Trigonometry Questions in English

(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 $\sin \frac{5\pi}{8} = \sin \frac{3\pi}{8}$.
Thus,$S = 2 \left( \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} \right)$.
Using the identity $2 \sin^2 \theta = 1 - \cos 2\theta$,we have $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$.
So,$\sin^4 \theta = \left( \frac{1 - \cos 2\theta}{2} \right)^2 = \frac{1}{4} (1 - 2 \cos 2\theta + \cos^2 2\theta)$.
$S = 2 \left[ \frac{1}{4} (1 - 2 \cos \frac{\pi}{4} + \cos^2 \frac{\pi}{4}) + \frac{1}{4} (1 - 2 \cos \frac{3\pi}{4} + \cos^2 \frac{3\pi}{4}) \right]$.
$S = \frac{1}{2} \left[ (1 - 2(\frac{1}{\sqrt{2}}) + \frac{1}{2}) + (1 - 2(-\frac{1}{\sqrt{2}}) + \frac{1}{2}) \right]$.
$S = \frac{1}{2} \left[ (1 - \sqrt{2} + 0.5) + (1 + \sqrt{2} + 0.5) \right] = \frac{1}{2} [1.5 + 1.5] = \frac{1}{2} (3) = \frac{3}{2}$.

(C) Let the given expression be $E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 + \cos \frac{{5\pi }}{8}} \right)\left( {1 + \cos \frac{{7\pi }}{8}} \right)$.
Since $\cos \frac{{7\pi }}{8} = \cos \left( \pi - \frac{\pi }{8} \right) = -\cos \frac{\pi }{8}$ and $\cos \frac{{5\pi }}{8} = \cos \left( \pi - \frac{{3\pi }}{8} \right) = -\cos \frac{{3\pi }}{8}$,we can rewrite the expression as:
$E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 - \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{{3\pi }}{8}} \right)$
Using the identity $(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta$,we get:
$E = \left( {1 - \cos^2 \frac{\pi }{8}} \right)\left( {1 - \cos^2 \frac{{3\pi }}{8}} \right) = \sin^2 \frac{\pi }{8} \sin^2 \frac{{3\pi }}{8}$
Using the identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$:
$E = \left( \sin \frac{\pi }{8} \sin \frac{{3\pi }}{8} \right)^2 = \left( \frac{1}{2} \left( \cos \left( \frac{3\pi }{8} - \frac{\pi }{8} \right) - \cos \left( \frac{3\pi }{8} + \frac{\pi }{8} \right) \right) \right)^2$
$E = \left( \frac{1}{2} \left( \cos \frac{\pi }{4} - \cos \frac{\pi }{2} \right) \right)^2 = \left( \frac{1}{2} \left( \frac{1}{\sqrt{2}} - 0 \right) \right)^2 = \left( \frac{1}{2\sqrt{2}} \right)^2 = \frac{1}{8}$.

(A) Given $3\tan A - 4 = 0$,so $\tan A = \frac{4}{3}$.
Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.
Using $\tan A = \frac{4}{3}$,we can represent this as a right triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.
Thus,$\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.
Now,substitute these values into the expression $5\sin 2A + 3\sin A + 4\cos A$:
$5(2\sin A \cos A) + 3\sin A + 4\cos A$
$= 10\sin A \cos A + 3\sin A + 4\cos A$
$= 10\left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right) + 3\left(-\frac{4}{5}\right) + 4\left(-\frac{3}{5}\right)$
$= 10\left(\frac{12}{25}\right) - \frac{12}{5} - \frac{12}{5}$
$= \frac{120}{25} - \frac{24}{5} = \frac{24}{5} - \frac{24}{5} = 0$.

(A) We know the identity $\cot(A) = \frac{1 + \cos(2A)}{\sin(2A)}$.
Let $A = 7.5^{\circ}$,then $2A = 15^{\circ}$.
Substituting $A = 7.5^{\circ}$ in the identity,we get $\cot(7.5^{\circ}) = \frac{1 + \cos(15^{\circ})}{\sin(15^{\circ})}$.
We know that $\cos(15^{\circ}) = \cos(45^{\circ} - 30^{\circ}) = \cos(45^{\circ})\cos(30^{\circ}) + \sin(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Similarly,$\sin(15^{\circ}) = \sin(45^{\circ} - 30^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) - \cos(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.
Substituting these values: $\cot(7.5^{\circ}) = \frac{1 + \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}$.
Rationalizing the denominator: $\frac{(2\sqrt{2} + \sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{3 - 1} = \frac{2\sqrt{6} + 2\sqrt{2} + 2\sqrt{3} + 4}{2} = \sqrt{6} + \sqrt{2} + \sqrt{3} + 2$.
Since $\sqrt{4} = 2$,the expression is $\sqrt{6} + \sqrt{2} + \sqrt{3} + \sqrt{4}$.

(C) We know that $2A = (A + B) + (A - B).$
Taking the tangent on both sides,we get $\tan 2A = \tan((A + B) + (A - B)).$
Using the formula $\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y},$
we substitute $x = (A + B)$ and $y = (A - B).$
Thus,$\tan 2A = \frac{\tan(A + B) + \tan(A - B)}{1 - \tan(A + B)\tan(A - B)}.$
Substituting the given values $\tan(A + B) = p$ and $\tan(A - B) = q,$
we get $\tan 2A = \frac{p + q}{1 - pq}.$

(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 $\cos 2(\alpha + \beta) = 2\cos^2(\alpha + \beta) - 1$:
$E = (1 - \cos 2\beta) + 4\cos(\alpha + \beta)\sin \alpha \sin \beta + (2\cos^2(\alpha + \beta) - 1)$
$E = 2\cos^2(\alpha + \beta) - \cos 2\beta + 4\cos(\alpha + \beta)\sin \alpha \sin \beta$
$E = 2\cos(\alpha + \beta) [\cos(\alpha + \beta) + 2\sin \alpha \sin \beta] - \cos 2\beta$
Using $2\sin \alpha \sin \beta = \cos(\alpha - \beta) - \cos(\alpha + \beta)$:
$E = 2\cos(\alpha + \beta) [\cos(\alpha + \beta) + \cos(\alpha - \beta) - \cos(\alpha + \beta)] - \cos 2\beta$
$E = 2\cos(\alpha + \beta)\cos(\alpha - \beta) - \cos 2\beta$
Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = [\cos 2\alpha + \cos 2\beta] - \cos 2\beta$
$E = \cos 2\alpha$.

(B) Let $f(\theta) = \sin \theta + \cos \theta$.
To find the maximum value,we can rewrite the expression as $f(\theta) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta \right)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f(\theta) = \sqrt{2} \sin(\theta + 45^\circ)$.
The maximum value of the sine function is $1$,which occurs when the angle is $90^\circ$.
Therefore,$\sin(\theta + 45^\circ) = 1$ implies $\theta + 45^\circ = 90^\circ$.
Solving for $\theta$,we get $\theta = 90^\circ - 45^\circ = 45^\circ$.

(D) We know that for any real number $a$,the square of a real number is always non-negative,i.e.,$(a - b)^2 \ge 0$.
This implies $a^2 + b^2 \ge 2ab$.
Let $a = \cos x$ and $b = \sec x = \frac{1}{\cos x}$.
Then,$f(x) = \cos^2 x + \sec^2 x = \cos^2 x + \frac{1}{\cos^2 x}$.
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that for positive real numbers,the arithmetic mean is greater than or equal to the geometric mean:
$\frac{\cos^2 x + \sec^2 x}{2} \ge \sqrt{\cos^2 x \cdot \sec^2 x}$.
Since $\cos^2 x \cdot \sec^2 x = \cos^2 x \cdot \frac{1}{\cos^2 x} = 1$,we have:
$\frac{f(x)}{2} \ge \sqrt{1} = 1$.
Therefore,$f(x) \ge 2$.

(A) Let $y = \frac{\tan x}{\tan 3x}$.
Using the formula $\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$,we get:
$y = \frac{\tan x}{\frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}} = \frac{1 - 3\tan^2 x}{3 - \tan^2 x}$.
Let $t = \tan^2 x$,where $t \ge 0$ and $t \neq 3$ (since $\tan 3x$ must be defined).
Then $y = \frac{1 - 3t}{3 - t}$.
To find the range,we solve for $t$ in terms of $y$:
$y(3 - t) = 1 - 3t \implies 3y - yt = 1 - 3t \implies 3t - yt = 1 - 3y \implies t(3 - y) = 1 - 3y$.
$t = \frac{1 - 3y}{3 - y} = \frac{3y - 1}{y - 3}$.
Since $t = \tan^2 x \ge 0$,we have $\frac{3y - 1}{y - 3} \ge 0$.
Using the wavy curve method,the expression is $\ge 0$ when $y \le 1/3$ or $y \ge 3$.
Thus,$y$ never lies between $1/3$ and $3$.

(C) Let $f(\theta) = \cos 2\theta + 2\cos \theta$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$f(\theta) = 2\cos^2 \theta - 1 + 2\cos \theta = 2(\cos^2 \theta + \cos \theta) - 1$.
Completing the square:
$f(\theta) = 2(\cos^2 \theta + \cos \theta + \frac{1}{4}) - 1 - \frac{1}{2} = 2(\cos \theta + \frac{1}{2})^2 - \frac{3}{2}$.
Since $-1 \le \cos \theta \le 1$,the range of $(\cos \theta + \frac{1}{2})$ is $[-\frac{1}{2}, \frac{3}{2}]$.
Thus,the range of $(\cos \theta + \frac{1}{2})^2$ is $[0, \frac{9}{4}]$.
Therefore,the range of $2(\cos \theta + \frac{1}{2})^2 - \frac{3}{2}$ is $[2(0) - \frac{3}{2}, 2(\frac{9}{4}) - \frac{3}{2}] = [-\frac{3}{2}, \frac{9}{2} - \frac{3}{2}] = [-\frac{3}{2}, 3]$.
So,the expression is always greater than or equal to $-\frac{3}{2}$ and less than or equal to $3$.

(A) In a triangle $ABC$,the sum of angles is $A + B + C = \pi$.
Dividing by $2$,we get $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.
Taking the tangent on both sides,we have $\tan \left( \frac{A+B}{2} \right) = \tan \left( \frac{\pi}{2} - \frac{C}{2} \right) = \cot \frac{C}{2}$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get $\frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \cot \frac{C}{2}$.
Substituting the given values $\tan \frac{A}{2} = \frac{1}{3}$ and $\tan \frac{B}{2} = \frac{2}{3}$:
$\cot \frac{C}{2} = \frac{\frac{1}{3} + \frac{2}{3}}{1 - (\frac{1}{3})(\frac{2}{3})} = \frac{1}{1 - \frac{2}{9}} = \frac{1}{\frac{7}{9}} = \frac{9}{7}$.
Since $\cot \frac{C}{2} = \frac{9}{7}$,we have $\tan \frac{C}{2} = \frac{7}{9}$.

(B) Given $A + B + C = \pi$,so $A = \pi - (B + C)$.
Substituting this into the given equation $\cos A = \cos B \cos C$:
$\cos(\pi - (B + C)) = \cos B \cos C$
Using the identity $\cos(\pi - \theta) = -\cos \theta$:
$-\cos(B + C) = \cos B \cos C$
Expanding $\cos(B + C)$:
$-(\cos B \cos C - \sin B \sin C) = \cos B \cos C$
$-\cos B \cos C + \sin B \sin C = \cos B \cos C$
$\sin B \sin C = 2 \cos B \cos C$
Dividing both sides by $\cos B \cos C$ (assuming $\cos B, \cos C \neq 0$):
$\frac{\sin B \sin C}{\cos B \cos C} = 2$
$\tan B \tan C = 2$.

(B) Given the relation $A + C = B.$
Taking the tangent on both sides,we get $\tan(A + C) = \tan B.$
Using the trigonometric identity $\tan(A + C) = \frac{\tan A + \tan C}{1 - \tan A \tan C},$
we have $\frac{\tan A + \tan C}{1 - \tan A \tan C} = \tan B.$
Cross-multiplying gives $\tan A + \tan C = \tan B(1 - \tan A \tan C).$
Expanding the right side,we get $\tan A + \tan C = \tan B - \tan A \tan B \tan C.$
Rearranging the terms to solve for $\tan A \tan B \tan C,$
we obtain $\tan A \tan B \tan C = \tan B - \tan A - \tan C.$

(B) Let $u = \cos \theta \{\sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha\}}$.
Rearranging the terms,we have $u - \sin \theta \cos \theta = \cos \theta \sqrt{\sin^2 \theta + \sin^2 \alpha}$.
Squaring both sides:
$(u - \sin \theta \cos \theta)^2 = \cos^2 \theta (\sin^2 \theta + \sin^2 \alpha)$.
$u^2 - 2u \sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta = \sin^2 \theta \cos^2 \theta + \cos^2 \theta \sin^2 \alpha$.
$u^2 - 2u \sin \theta \cos \theta - \cos^2 \theta \sin^2 \alpha = 0$.
Dividing by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$):
$u^2 \sec^2 \theta - 2u \tan \theta - \sin^2 \alpha = 0$.
Using $\sec^2 \theta = 1 + \tan^2 \theta$:
$u^2 (1 + \tan^2 \theta) - 2u \tan \theta - \sin^2 \alpha = 0$.
$u^2 \tan^2 \theta - 2u \tan \theta + (u^2 - \sin^2 \alpha) = 0$.
Since $\tan \theta$ is real,the discriminant $D \ge 0$:
$(-2u)^2 - 4(u^2)(u^2 - \sin^2 \alpha) \ge 0$.
$4u^2 - 4u^4 + 4u^2 \sin^2 \alpha \ge 0$.
Dividing by $4u^2$ (assuming $u \neq 0$):
$1 - u^2 + \sin^2 \alpha \ge 0$.
$u^2 \le 1 + \sin^2 \alpha$.
Therefore,$|u| \le \sqrt{1 + \sin^2 \alpha}$.
Thus,$k = \sqrt{1 + \sin^2 \alpha}$.

(B) Given $\cos x + \sin x = \frac{1}{2}$.
Squaring both sides,we get $(\cos x + \sin x)^2 = (\frac{1}{2})^2$.
$\cos^2 x + \sin^2 x + 2 \sin x \cos x = \frac{1}{4}$.
Since $\cos^2 x + \sin^2 x = 1$,we have $1 + \sin 2x = \frac{1}{4}$,which implies $\sin 2x = -\frac{3}{4}$.
Since $\sin 2x < 0$ and $0 < x < \pi$,$x$ must be in the second quadrant,so $\tan x < 0$.
Using the identity $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we have $\frac{2 \tan x}{1 + \tan^2 x} = -\frac{3}{4}$.
$8 \tan x = -3 - 3 \tan^2 x \Rightarrow 3 \tan^2 x + 8 \tan x + 3 = 0$.
Using the quadratic formula $\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $\tan x = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm 2\sqrt{7}}{6} = \frac{-4 \pm \sqrt{7}}{3}$.
Since $\tan x < 0$ and both values are negative,we check the condition $\cos x + \sin x = \frac{1}{2}$.
For $\tan x = \frac{-4 - \sqrt{7}}{3}$,$\cos x + \sin x$ is negative,which contradicts the given condition. Thus,$\tan x = \frac{-4 + \sqrt{7}}{3}$.

(B) Given: $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$
Multiply by $2$: $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] = -3$
$\Rightarrow 2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] + 3 = 0$
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 $\cos(\alpha + \beta) = \frac{4}{5}$. Since $0 \le \alpha, \beta \le \frac{\pi}{4}$,we have $0 \le \alpha + \beta \le \frac{\pi}{2}$,so $\tan(\alpha + \beta) = \frac{3}{4}$.
Given $\sin(\alpha - \beta) = \frac{5}{13}$. Since $0 \le \alpha, \beta \le \frac{\pi}{4}$,we have $-\frac{\pi}{4} \le \alpha - \beta \le \frac{\pi}{4}$,so $\tan(\alpha - \beta) = \frac{5}{12}$.
We know that $2\alpha = (\alpha + \beta) + (\alpha - \beta)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 2\alpha = \tan((\alpha + \beta) + (\alpha - \beta)) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta)\tan(\alpha - \beta)}$
$\tan 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} \times \frac{5}{12})} = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} \times \frac{48}{33} = \frac{14 \times 4}{33} = \frac{56}{33}$.

(C) Given $A = \sin^2 x + \cos^4 x$.
We can write this as $A = \sin^2 x + (\cos^2 x)^2$.
Since $\sin^2 x = 1 - \cos^2 x$,let $t = \cos^2 x$,where $0 \le t \le 1$.
Then $A = (1 - t) + t^2 = t^2 - t + 1$.
This is a quadratic expression in $t$. The vertex of the parabola $f(t) = t^2 - t + 1$ occurs at $t = -b/(2a) = -(-1)/(2 \times 1) = 1/2$.
Since $1/2$ is 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 $[0, 1]$.
At $t = 0$,$f(0) = 0^2 - 0 + 1 = 1$.
At $t = 1$,$f(1) = 1^2 - 1 + 1 = 1$.
Thus,the range of $A$ is $\frac{3}{4} \le A \le 1$.

(B) Given expression: $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$
Convert $\tan A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:
$= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)}$
$= \frac{1}{\sin A - \cos A} \left[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} \right]$
Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:
$= \frac{1}{\sin A - \cos A} \left[ \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A \cos A} \right]$
$= \frac{\sin^2 A + \cos^2 A + \sin A \cos A}{\sin A \cos A}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1 = \sec A \csc A + 1$

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

(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 $\cos^2 x = t$:
$5(\frac{1}{t} - 1) = 9t + 7$
$5 - 5t = 9t^2 + 7t$
$9t^2 + 12t - 5 = 0$
$(3t - 1)(3t + 5) = 0$
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,$\cos 4x = 2\cos^2 2x - 1 = 2(-\frac{1}{3})^2 - 1 = 2(\frac{1}{9}) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.

(D) For the quadratic equation $x^2 + px + q = 0$,the sum of the roots is $-p$ and the product of the roots is $q$.
Given roots are $\tan 30^\circ$ and $\tan 15^\circ$.
So,$-p = \tan 30^\circ + \tan 15^\circ$ and $q = \tan 30^\circ \cdot \tan 15^\circ$.
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan 45^\circ = \frac{\tan 30^\circ + \tan 15^\circ}{1 - \tan 30^\circ \tan 15^\circ} = 1$.
Substituting the values of $p$ and $q$:
$1 = \frac{-p}{1 - q} \Rightarrow 1 - q = -p \Rightarrow p - q = -1$.
We need to find the value of $2 + q - p$.
Since $p - q = -1$,then $q - p = 1$.
Therefore,$2 + q - p = 2 + 1 = 3$.

(B) Given the equation is an identity,it must hold true for all values of $x$.
Expanding the expression: $(\cos^2 x + \sin^2 x) + 2 \sin x \cos x + k \sin x \cos x - 1 = 0$.
Using the trigonometric identity $\cos^2 x + \sin^2 x = 1$,we substitute it into the equation:
$1 + 2 \sin x \cos x + k \sin x \cos x - 1 = 0$.
Simplifying the equation:
$(2 + k) \sin x \cos x = 0$.
For this to be an identity (true for all $x$),the coefficient of $\sin x \cos x$ must be zero.
Therefore,$2 + k = 0$,which gives $k = -2$.

(B) Given $\tan (A - B) = 1$. Since $\tan(\pi/4) = 1$,we have $A - B = \frac{\pi}{4} + n\pi$. For the smallest positive value,consider $A - B = \frac{\pi}{4}$ $(i)$.
Given $\sec (A + B) = \frac{2}{\sqrt{3}}$. Since $\sec(\pi/6) = \frac{2}{\sqrt{3}}$,we have $A + B = 2n\pi \pm \frac{\pi}{6}$.
To find the smallest positive $B$,we solve for $B$ using $B = \frac{(A+B) - (A-B)}{2}$.
Using $A + B = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ and $A - B = \frac{\pi}{4}$:
$2B = \frac{11\pi}{6} - \frac{\pi}{4} = \frac{22\pi - 3\pi}{12} = \frac{19\pi}{12}$.
Therefore,$B = \frac{19\pi}{24}$.

(A) Given: $\frac{{\sin^4 A}}{a} + \frac{{\cos^4 A}}{b} = \frac{1}{{a + b}}$
Using the identity $\sin^2 A = \frac{{1 - \cos 2A}}{2}$ 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 $4ab(a + b)$:
$b(a + b)(1 - 2\cos 2A + \cos^2 2A) + a(a + b)(1 + 2\cos 2A + \cos^2 2A) = 4ab$
Expanding and rearranging terms:
$(a + b)^2 \cos^2 2A + 2(a + b)(a - b) \cos 2A + (a - b)^2 = 0$
This is a perfect square:
$((a + b) \cos 2A + (a - b))^2 = 0$
Thus,$\cos 2A = \frac{{b - a}}{{a + b}}$.
Now,calculate $\frac{{\sin^8 A}}{{a^3}} + \frac{{\cos^8 A}}{{b^3}}$:
$= \frac{{(1 - \cos 2A)^4}}{{16a^3}} + \frac{{(1 + \cos 2A)^4}}{{16b^3}}$
Substituting $\cos 2A = \frac{{b - a}}{{a + b}}$:
$= \frac{1}{{16a^3}} \left( 1 - \frac{{b - a}}{{a + b}} \right)^4 + \frac{1}{{16b^3}} \left( 1 + \frac{{b - a}}{{a + b}} \right)^4$
$= \frac{1}{{16a^3}} \left( \frac{{2a}}{{a + b}} \right)^4 + \frac{1}{{16b^3}} \left( \frac{{2b}}{{a + b}} \right)^4$
$= \frac{{16a^4}}{{16a^3(a + b)^4}} + \frac{{16b^4}}{{16b^3(a + b)^4}} = \frac{a + b}{{(a + b)^4}} = \frac{1}{{(a + b)^3}}$.

(B) We know that $\sin 3x = 3 \sin x - 4 \sin^3 x$,so $\sin^3 x = \frac{1}{4}(3 \sin x - \sin 3x)$.
Substituting this into the expression:
${\sin ^3}x \sin 3x = \frac{1}{4}(3 \sin x - \sin 3x) \sin 3x$
$= \frac{3}{4} \sin x \sin 3x - \frac{1}{4} \sin^2 3x$
Using the product-to-sum formulas $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$= \frac{3}{8}(2 \sin x \sin 3x) - \frac{1}{8}(2 \sin^2 3x)$
$= \frac{3}{8}(\cos 2x - \cos 4x) - \frac{1}{8}(1 - \cos 6x)$
$= -\frac{1}{8} + \frac{3}{8} \cos 2x - \frac{3}{8} \cos 4x + \frac{1}{8} \cos 6x$
Comparing this with $\sum_{m=0}^n c_m \cos mx = c_0 + c_1 \cos x + c_2 \cos 2x + \dots + c_n \cos nx$,we see that the highest frequency term is $\cos 6x$,which implies $n = 6$.

(D) Given the infinite geometric series:
$x = 1 + \cos^2\phi + \cos^4\phi + \dots = \frac{1}{1 - \cos^2\phi} = \frac{1}{\sin^2\phi}$
$y = 1 + \sin^2\phi + \sin^4\phi + \dots = \frac{1}{1 - \sin^2\phi} = \frac{1}{\cos^2\phi}$
$z = 1 + \cos^2\phi \sin^2\phi + \cos^4\phi \sin^4\phi + \dots = \frac{1}{1 - \cos^2\phi \sin^2\phi}$
Now,calculate $xy$:
$xy = \frac{1}{\sin^2\phi} \cdot \frac{1}{\cos^2\phi} = \frac{1}{\sin^2\phi \cos^2\phi}$
Check option $(b)$:
$xy + z = \frac{1}{\sin^2\phi \cos^2\phi} + \frac{1}{1 - \cos^2\phi \sin^2\phi} = \frac{1 - \cos^2\phi \sin^2\phi + \sin^2\phi \cos^2\phi}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)} = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$
$xyz = \left(\frac{1}{\sin^2\phi \cos^2\phi}\right) \left(\frac{1}{1 - \cos^2\phi \sin^2\phi}\right) = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$
Thus,$xyz = xy + z$ is correct.
Check option $(c)$:
$x + y + z = \frac{1}{\sin^2\phi} + \frac{1}{\cos^2\phi} + z = \frac{\cos^2\phi + \sin^2\phi}{\sin^2\phi \cos^2\phi} + z = \frac{1}{\sin^2\phi \cos^2\phi} + z = xy + z = xyz$.
Since both $(b)$ and $(c)$ are correct,option $(d)$ is the correct choice.

(D) Given ${f_n}(\theta ) = \tan \frac{\theta }{2} (1 + \sec \theta ) (1 + \sec 2\theta ) \dots (1 + \sec {2^n}\theta ).$
Using $1 + \sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2\cos^2(x/2)}{\cos x}$,we have:
${f_n}(\theta ) = \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot \frac{2\cos^2(\theta/2)}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \cdot \frac{2\cos^2 2\theta}{\cos 4\theta} \dots \frac{2\cos^2(2^{n-1}\theta)}{\cos(2^n\theta)}.$
Simplifying the terms 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 ${f_n}(\theta ) = \tan(2^n\theta).$
For option $A$: ${f_2}(\pi/16) = \tan(2^2 \cdot \pi/16) = \tan(4\pi/16) = \tan(\pi/4) = 1.$
For option $B$: ${f_3}(\pi/32) = \tan(2^3 \cdot \pi/32) = \tan(8\pi/32) = \tan(\pi/4) = 1.$
For option $C$: ${f_4}(\pi/64) = \tan(2^4 \cdot \pi/64) = \tan(16\pi/64) = \tan(\pi/4) = 1.$
Thus,all the above are correct.

(B) Given equation: $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.
Multiply both sides by $\sqrt{2}$:
$\sqrt{2} \left( \sin \frac{\pi }{2^n} \cdot \frac{1}{\sqrt{2}} + \cos \frac{\pi }{2^n} \cdot \frac{1}{\sqrt{2}} \right) = \frac{\sqrt{n}}{2}$.
Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get:
$\sqrt{2} \sin \left( \frac{\pi }{4} + \frac{\pi }{2^n} \right) = \frac{\sqrt{n}}{2}$.
Since the maximum value of $\sin \theta$ is $1$,we have $\sin \left( \frac{\pi }{4} + \frac{\pi }{2^n} \right) \le 1$.
Therefore,$\frac{\sqrt{n}}{2} \le \sqrt{2} \implies \sqrt{n} \le 2\sqrt{2} \implies n \le 8$.
For $n=4$,$\sin \frac{\pi}{16} + \cos \frac{\pi}{16} = \sqrt{2} \sin(\frac{\pi}{4} + \frac{\pi}{16}) = \sqrt{2} \sin(\frac{5\pi}{16}) \approx \sqrt{2} \times 0.831 = 1.175$,while $\frac{\sqrt{4}}{2} = 1$. Since $1.175 > 1$,the equation holds for $n > 4$.
Thus,$4 < n \le 8$.

(B) Given $k = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$.
Using the identity $\sin \theta = \cos \left( \frac{\pi}{2} - \theta \right)$:
$k = \cos \left( \frac{\pi}{2} - \frac{\pi}{18} \right) \cos \left( \frac{\pi}{2} - \frac{5\pi}{18} \right) \cos \left( \frac{\pi}{2} - \frac{7\pi}{18} \right)$
$k = \cos \frac{8\pi}{18} \cos \frac{4\pi}{18} \cos \frac{2\pi}{18} = \cos \frac{4\pi}{9} \cos \frac{2\pi}{9} \cos \frac{\pi}{9}$.
Using the formula $\cos \theta \cos 2\theta \cos 4\theta = \frac{\sin 8\theta}{8 \sin \theta}$ where $\theta = \frac{\pi}{9}$:
$k = \frac{\sin \left( 8 \cdot \frac{\pi}{9} \right)}{8 \sin \frac{\pi}{9}} = \frac{\sin \frac{8\pi}{9}}{8 \sin \frac{\pi}{9}}$.
Since $\sin \frac{8\pi}{9} = \sin \left( \pi - \frac{\pi}{9} \right) = \sin \frac{\pi}{9}$:
$k = \frac{\sin \frac{\pi}{9}}{8 \sin \frac{\pi}{9}} = \frac{1}{8}$.

(C) Given $\alpha < \beta < \gamma < \delta$ and $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = k$. Since these are the smallest positive angles,we have $\beta = \pi - \alpha$,$\gamma = 2\pi + \alpha$,and $\delta = 3\pi - \alpha$.
Substituting these into the expression:
$E = 4\sin \frac{\alpha}{2} + 3\sin \frac{\pi - \alpha}{2} + 2\sin \frac{2\pi + \alpha}{2} + \sin \frac{3\pi - \alpha}{2}$
$E = 4\sin \frac{\alpha}{2} + 3\sin \left( \frac{\pi}{2} - \frac{\alpha}{2} \right) + 2\sin \left( \pi + \frac{\alpha}{2} \right) + \sin \left( \frac{3\pi}{2} - \frac{\alpha}{2} \right)$
Using trigonometric identities:
$E = 4\sin \frac{\alpha}{2} + 3\cos \frac{\alpha}{2} - 2\sin \frac{\alpha}{2} - \cos \frac{\alpha}{2}$
$E = 2\sin \frac{\alpha}{2} + 2\cos \frac{\alpha}{2} = 2 \left( \sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \right)$
Squaring the expression inside the root:
$E = 2 \sqrt{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = 2 \sqrt{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}$
$E = 2 \sqrt{1 + \sin \alpha} = 2 \sqrt{1 + k}$.

(D) Let $L = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C)$ and $M = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)$.
Given that $L = M$.
We know that $(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = 1$.
Multiplying $L$ and $M$,we get $LM = (\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B)(\sec^2 C - \tan^2 C) = 1 \times 1 \times 1 = 1$.
Since $L = M$,we can substitute $M$ with $L$ in the equation $LM = 1$,which gives $L^2 = 1$.
This implies $L = 1$ or $L = -1$.
Since $L = M$,both sides are equal to $1$ or $-1$.

(D) Let the given series be $S = 1 + 2x + 3x^2 + 4x^3 + \dots \infty$,where $x = (1 - \cos \theta)$.
This is an arithmetico-geometric series of the form $\sum_{n=1}^{\infty} n x^{n-1}$.
The sum of this infinite series is given by $S = (1 - x)^{-2}$ for $|x| < 1$.
Substituting $x = 1 - \cos \theta$ into the formula:
$S = (1 - (1 - \cos \theta))^{-2} = (\cos \theta)^{-2} = \sec^2 \theta$.
Using the trigonometric identity $\sec^2 \theta = 1 + \tan^2 \theta$:
$S = 1 + \tan^2 \theta$.
Given $\tan \theta = \sqrt{\frac{3}{2}}$,then $\tan^2 \theta = \frac{3}{2}$.
Therefore,$S = 1 + \frac{3}{2} = \frac{5}{2}$.

(A) Given that ${x_1}, {x_2}, ..., {x_n}$ are in $A.P.$ with common difference $\alpha$,so ${x_{k+1}} - {x_k} = \alpha$ for all $k$.
We can write $\sin \alpha = \sin({x_{k+1}} - {x_k})$.
Thus,the expression becomes $\sum_{k=1}^{n-1} \sin \alpha \sec {x_k} \sec {x_{k+1}} = \sum_{k=1}^{n-1} \frac{\sin({x_{k+1}} - {x_k})}{\cos {x_k} \cos {x_{k+1}}}$.
Using the identity $\tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B}$,we get $\sum_{k=1}^{n-1} (\tan {x_{k+1}} - \tan {x_k})$.
This is a telescoping series: $(\tan {x_2} - \tan {x_1}) + (\tan {x_3} - \tan {x_2}) + ... + (\tan {x_n} - \tan {x_{n-1}})$.
After cancellation,we are left with $\tan {x_n} - \tan {x_1} = \frac{\sin({x_n} - {x_1})}{\cos {x_n} \cos {x_1}}$.
Since ${x_n} = {x_1} + (n-1)\alpha$,then ${x_n} - {x_1} = (n-1)\alpha$.
Therefore,the value is $\frac{\sin(n - 1)\alpha}{\cos {x_1} \cos {x_n}}$.

(D) Given $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$.
Squaring and adding both equations:
$(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$
$(\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2(\sin \alpha \sin \beta + \cos \alpha \cos \beta) = \frac{441 + 729}{4225}$
$1 + 1 + 2\cos(\alpha - \beta) = \frac{1170}{4225}$
$2 + 2\cos(\alpha - \beta) = \frac{1170}{4225} = \frac{18}{65}$
$2(1 + \cos(\alpha - \beta)) = \frac{18}{65}$
$4\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{65}$
$\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{260} = \frac{9}{130}$
$\cos\left(\frac{\alpha - \beta}{2}\right) = \pm \frac{3}{\sqrt{130}}$
Since $\pi < \alpha - \beta < 3\pi$,dividing by $2$ gives $\frac{\pi}{2} < \frac{\alpha - \beta}{2} < \frac{3\pi}{2}$.
In this interval,the cosine function is negative.
Therefore,$\cos\left(\frac{\alpha - \beta}{2}\right) = -\frac{3}{\sqrt{130}}$.

(C) Let $x \cos \theta = y \cos \left( \theta + \frac{2\pi}{3} \right) = z \cos \left( \theta + \frac{4\pi}{3} \right) = k$.
Then,$\frac{k}{x} = \cos \theta$,$\frac{k}{y} = \cos \left( \theta + \frac{2\pi}{3} \right)$,and $\frac{k}{z} = \cos \left( \theta + \frac{4\pi}{3} \right)$.
Adding these three equations,we get:
$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \cos \theta + \cos \left( \theta + \frac{2\pi}{3} \right) + \cos \left( \theta + \frac{4\pi}{3} \right)$.
Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$:
$\cos \left( \theta + \frac{2\pi}{3} \right) + \cos \left( \theta + \frac{4\pi}{3} \right) = 2 \cos \left( \frac{2\theta + 2\pi}{2} \right) \cos \left( \frac{-2\pi/3}{2} \right) = 2 \cos (\theta + \pi) \cos \left( -\frac{\pi}{3} \right)$.
Since $\cos (\theta + \pi) = -\cos \theta$ and $\cos (-\pi/3) = 1/2$,the sum becomes:
$2(-\cos \theta)(1/2) = -\cos \theta$.
Thus,$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \cos \theta - \cos \theta = 0$.
Since $k \neq 0$,we have $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.

(C) The regular hexagon $A_0 A_1 A_2 A_3 A_4 A_5$ is inscribed in a circle of radius $R = 1$.
In a regular hexagon,the side length is equal to the radius of the circumscribed circle.
Thus,$A_0 A_1 = 1$.
To find $A_0 A_2$,consider the triangle $A_0 A_1 A_2$. The interior angle of a regular hexagon is $120^\circ$.
Using the Law of Cosines in $\triangle A_0 A_1 A_2$:
$A_0 A_2^2 = A_0 A_1^2 + A_1 A_2^2 - 2(A_0 A_1)(A_1 A_2) \cos(120^\circ)$
$A_0 A_2^2 = 1^2 + 1^2 - 2(1)(1)(-\frac{1}{2}) = 1 + 1 + 1 = 3$.
So,$A_0 A_2 = \sqrt{3}$.
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 $a^2+b^2 = (a+b)^2 - 2ab$ and $a^3+b^3 = (a+b)^3 - 3ab(a+b)$:
$= 3[(\cos^2 \alpha + \sin^2 \alpha)^2 - 2\sin^2 \alpha \cos^2 \alpha] - 2[(\cos^2 \alpha + \sin^2 \alpha)^3 - 3\sin^2 \alpha \cos^2 \alpha(\cos^2 \alpha + \sin^2 \alpha)]$
Since $\sin^2 \alpha + \cos^2 \alpha = 1$:
$= 3[1 - 2\sin^2 \alpha \cos^2 \alpha] - 2[1 - 3\sin^2 \alpha \cos^2 \alpha]$
$= 3 - 6\sin^2 \alpha \cos^2 \alpha - 2 + 6\sin^2 \alpha \cos^2 \alpha$
$= 3 - 2 = 1$

(A) We are given the expression $E = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}.$
Since $\alpha \in \left( 0, \frac{\pi}{2} \right),$ $\tan \alpha > 0,$ and for $x^2 + x > 0,$ both terms are positive.
By the Arithmetic Mean-Geometric Mean Inequality $(AM \ge GM)$,for any two positive real numbers $a$ and $b,$ $a + b \ge 2\sqrt{ab}.$
Let $a = \sqrt{x^2 + x}$ and $b = \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}.$
Then $a + b \ge 2 \sqrt{\sqrt{x^2 + x} \cdot \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}}.$
$a + b \ge 2 \sqrt{\tan^2 \alpha}.$
Since $\tan \alpha > 0$ for $\alpha \in \left( 0, \frac{\pi}{2} \right),$ we have $a + b \ge 2 \tan \alpha.$
Thus,the expression is always greater than or equal to $2 \tan \alpha.$

(A) Given that $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$.
This implies $\frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_2}{\sin \alpha_2} \cdots \frac{\cos \alpha_n}{\sin \alpha_n} = 1$,which means $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n = \sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n$.
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 $i$,the maximum value of $P^2$ is $\frac{1}{2^n}$.
Therefore,the maximum value of $P$ is $\sqrt{\frac{1}{2^n}} = \frac{1}{2^{n/2}}$.

(B) Given that $A = \sin^8 \theta + \cos^{14} \theta$.
Since $0 \le \sin^2 \theta \le 1$ and $0 \le \cos^2 \theta \le 1$ for all real $\theta$,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 $\sin^8 \theta > 0$ and $\cos^{14} \theta > 0$ for all $\theta$ (excluding cases where one is zero,but they cannot both be zero simultaneously),$A > 0$.
Therefore,$0 < A \le 1$.

(B) Since $\theta$ is an acute angle,we have $0^\circ < \theta < 90^\circ$.
For an acute angle,the value of $\sin\theta$ must satisfy $0 \le \sin\theta < 1$.
Substituting the given expression: $0 \le \frac{p - 6}{8 - p} < 1$.
First,consider $0 \le \frac{p - 6}{8 - p}$. Since the denominator $8 - p$ must be positive for $\sin\theta$ to be defined and positive,we have $8 - p > 0$,which implies $p < 8$. For the numerator,$p - 6 \ge 0$,so $p \ge 6$.
Next,consider $\frac{p - 6}{8 - p} < 1$. Since $8 - p > 0$,we can multiply both sides by $(8 - p)$ without changing the inequality sign: $p - 6 < 8 - p$.
Adding $p$ to both sides: $2p - 6 < 8$.
Adding $6$ to both sides: $2p < 14$.
Dividing by $2$: $p < 7$.
Combining the conditions $p \ge 6$ and $p < 7$,we get $6 \le p < 7$.

(A) Given the expression: $\sum \frac{\cot A + \cot B}{\tan A + \tan B}$.
We know that $\cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin B \cos A + \cos B \sin A}{\sin A \sin B} = \frac{\sin(A+B)}{\sin A \sin B}$.
Similarly,$\tan A + \tan B = \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \frac{\sin(A+B)}{\cos A \cos B}$.
Substituting these into the expression:
$\frac{\cot A + \cot B}{\tan A + \tan B} = \frac{\sin(A+B)}{\sin A \sin B} \times \frac{\cos A \cos B}{\sin(A+B)} = \frac{\cos A \cos B}{\sin A \sin B} = \cot A \cot B$.
Thus,the sum is $\sum \cot A \cot B = \cot A \cot B + \cot B \cot C + \cot C \cot A$.
For any triangle where $A + B + C = \pi$,the identity $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$ holds true.
Therefore,the final answer is $1$.

(C) Given $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$,so $\tan \beta = \tan(\frac{\pi}{2} - \alpha) = \cot \alpha$.
Also,given $\beta + \gamma = \alpha$,taking $\tan$ on both sides gives $\tan(\beta + \gamma) = \tan \alpha$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\frac{\tan \beta + \tan \gamma}{1 - \tan \beta \tan \gamma} = \tan \alpha$.
Substituting $\tan \beta = \cot \alpha = \frac{1}{\tan \alpha}$,we have $\frac{\frac{1}{\tan \alpha} + \tan \gamma}{1 - \frac{1}{\tan \alpha} \tan \gamma} = \tan \alpha$.
Multiplying the numerator and denominator by $\tan \alpha$,we get $\frac{1 + \tan \alpha \tan \gamma}{\tan \alpha - \tan \gamma} = \tan \alpha$.
$1 + \tan \alpha \tan \gamma = \tan^2 \alpha - \tan \alpha \tan \gamma$.
$1 + 2\tan \alpha \tan \gamma = \tan^2 \alpha$.
Dividing by $\tan \alpha$,we get $\cot \alpha + 2\tan \gamma = \tan \alpha$.
Since $\cot \alpha = \tan \beta$,we have $\tan \alpha = \tan \beta + 2\tan \gamma$.

(A) Let $E = a\sin^2 \theta + b\sin \theta \cos \theta + c\cos^2 \theta - \frac{1}{2}(a + c)$.
Using the identities $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:
$E = a\left( \frac{1 - \cos 2\theta}{2} \right) + b\left( \frac{\sin 2\theta}{2} \right) + c\left( \frac{1 + \cos 2\theta}{2} \right) - \frac{1}{2}(a + c)$
$E = \frac{1}{2} [a - a\cos 2\theta + b\sin 2\theta + c + c\cos 2\theta - a - c]$
$E = \frac{1}{2} [b\sin 2\theta - (a - c)\cos 2\theta]$
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,$|E| = \left| \frac{1}{2} [b\sin 2\theta - (a - c)\cos 2\theta] \right| \le \frac{1}{2} \sqrt{b^2 + (a - c)^2}$.
Comparing this with the given inequality $|E| \le \frac{1}{2}k$,we get $k = \sqrt{b^2 + (a - c)^2}$.
Squaring both sides,$k^2 = b^2 + (a - c)^2$.

(D) Given the inequality: $\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5} < \frac{1}{32}$.
Since the expression must hold for all real $x$,the denominator $64x^2 - 96x \sin \alpha + 5$ must be positive for all $x$ (as the numerator is always positive).
Cross-multiplying: $32(4x^2 + 1) < 64x^2 - 96x \sin \alpha + 5$.
$128x^2 + 32 < 64x^2 - 96x \sin \alpha + 5$.
$64x^2 + 96x \sin \alpha + 27 < 0$.
For this quadratic in $x$ to be negative for all $x$,the discriminant $D$ must be positive,but wait,the inequality must hold for all $x$. Actually,for the quadratic $ax^2 + bx + c < 0$ to hold for all $x$,it is impossible as $a > 0$. Re-evaluating: the expression must be negative for all $x$. This implies the quadratic $64x^2 + 96x \sin \alpha + 27$ must be less than $0$ for all $x$,which is impossible for a parabola opening upwards. The condition implies the range of the function must be less than $1/32$. The discriminant of $64x^2 + 96x \sin \alpha + 27 = 0$ must be positive for real roots to exist,allowing the expression to be negative between the roots. Thus,$D = (96 \sin \alpha)^2 - 4(64)(27) > 0$.
$9216 \sin^2 \alpha - 6912 > 0$.
$\sin^2 \alpha > \frac{6912}{9216} = \frac{3}{4}$.
$|\sin \alpha| > \frac{\sqrt{3}}{2}$.
This implies $\sin \alpha > \frac{\sqrt{3}}{2}$ or $\sin \alpha < -\frac{\sqrt{3}}{2}$.
For $\sin \alpha > \frac{\sqrt{3}}{2}$,$\alpha \in (\pi/3, 2\pi/3)$.
For $\sin \alpha < -\frac{\sqrt{3}}{2}$,$\alpha \in (4\pi/3, 5\pi/3)$.

(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:
$2x = 2 + 2 \sin^2 2\theta + 4 \sin 2\theta = 2(1 + 2 \sin 2\theta + \sin^2 2\theta) = 2(1 + \sin 2\theta)^2$
$x = (1 + \sin 2\theta)^2 \implies \sqrt{x} = 1 + \sin 2\theta$
Subtracting the two equations:
$2y = 2 + 2 \sin^2 2\theta - 4 \sin 2\theta = 2(1 - 2 \sin 2\theta + \sin^2 2\theta) = 2(1 - \sin 2\theta)^2$
$y = (1 - \sin 2\theta)^2 \implies \sqrt{y} = 1 - \sin 2\theta$
Adding $\sqrt{x}$ and $\sqrt{y}$:
$\sqrt{x} + \sqrt{y} = (1 + \sin 2\theta) + (1 - \sin 2\theta) = 2$
Thus,the correct option is $D$.

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

(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 can rewrite the equation as $(a + 1)^2 + 1 + \cot^2 \left( \frac{\pi}{2}(a + x) \right) = 1$.
Subtracting $1$ from both sides,we get $(a + 1)^2 + \cot^2 \left( \frac{\pi}{2}(a + x) \right) = 0$.
Since both terms are squares of real numbers,their sum can be zero only if each term is zero individually.
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 $\cot \left( \frac{\pi}{2}(x - 1) \right) = 0$.
This occurs when the argument is an odd multiple of $\frac{\pi}{2}$,i.e.,$\frac{\pi}{2}(x - 1) = (2n + 1) \frac{\pi}{2}$ for some integer $n$.
$x - 1 = 2n + 1 \Rightarrow x = 2n + 2 = 2(n + 1)$.
Thus,$\frac{x}{2} = n + 1$,which is an integer $(I)$.

(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 $T_k = \frac{\cos(2 \cdot 3^{k-1} x)}{\sin(3^k x)}$ for $k = 1, 2, 3$.
Using the identity $\cos(A) \csc(B) = \frac{\cos A}{\sin B}$,we multiply numerator and denominator by $\sin(3^{k-1} x)$:
$T_k = \frac{\cos(2 \cdot 3^{k-1} x) \sin(3^{k-1} x)}{\sin(3^k x) \sin(3^{k-1} x)}$.
Using $2 \sin A \cos B = \sin(A+B) - \sin(B-A)$,we get:
$T_k = \frac{1}{2} \frac{\sin(3^k x) - \sin(3^{k-1} x)}{\sin(3^k x) \sin(3^{k-1} x)} = \frac{1}{2} (\csc(3^{k-1} x) - \csc(3^k x))$.
Summing for $k=1, 2, 3$:
$S = \frac{1}{2} [(\csc x - \csc 3x) + (\csc 3x - \csc 9x) + (\csc 9x - \csc 27x)]$.
$S = \frac{1}{2} (\csc x - \csc 27x)$.
Since $27x = 27\frac{\pi}{28} = \pi - \frac{\pi}{28} = \pi - x$,we have $\csc(27x) = \csc(\pi - x) = \csc x$.
Thus,$S = \frac{1}{2} (\csc x - \csc x) = 0$.