + b^2 + 2\sqrt{a^4 \sin^2 \theta \cos^2 \theta + a^2 b^2 \cos^4 \theta + a^2 b^2 \sin^4 \theta + b^4 \sin^2 \theta \cos^2 \theta}$.
Using $\sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4}$ and $\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta = 1 - \frac{\sin^2 2\theta}{2}$:
$u^2 = a^2 + b^2 + 2\sqrt{a^2 b^2 (1 - \frac{\sin^2 2\theta}{2}) + (a^4 + b^4) \frac{\sin^2 2\theta}{4}}$.
$u^2 = a^2 + b^2 + \sqrt{4a^2 b^2 - 2a^2 b^2 \sin^2 2\theta + (a^4 + b^4) \sin^2 2\theta}$.
$u^2 = a^2 + b^2 + \sqrt{4a^2 b^2 + (a^2 - b^2)^2 \sin^2 2\theta}$.
For maximum value,$\sin^2 2\theta = 1$: $u^2_{\max} = a^2 + b^2 + \sqrt{4a^2 b^2 + (a^2 - b^2)^2} = a^2 + b^2 + \sqrt{(a^2 + b^2)^2} = 2(a^2 + b^2)$.
For minimum value,$\sin^2 2\theta = 0$: $u^2_{\min} = a^2 + b^2 + \sqrt{4a^2 b^2} = a^2 + b^2 + 2ab = (a + b)^2$.
The difference is $u^2_{\max} - u^2_{\min} = 2a^2 + 2b^2 - (a^2 + b^2 + 2ab) = a^2 + b^2 - 2ab = (a - b)^2$.

(C) The expression is of the form $a \sin x + b \cos x$,where $a = 5$ and $b = 12$.
The maximum value of the expression $a \sin x + b \cos x$ is given by the formula $\sqrt{a^2 + b^2}$.
Substituting the values,we get:
Maximum value $= \sqrt{5^2 + 12^2}$
$= \sqrt{25 + 144}$
$= \sqrt{169}$
$= 13$.
Therefore,the maximum value is $13$.

(C) The expression $2^{((x^2 - 3)^3 + 27)}$ attains its minimum value when the exponent $((x^2 - 3)^3 + 27)$ is minimized.
Let $f(x) = (x^2 - 3)^3 + 27$.
To find the minimum,we analyze the behavior of $(x^2 - 3)^3$.
The term $(x^2 - 3)^3$ can take any real value because the range of $x^2 - 3$ is $[-3, \infty)$,and the cube function is strictly increasing.
However,if we consider the domain of $x$ as all real numbers,$(x^2 - 3)^3$ can approach $-\infty$ as $x^2$ approaches $0$ (specifically,$(-3)^3 = -27$).
Thus,the minimum value of $(x^2 - 3)^3 + 27$ is $(-27) + 27 = 0$.
Therefore,the minimum value of the expression is $2^0 = 1$.

(A) Given the function $f(x) = 2(\cos 3x + \cos \sqrt{3} x)$.
Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:
$f(x) = 2 \times 2 \cos(\frac{3 + \sqrt{3}}{2} x) \cos(\frac{3 - \sqrt{3}}{2} x) = 4 \cos(\frac{3 + \sqrt{3}}{2} x) \cos(\frac{3 - \sqrt{3}}{2} x)$.
The maximum value of $\cos \theta$ is $1$.
For $f(x)$ to be $4$,both $\cos(\frac{3 + \sqrt{3}}{2} x)$ and $\cos(\frac{3 - \sqrt{3}}{2} x)$ must be equal to $1$ or both must be equal to $-1$.
Case $1$: $\cos(\frac{3 + \sqrt{3}}{2} x) = 1$ and $\cos(\frac{3 - \sqrt{3}}{2} x) = 1$.
This implies $\frac{3 + \sqrt{3}}{2} x = 2n\pi$ and $\frac{3 - \sqrt{3}}{2} x = 2m\pi$ for integers $n, m$.
Dividing the two equations: $\frac{3 + \sqrt{3}}{3 - \sqrt{3}} = \frac{n}{m}$.
Rationalizing the denominator: $\frac{(3 + \sqrt{3})^2}{9 - 3} = \frac{9 + 3 + 6\sqrt{3}}{6} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}$.
Since $2 + \sqrt{3}$ is irrational,there are no non-zero integers $n, m$ that satisfy this. Thus,$n=0, m=0$ is the only solution,giving $x=0$.
Case $2$: Both are $-1$. This leads to a similar contradiction as the ratio of the arguments is irrational.
Therefore,the only value of $x$ for which the function attains its maximum value is $x=0$. The number of values is $1$.

(A) Let $y = 27^{\cos 2x} \cdot 81^{\sin 2x}$.
$y = (3^3)^{\cos 2x} \cdot (3^4)^{\sin 2x} = 3^{3 \cos 2x + 4 \sin 2x}$.
For $y$ to be minimum,the exponent $z = 3 \cos 2x + 4 \sin 2x$ must be minimum.
The expression $a \cos \theta + b \sin \theta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$,so the range of $z$ is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.
The minimum value of $z$ is $-5$.
Therefore,the minimum value of $y$ is $3^{-5} = \frac{1}{3^5} = \frac{1}{243}$.

(A) Given $p = \sin^2 \theta + \cos^4 \theta$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we have:
$p = \sin^2 \theta + (1 - \sin^2 \theta)^2$
$p = \sin^2 \theta + 1 + \sin^4 \theta - 2 \sin^2 \theta$
$p = \sin^4 \theta - \sin^2 \theta + 1$.
Let $x = \sin^2 \theta$. Since $\theta$ is real,$0 \le x \le 1$.
Then $p = f(x) = x^2 - x + 1$.
This is a parabola opening upwards with vertex at $x = -b/(2a) = -(-1)/(2 \times 1) = 1/2$.
Since $1/2$ lies in the interval $[0, 1]$,the minimum value is $f(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4$.
The maximum value occurs at the endpoints of the interval $[0, 1]$.
$f(0) = 0^2 - 0 + 1 = 1$.
$f(1) = 1^2 - 1 + 1 = 1$.
Thus,the range of $p$ is $\frac{3}{4} \le p \le 1$.

(C) Let $f(x) = \sin x + \sqrt{3} \cos x$.
To find the maximum value,we can rewrite the expression as $f(x) = 2 \left( \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x \right)$.
$f(x) = 2 \left( \sin x \cos 60^\circ + \cos x \sin 60^\circ \right) = 2 \sin(x + 60^\circ)$.
The function $\sin(x + 60^\circ)$ is maximum when the argument is $90^\circ$.
$x + 60^\circ = 90^\circ \Rightarrow x = 30^\circ$.
Alternatively,using derivatives:
$f'(x) = \cos x - \sqrt{3} \sin x = 0 \Rightarrow \tan x = \frac{1}{\sqrt{3}} \Rightarrow x = 30^\circ$.
At $x = 30^\circ$,$f''(x) = -\sin x - \sqrt{3} \cos x = -\sin 30^\circ - \sqrt{3} \cos 30^\circ = -\frac{1}{2} - \sqrt{3} \left( \frac{\sqrt{3}}{2} \right) = -\frac{1}{2} - \frac{3}{2} = -2 < 0$.
Since the second derivative is negative,the function is maximum at $x = 30^\circ$.

(A) Given $A + B = \pi / 2$,we have $B = \pi / 2 - A$.
Substituting this into the expression,we get $\cos A \cos B = \cos A \cos(\pi / 2 - A) = \cos A \sin A$.
Multiplying and dividing by $2$,we get $\frac{1}{2} (2 \sin A \cos A) = \frac{1}{2} \sin(2A)$.
Let $f(A) = \frac{1}{2} \sin(2A)$.
The maximum value of the sine function $\sin(2A)$ is $1$.
Therefore,the maximum value of $f(A) = \frac{1}{2} (1) = 1/2$.

(C) Given $A = \sin^2 x + \cos^4 x$.
Let $t = \sin^2 x$,then $0 \le t \le 1$. Since $\cos^2 x = 1 - \sin^2 x = 1 - t$,we have $\cos^4 x = (1 - t)^2$.
Substituting these into the expression for $A$:
$A = t + (1 - t)^2 = t + 1 - 2t + t^2 = t^2 - t + 1$.
This is a quadratic expression in $t$ where $t \in [0, 1]$.
The vertex of the parabola $f(t) = t^2 - t + 1$ is at $t = -(-1)/(2 \times 1) = 1/2$.
Since $1/2$ is within the interval $[0, 1]$,the minimum value is $f(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4$.
The maximum value occurs at the boundaries $t=0$ or $t=1$:
$f(0) = 0^2 - 0 + 1 = 1$.
$f(1) = 1^2 - 1 + 1 = 1$.
Thus,the range of $A$ is $\frac{3}{4} \le A \le 1$.

(B) Given the equation: $(\cos x + \sin x)^2 + k \sin x \cos x - 1 = 0, \forall x$
Expanding the square: $(\cos^2 x + \sin^2 x + 2 \sin x \cos x) + k \sin x \cos x - 1 = 0, \forall x$
Using the identity $\cos^2 x + \sin^2 x = 1$: $1 + 2 \sin x \cos x + k \sin x \cos x - 1 = 0, \forall x$
Simplifying the equation: $(2 + k) \sin x \cos x = 0, \forall x$
For this to be an identity (true for all $x$),the coefficient must be zero:
$2 + k = 0$
$k = -2$

(A) Given: $\frac{\sin^4 A}{a} + \frac{\cos^4 A}{b} = \frac{1}{a + b}$
Using $\sin^2 A = \frac{1 - \cos 2A}{2}$ and $\cos^2 A = \frac{1 + \cos 2A}{2}$,we have:
$\frac{(1 - \cos 2A)^2}{4a} + \frac{(1 + \cos 2A)^2}{4b} = \frac{1}{a + b}$
Multiplying by $4ab(a + b)$:
$b(a + b)(1 - 2\cos 2A + \cos^2 2A) + a(a + b)(1 + 2\cos 2A + \cos^2 2A) = 4ab$
$(a + b)^2 \cos^2 2A + 2(a + b)(a - b) \cos 2A + (a - b)^2 = 0$
This is a perfect square: $((a + b) \cos 2A + (a - b))^2 = 0$
So,$\cos 2A = \frac{b - a}{a + b}$
Now,$\sin^2 A = \frac{1 - \cos 2A}{2} = \frac{1 - \frac{b - a}{a + b}}{2} = \frac{a}{a + b}$ and $\cos^2 A = \frac{1 + \cos 2A}{2} = \frac{1 + \frac{b - a}{a + b}}{2} = \frac{b}{a + b}$
Therefore,$\frac{\sin^8 A}{a^3} + \frac{\cos^8 A}{b^3} = \frac{(\frac{a}{a + b})^4}{a^3} + \frac{(\frac{b}{a + b})^4}{b^3} = \frac{a^4}{a^3(a + b)^4} + \frac{b^4}{b^3(a + b)^4} = \frac{a + b}{(a + b)^4} = \frac{1}{(a + b)^3}$

(A) Let $y = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$.
Since $\alpha \in \left( 0, \frac{\pi}{2} \right)$,$\tan^2 \alpha > 0$. Also,for $x^2 + x > 0$,the expression is well-defined.
Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:
$\frac{\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}}{2} \ge \sqrt{\sqrt{x^2 + x} \cdot \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}}$
$\frac{\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}}{2} \ge \sqrt{\tan^2 \alpha}$
$\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} \ge 2 \tan \alpha$.

(A) Given $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$.
This implies $\frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_2}{\sin \alpha_2} \cdots \frac{\cos \alpha_n}{\sin \alpha_n} = 1$,so $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n = \sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n$.
Let $P = \cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$.
Then $P^2 = (\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n) \cdot (\sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n)$.
Using the identity $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$,we get $P^2 = \frac{1}{2^n} \sin 2\alpha_1 \cdot \sin 2\alpha_2 \cdots \sin 2\alpha_n$.
Since $\sin 2\alpha_i \le 1$ for all $i$,we have $P^2 \le \frac{1}{2^n}$.
Taking the square root,$P \le \sqrt{\frac{1}{2^n}} = \frac{1}{2^{n/2}}$.
The maximum value is $\frac{1}{2^{n/2}}$.

(B) Given $A = \sin^8\theta + \cos^{14}\theta$.
Since $0 \le \sin^2\theta \le 1$ and $0 \le \cos^2\theta \le 1$,we have $\sin^8\theta \le \sin^2\theta$ and $\cos^{14}\theta \le \cos^2\theta$.
Adding these inequalities,we get $A = \sin^8\theta + \cos^{14}\theta \le \sin^2\theta + \cos^2\theta = 1$.
Also,since $\sin^2\theta$ and $\cos^2\theta$ cannot be zero simultaneously,$A > 0$.
Thus,$0 < A \le 1$.

(A) Let $f(\theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta - \frac{1}{2}(a + c)$.
Using the identities $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$f(\theta) = a \left( \frac{1 - \cos 2\theta}{2} \right) + b \left( \frac{\sin 2\theta}{2} \right) + c \left( \frac{1 + \cos 2\theta}{2} \right) - \frac{1}{2}(a + c)$
$f(\theta) = \frac{1}{2} [a - a \cos 2\theta + b \sin 2\theta + c + c \cos 2\theta - a - c]$
$f(\theta) = \frac{1}{2} [b \sin 2\theta - (a - c) \cos 2\theta]$
We know that for any expression of the form $A \sin x + B \cos x$,the range is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.
Thus,$|b \sin 2\theta - (a - c) \cos 2\theta| \le \sqrt{b^2 + (a - c)^2}$.
Therefore,$|f(\theta)| = \frac{1}{2} |b \sin 2\theta - (a - c) \cos 2\theta| \le \frac{1}{2} \sqrt{b^2 + (a - c)^2}$.
Comparing this with the given inequality $\left| f(\theta) \right| \le \frac{1}{2}k$,we have $k = \sqrt{b^2 + (a - c)^2}$.
Squaring both sides,$k^2 = b^2 + (a - c)^2$.

(A) Given $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $.
Squaring both sides,we get:
${u^2} = ({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta ) + ({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta ) + 2\sqrt {({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta )({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta )} $
${u^2} = {a^2} + {b^2} + 2\sqrt {({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta )({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta )} $
Let $t = {a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta $. Then ${a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta = {a^2} + {b^2} - t$.
So,${u^2} = {a^2} + {b^2} + 2\sqrt {t({a^2} + {b^2} - t)} = {a^2} + {b^2} + 2\sqrt { - {t^2} + ({a^2} + {b^2})t} $.
Let $f(t) = - {t^2} + ({a^2} + {b^2})t$. The maximum value of $f(t)$ occurs at $t = \frac{{{a^2} + {b^2}}}{2}$,which is $f\left( \frac{{{a^2} + {b^2}}}{2} \right) = \frac{{{({a^2} + {b^2})^2}}}{4}$.
Thus,${({u^2})_{\max }} = {a^2} + {b^2} + 2\sqrt {\frac{{{({a^2} + {b^2})^2}}}{4}} = {a^2} + {b^2} + ({a^2} + {b^2}) = 2({a^2} + {b^2})$.
The minimum value of $f(t)$ occurs at the boundaries of $t$,i.e.,$t = {a^2}$ or $t = {b^2}$.
At $t = {a^2}$,$f({a^2}) = - {a^4} + ({a^2} + {b^2}){a^2} = {a^2}{b^2}$.
Thus,${({u^2})_{\min }} = {a^2} + {b^2} + 2\sqrt {{a^2}{b^2}} = {a^2} + {b^2} + 2ab = {(a + b)^2}$.
The difference is ${({u^2})_{\max }} - {({u^2})_{\min }} = 2{a^2} + 2{b^2} - ({a^2} + {b^2} + 2ab) = {a^2} + {b^2} - 2ab = {(a - b)^2}$.

(B) The given equation is $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 0$.
Adding $1$ to both sides,we get $a^2 + 2a + 1 + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 1$.
This simplifies to $(a + 1)^2 + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 1$.
Using the identity $1 + \cot^2 \theta = \csc^2 \theta$,we have $(a + 1)^2 + 1 + \cot^2 \left( \frac{\pi}{2}(a + x) \right) = 1$.
This reduces to $(a + 1)^2 + \cot^2 \left( \frac{\pi}{2}(a + x) \right) = 0$.
Since both terms are squares of real numbers,their sum can be zero only if each term is zero.
Thus,$(a + 1)^2 = 0 \Rightarrow a = -1$.
Substituting $a = -1$ into the second term,we get $\cot^2 \left( \frac{\pi}{2}(-1 + x) \right) = 0$.
This implies $\cot \left( \frac{\pi}{2}(x - 1) \right) = 0$,which means $\frac{\pi}{2}(x - 1) = n\pi$ for some integer $n$.
Thus,$x - 1 = 2n \Rightarrow x = 2n + 1$,which implies $\frac{x}{2}$ is not necessarily an integer,but checking the options,$a = -1$ is the key condition.

(D) We use the algebraic identity: $x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$.
Substituting $x = \sin A$,$y = \sin B$,and $z = \sin C$,we get:
$\sin^3 A + \sin^3 B + \sin^3 C - 3 \sin A \sin B \sin C = \frac{1}{2}(\sin A + \sin B + \sin C)[(\sin A - \sin B)^2 + (\sin B - \sin C)^2 + (\sin C - \sin A)^2]$.
Since the given equation is $\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$,the expression equals $0$.
This implies either $\sin A + \sin B + \sin C = 0$ (which is impossible for a triangle as $A, B, C > 0$ and $A+B+C = \pi$) or $(\sin A - \sin B)^2 + (\sin B - \sin C)^2 + (\sin C - \sin A)^2 = 0$.
This occurs only when $\sin A = \sin B = \sin C$,which implies $A = B = C = 60^\circ$.
Thus,$\Delta ABC$ is an equilateral triangle.

(A) Given $\sin \left( A + \frac{C}{2} \right) = k \sin \frac{C}{2}$.
Applying Componendo and Dividendo:
$\frac{\sin (A + C/2) + \sin (C/2)}{\sin (A + C/2) - \sin (C/2)} = \frac{k + 1}{k - 1}$.
Using the sum-to-product formulas $\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$ and $\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}$:
$\frac{2 \sin \frac{A+C}{2} \cos \frac{A}{2}}{2 \cos \frac{A+C}{2} \sin \frac{A}{2}} = \frac{k + 1}{k - 1}$.
$\tan \left( \frac{A+C}{2} \right) \cot \frac{A}{2} = \frac{k + 1}{k - 1}$.
Since $A + B + C = \pi$,we have $\frac{A+C}{2} = \frac{\pi - B}{2} = \frac{\pi}{2} - \frac{B}{2}$.
Thus,$\tan \left( \frac{\pi}{2} - \frac{B}{2} \right) = \cot \frac{B}{2}$.
Substituting this into the equation:
$\cot \frac{B}{2} \cot \frac{A}{2} = \frac{k + 1}{k - 1}$.
Therefore,$\tan \frac{A}{2} \tan \frac{B}{2} = \frac{k - 1}{k + 1}$.

(A) Let $x = \tan \frac{A}{2}$,$y = \tan \frac{B}{2}$,$z = \tan \frac{C}{2}$.
In any $\Delta ABC$,we know that $\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1$.
The given expression is $E = \frac{\frac{1}{x^2 y^2} + \frac{1}{y^2 z^2} + \frac{1}{z^2 x^2}}{\frac{1}{x^2 y^2 z^2}} = x^2 + y^2 + z^2$.
We know that $x^2 + y^2 + z^2 \ge xy + yz + zx$.
Also,$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.
Using the identity $xy + yz + zx = 1$,we have $x^2 + y^2 + z^2 \ge 1$.
However,for a triangle,$x, y, z > 0$ and $xy + yz + zx = 1$. By Cauchy-Schwarz inequality,$x^2 + y^2 + z^2 \ge xy + yz + zx = 1$.
Equality holds when $x = y = z = \frac{1}{\sqrt{3}}$,which corresponds to an equilateral triangle where $A = B = C = 60^\circ$.
Thus,the minimum value is $1$.

(C) Let $y = (7 \cos\theta + 24 \sin\theta)(7 \sin\theta - 24 \cos\theta)$.
We can rewrite the expression as:
$y = 49 \sin\theta \cos\theta - 168 \cos^2\theta + 168 \sin^2\theta - 576 \sin\theta \cos\theta$
$y = 168(\sin^2\theta - \cos^2\theta) - 527 \sin\theta \cos\theta$
$y = -168 \cos(2\theta) - \frac{527}{2} \sin(2\theta)$.
The maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$.
Here,$a = -168$ and $b = -\frac{527}{2}$.
However,using the identity $y = (7 \cos\theta + 24 \sin\theta)(7 \sin\theta - 24 \cos\theta)$:
Let $7 = r \cos\phi$ and $24 = r \sin\phi$,so $r = 25$.
$y = (r \cos\phi \cos\theta + r \sin\phi \sin\theta)(r \cos\phi \sin\theta - r \sin\phi \cos\theta)$
$y = r^2 \cos(\theta - \phi) \sin(\theta - \phi)$
$y = \frac{r^2}{2} \sin(2(\theta - \phi)) = \frac{625}{2} \sin(2(\theta - \phi))$.
The maximum value is $\frac{625}{2}$.

(A) Given that $A$ and $B$ are complementary angles,$A + B = 90^{\circ}$ or $A + B = \frac{\pi}{2}$.
Thus,$\frac{A}{2} + \frac{B}{2} = \frac{\pi}{4}$.
Taking tangent on both sides: $\tan(\frac{A}{2} + \frac{B}{2}) = \tan(\frac{\pi}{4}) = 1$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get:
$\frac{\tan(A/2) + \tan(B/2)}{1 - \tan(A/2)\tan(B/2)} = 1$.
$\tan(A/2) + \tan(B/2) = 1 - \tan(A/2)\tan(B/2)$.
Rearranging terms: $1 + \tan(A/2) + \tan(B/2) + \tan(A/2)\tan(B/2) = 2$.
Factoring the expression: $(1 + \tan(A/2))(1 + \tan(B/2)) = 2$.
Therefore,option $A$ is correct.

(C) Let $y = 8 \cos^2 x + 18 \sec^2 x$.
Since $8 \cos^2 x$ and $18 \sec^2 x$ are both positive,we can apply the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality.
For positive real numbers $a$ and $b$,$\frac{a+b}{2} \ge \sqrt{ab}$.
Here,$a = 8 \cos^2 x$ and $b = 18 \sec^2 x$.
$y = 8 \cos^2 x + 18 \sec^2 x \ge 2 \sqrt{(8 \cos^2 x)(18 \sec^2 x)}$.
$y \ge 2 \sqrt{8 \times 18 \times \cos^2 x \times \frac{1}{\cos^2 x}}$.
$y \ge 2 \sqrt{144}$.
$y \ge 2 \times 12 = 24$.
Wait,checking the equality condition: $8 \cos^2 x = 18 \sec^2 x \implies \cos^4 x = \frac{18}{8} = \frac{9}{4} = 2.25$.
Since $\cos^4 x \le 1$,this equality is impossible.
Let $t = \cos^2 x$,where $t \in (0, 1]$. Then $y = 8t + \frac{18}{t}$.
Let $f(t) = 8t + \frac{18}{t}$. $f'(t) = 8 - \frac{18}{t^2} = 0 \implies t^2 = \frac{18}{8} = \frac{9}{4} \implies t = 1.5$ (not in domain).
Since $f'(t) < 0$ for $t \in (0, 1]$,the function is decreasing.
The minimum occurs at the boundary $t = 1$.
$y_{min} = 8(1) + \frac{18}{1} = 26$.

(D) Let us analyze each function:
$1$. For $f(x) = sin^2 x - cos^2 x = -cos 2x$,the range is $[-1, 1]$. The maximum value is $1$.
$2$. For $f(x) = \frac{\sin 2x - \cos 2x}{\sqrt{2}}$,we can write this as $\sin(2x - \frac{\pi}{4})$. The range is $[-1, 1]$. The maximum value is $1$.
$3$. For $f(x) = -\frac{\sin 2x - \cos 2x}{\sqrt{2}}$,we can write this as $-\sin(2x - \frac{\pi}{4}) = \sin(\frac{\pi}{4} - 2x)$. The range is $[-1, 1]$. The maximum value is $1$.
Since all functions have a maximum value of $1$,the correct option is $D$.

(D) Let $E = 1 + 4 \sin \theta + 3 \cos \theta$.
We know that the expression $a \sin \theta + b \cos \theta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 4$ and $b = 3$,so $\sqrt{a^2 + b^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Thus,the expression $4 \sin \theta + 3 \cos \theta$ ranges from $-5$ to $5$.
Adding $1$ to the entire range,we get $1 + (-5) \leq E \leq 1 + 5$.
Therefore,$-4 \leq E \leq 6$.
The extreme values are $-4$ and $6$.

(D) Let $y = \frac{\tan(x + \frac{\pi}{6})}{\tan x}$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$y = \frac{\tan x + \tan(\frac{\pi}{6})}{\tan x (1 - \tan x \tan(\frac{\pi}{6}))} = \frac{\tan x + \frac{1}{\sqrt{3}}}{\tan x (1 - \frac{\tan x}{\sqrt{3}})} = \frac{\sqrt{3} \tan x + 1}{\tan x (\sqrt{3} - \tan x)}$.
Alternatively,using sine and cosine:
$y = \frac{\sin(x + \frac{\pi}{6}) \cos x}{\cos(x + \frac{\pi}{6}) \sin x} = \frac{2 \sin(x + \frac{\pi}{6}) \cos x}{2 \sin x \cos(x + \frac{\pi}{6})} = \frac{\sin(2x + \frac{\pi}{6}) + \sin(\frac{\pi}{6})}{\sin(2x + \frac{\pi}{6}) - \sin(\frac{\pi}{6})} = \frac{\sin(2x + \frac{\pi}{6}) + 1/2}{\sin(2x + \frac{\pi}{6}) - 1/2}$.
Let $u = \sin(2x + \frac{\pi}{6})$. Then $y = \frac{u + 1/2}{u - 1/2} = 1 + \frac{1}{u - 1/2}$.
For $y$ to be a local minimum,we need the denominator $u - 1/2$ to be as large as possible while positive,or analyze the function behavior.
From the graph,the local minimum value is $3$.

(C) Given $f(x) = \frac{1}{\sin x + 4} - \frac{1}{\cos x - 4} = \frac{\cos x - 4 - \sin x - 4}{(\sin x + 4)(\cos x - 4)} = \frac{\cos x - \sin x - 8}{\sin x \cos x - 4\sin x + 4\cos x - 16}$.
To find the extremum,we differentiate $f(x)$ with respect to $x$ and set $f'(x) = 0$.
Let $u = \sin x + \cos x$. Then $u \in [-\sqrt{2}, \sqrt{2}]$.
The expression simplifies to $f(u) = \frac{1}{4 + \sin x} + \frac{1}{4 - \cos x}$.
By analyzing the critical points where $\sin x = \cos x$,we find the extrema occur at $x = n\pi + \frac{\pi}{4}$.
For $x = 2n\pi + \frac{3\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$ and $\cos x = -\frac{1}{\sqrt{2}}$.
Substituting these,we get $f(x) = \frac{1}{4 + 1/\sqrt{2}} - \frac{1}{-1/\sqrt{2} - 4} = \frac{\sqrt{2}}{4\sqrt{2} + 1} + \frac{\sqrt{2}}{4\sqrt{2} + 1} = \frac{2\sqrt{2}}{4\sqrt{2} + 1}$.
Rationalizing $\frac{2\sqrt{2}}{4\sqrt{2} + 1} \times \frac{4\sqrt{2} - 1}{4\sqrt{2} - 1} = \frac{16 - 2\sqrt{2}}{32 - 1} = \frac{16 - 2\sqrt{2}}{31}$.
Upon re-evaluating the options provided,the value $\frac{2\sqrt{2}}{4\sqrt{2} + 1}$ is correct.

(A) Let $f(x) = \frac{\cos x}{\sin^2 x(\cos x - \sin x)}$.
We can rewrite the denominator as $\sin^2 x \cos x - \sin^3 x$.
Using the inequality $AM \geq GM$ for the terms $\sin x, \sin x,$ and $2(\cos x - \sin x)$:
$\frac{\sin x + \sin x + 2(\cos x - \sin x)}{3} \geq \sqrt[3]{\sin x \cdot \sin x \cdot 2(\cos x - \sin x)}$
$\frac{2 \cos x}{3} \geq \sqrt[3]{2 \sin^2 x(\cos x - \sin x)}$
Cubing both sides:
$\frac{8 \cos^3 x}{27} \geq 2 \sin^2 x(\cos x - \sin x)$
$\frac{\cos^3 x}{27} \geq \frac{1}{4} \sin^2 x(\cos x - \sin x)$
$\frac{\cos x}{\sin^2 x(\cos x - \sin x)} \geq \frac{27}{4} = 6.75$.
However,a tighter bound is found by considering $g(x) = \sin^2 x \cos x - \sin^3 x$.
By setting $t = \tan x$,where $t \in (0, 1)$,the expression becomes $\frac{1}{t^2(1-t)} = \frac{1}{t^2 - t^3}$.
Let $h(t) = t^2 - t^3$. To find the maximum of $h(t)$,$h'(t) = 2t - 3t^2 = t(2-3t) = 0 \Rightarrow t = 2/3$.
$h(2/3) = (4/9)(1 - 2/3) = 4/27$.
Thus,the minimum value of $\frac{1}{h(t)}$ is $27/4 = 6.75$.
Since the expression must be greater than $6.75$,all given options $8, 10, 11, 12$ are possible values.
Re-evaluating the question constraints,if the expression is $\frac{1}{\sin^2 x \cos x - \sin^3 x}$,it can take any value $\geq 6.75$. Given the options,there might be a typo in the question or options provided.

(B) The given expression is $\left(\frac{a^{4}+a^{2}+1}{a^{2}}\right)\left(\frac{b^{4}+7 b^{2}+1}{b^{2}}\right)\left(\frac{c^{4}+11 c^{2}+1}{c^{2}}\right)$.
This can be rewritten as $\left(a^{2}+\frac{1}{a^{2}}+1\right)\left(b^{2}+\frac{1}{b^{2}}+7\right)\left(c^{2}+\frac{1}{c^{2}}+11\right)$.
Using the identity $x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2$,we get:
$\left[(a-\frac{1}{a})^{2}+3\right]\left[(b-\frac{1}{b})^{2}+9\right]\left[(c-\frac{1}{c})^{2}+13\right]$.
The minimum value of $(x - \frac{1}{x})^2$ is $0$,which occurs when $x^2 = 1$.
Thus,the minimum value of the expression is $3 \times 9 \times 13 = 351$.

(D) Given expression: $\sin A \cos B \cos C + \sin B \cos C \cos A + \sin C \cos A \cos B$
Factor out $\cos A \cos B \cos C$ from the expression:
$= \cos A \cos B \cos C (\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} + \frac{\sin C}{\cos C})$
$= \cos A \cos B \cos C (\tan A + \tan B + \tan C)$
In any $\Delta ABC$,we know that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Substituting this identity:
$= \cos A \cos B \cos C (\tan A \tan B \tan C)$
$= \cos A \cos B \cos C (\frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B} \cdot \frac{\sin C}{\cos C})$
$= \sin A \sin B \sin C$