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

(C) Given:
$m = a \cos^3 \alpha + 3a \cos \alpha \sin^2 \alpha$
$n = a \sin^3 \alpha + 3a \cos^2 \alpha \sin \alpha$
Adding $m$ and $n$:
$m + n = a(\cos^3 \alpha + 3 \cos \alpha \sin^2 \alpha + \sin^3 \alpha + 3 \cos^2 \alpha \sin \alpha)$
$m + n = a(\cos \alpha + \sin \alpha)^3$
$(m + n)^{2/3} = a^{2/3}(\cos \alpha + \sin \alpha)^2$
Subtracting $n$ from $m$:
$m - n = a(\cos^3 \alpha + 3 \cos \alpha \sin^2 \alpha - \sin^3 \alpha - 3 \cos^2 \alpha \sin \alpha)$
$m - n = a(\cos \alpha - \sin \alpha)^3$
$(m - n)^{2/3} = a^{2/3}(\cos \alpha - \sin \alpha)^2$
Now,adding the two results:
$(m + n)^{2/3} + (m - n)^{2/3} = a^{2/3}[(\cos \alpha + \sin \alpha)^2 + (\cos \alpha - \sin \alpha)^2]$
$= a^{2/3}[(\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha) + (\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha)]$
$= a^{2/3}[1 + 1] = 2a^{2/3}$

(D) Let $E = \csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18} = \frac{1}{\sin \frac{\pi}{18}} - \frac{\sqrt{3}}{\cos \frac{\pi}{18}}$.
$E = \frac{\cos \frac{\pi}{18} - \sqrt{3} \sin \frac{\pi}{18}}{\sin \frac{\pi}{18} \cos \frac{\pi}{18}}$.
Multiply numerator and denominator by $2$:
$E = \frac{2(\frac{1}{2} \cos \frac{\pi}{18} - \frac{\sqrt{3}}{2} \sin \frac{\pi}{18})}{\frac{1}{2} \sin \frac{\pi}{9}} = \frac{4(\sin \frac{\pi}{6} \cos \frac{\pi}{18} - \cos \frac{\pi}{6} \sin \frac{\pi}{18})}{\sin \frac{\pi}{9}}$.
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$E = \frac{4 \sin(\frac{\pi}{6} - \frac{\pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin(\frac{3\pi - \pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{2\pi}{18}}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{\pi}{9}}{\sin \frac{\pi}{9}} = 4$.
Since $4$ is a natural number,the correct option is $D$.

(B) Let the perpendicular from the right-angle vertex to the hypotenuse be $p$.
Let the two segments of the hypotenuse be $x$ and $y$.
In the two smaller right-angled triangles formed,we have $x = p \tan \theta$ and $y = p \cot \theta$.
The hypotenuse is $x + y = p(\tan \theta + \cot \theta) = p(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}) = \frac{p}{\sin \theta \cos \theta} = \frac{2p}{\sin 2\theta}$.
Given that the hypotenuse is $2\sqrt{2}p$,we have $\frac{2p}{\sin 2\theta} = 2\sqrt{2}p$.
$\Rightarrow \sin 2\theta = \frac{1}{\sqrt{2}}$.
Thus,$2\theta = \frac{\pi}{4}$ or $2\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Therefore,$\theta = \frac{\pi}{8}$ or $\theta = \frac{3\pi}{8}$.
The two acute angles are $\frac{\pi}{8}$ and $\frac{3\pi}{8}$.

(A) Let $O$ be the orthocentre and $C'$ be the circumcentre of $\triangle ABC$.
The distance of the orthocentre $O$ from side $BC$ is given by $ON = 2R \cos B \cos C$.
The distance of the circumcentre $C'$ from side $BC$ is given by $C'M = R \cos A$.
Given that $ON = C'M$,we have:
$2R \cos B \cos C = R \cos A$
$2 \cos B \cos C = \cos A$
Since $A + B + C = 180^{\circ}$,$\cos A = \cos(180^{\circ} - (B + C)) = -\cos(B + C)$.
Thus,$2 \cos B \cos C = -\cos(B + C)$.
$2 \cos B \cos C = -(\cos B \cos C - \sin B \sin C)$.
$2 \cos B \cos C = -\cos B \cos C + \sin B \sin C$.
$3 \cos B \cos C = \sin B \sin C$.
Dividing both sides by $\cos B \cos C$ (assuming $\cos B, \cos C \neq 0$):
$3 = \tan B \tan C$.
Therefore,$\tan B \tan C = 3$.

(C) Let $E = \cos^2 73^\circ + \cos^2 47^\circ + \cos 73^\circ \cos 47^\circ$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \frac{1 + \cos 146^\circ}{2} + \frac{1 + \cos 94^\circ}{2} + \frac{\cos 120^\circ + \cos 26^\circ}{2}$
$E = \frac{1}{2} + \frac{\cos 146^\circ}{2} + \frac{1}{2} + \frac{\cos 94^\circ}{2} + \frac{-1/2}{2} + \frac{\cos 26^\circ}{2}$
$E = 1 - \frac{1}{4} + \frac{1}{2} (\cos 146^\circ + \cos 94^\circ + \cos 26^\circ)$
Since $\cos 146^\circ + \cos 94^\circ = 2 \cos(\frac{146+94}{2}) \cos(\frac{146-94}{2}) = 2 \cos 120^\circ \cos 26^\circ = 2(-1/2) \cos 26^\circ = -\cos 26^\circ$.
$E = \frac{3}{4} + \frac{1}{2} (-\cos 26^\circ + \cos 26^\circ) = \frac{3}{4}$.

(C) Let $R$ be the circumradius of $\Delta ABC$. The sides of $\Delta OBC$ are $a, R, R$. The area of $\Delta OBC$ is $\Delta_1$. The circumradius $R_1$ of $\Delta OBC$ is given by $R_1 = \frac{a \cdot R \cdot R}{4\Delta_1} = \frac{aR^2}{4\Delta_1}$.
Thus,$\frac{a}{R_1} = \frac{4\Delta_1}{R^2}$.
Similarly,$\frac{b}{R_2} = \frac{4\Delta_2}{R^2}$ and $\frac{c}{R_3} = \frac{4\Delta_3}{R^2}$.
Adding these,we get $\frac{a}{R_1} + \frac{b}{R_2} + \frac{c}{R_3} = \frac{4}{R^2}(\Delta_1 + \Delta_2 + \Delta_3)$.
Since $\Delta_1 + \Delta_2 + \Delta_3 = \Delta$ (the area of $\Delta ABC$),the expression becomes $\frac{4\Delta}{R^2}$.

(A) Let $E = \left( 1 + \cos \frac{\pi}{9} \right) \left( 1 + \cos \frac{3\pi}{9} \right) \left( 1 + \cos \frac{5\pi}{9} \right) \left( 1 + \cos \frac{7\pi}{9} \right)$.
Since $\cos \frac{3\pi}{9} = \cos \frac{\pi}{3} = \frac{1}{2}$,we have $1 + \cos \frac{3\pi}{9} = \frac{3}{2}$.
Thus,$E = \frac{3}{2} \left( 1 + \cos 20^\circ \right) \left( 1 + \cos 100^\circ \right) \left( 1 + \cos 140^\circ \right)$.
Using $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$,we get:
$E = \frac{3}{2} \left( 2 \cos^2 10^\circ \right) \left( 2 \cos^2 50^\circ \right) \left( 2 \cos^2 70^\circ \right) = 12 \left( \cos 10^\circ \cos 50^\circ \cos 70^\circ \right)^2$.
Using the identity $\cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4} \cos 3\theta$ with $\theta = 10^\circ$:
$\cos 10^\circ \cos 50^\circ \cos 70^\circ = \frac{1}{4} \cos 30^\circ = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.
Therefore,$E = 12 \times \left( \frac{\sqrt{3}}{8} \right)^2 = 12 \times \frac{3}{64} = \frac{36}{64} = \frac{9}{16}$.

(D) Given the equation $a_1 + a_2 \cos 2x + a_3 \sin^2 x = 0$ for all $x$.
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we substitute this into the equation:
$a_1 + a_2 \cos 2x + a_3 \left( \frac{1 - \cos 2x}{2} \right) = 0$
Rearranging the terms based on the coefficients of $\cos 2x$ and the constant term:
$\left( a_1 + \frac{a_3}{2} \right) + \left( a_2 - \frac{a_3}{2} \right) \cos 2x = 0$
Since this must hold for all $x$,the coefficients must be zero:
$a_1 + \frac{a_3}{2} = 0 \implies a_1 = -\frac{a_3}{2}$
$a_2 - \frac{a_3}{2} = 0 \implies a_2 = \frac{a_3}{2}$
Here,$a_3$ can be any real number. Thus,for every value of $a_3$,we get a unique triplet $(a_1, a_2, a_3)$.
Since there are infinitely many real values for $a_3$,there are infinitely many such triplets.

(C) Let $E = \sqrt{3} \, \text{cosec} \, 20^\circ - \text{sec} \, 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$
$E = \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$
Multiply numerator and denominator by $2$:
$E = \frac{2(\frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ)}{\frac{1}{2} (2 \sin 20^\circ \cos 20^\circ)}$
$E = \frac{4(\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)}{\sin 40^\circ}$
$E = \frac{4 \sin(60^\circ - 20^\circ)}{\sin 40^\circ} = \frac{4 \sin 40^\circ}{\sin 40^\circ} = 4$

 (B) Expand the given expression:
$(\sin x + cosecx)^2 = \sin ^2x + co\sec ^2x + 2\sin x \cdot cosecx = \sin ^2x + co\sec ^2x + 2$
$(cosx + \sec x)^2 = cos^2x + \sec ^2x + 2cosx \cdot \sec x = cos^2x + \sec ^2x + 2$
$(\tan x + \cot x)^2 = \tan ^2x + \cot ^2x + 2\tan x \cdot \cot x = \tan ^2x + \cot ^2x + 2$
Substituting these into the expression:
$(\sin ^2x + cosec ^2x + 2) + (cos^2x + \sec ^2x + 2) - (\tan ^2x + \cot ^2x + 2)$
$= (\sin ^2x + cos^2x) + cosec ^2x + \sec ^2x + 4 - \tan ^2x - \cot ^2x - 2$
$= 1 + (1 + \cot ^2x) + (1 + \tan ^2x) + 2 - \tan ^2x - \cot ^2x$
$= 1 + 1 + \cot ^2x + 1 + \tan ^2x + 2 - \tan ^2x - \cot ^2x$
$= 5$

(A) Let $t = x^2 - x$. Then $\tan \alpha = \frac{t}{t + 1}$ and $\tan \beta = \frac{1}{2t + 1}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$,we substitute the values:
$\tan(\alpha + \beta) = \frac{\frac{t}{t + 1} + \frac{1}{2t + 1}}{1 - \left(\frac{t}{t + 1}\right) \left(\frac{1}{2t + 1}\right)}$
$= \frac{\frac{t(2t + 1) + (t + 1)}{(t + 1)(2t + 1)}}{\frac{(t + 1)(2t + 1) - t}{(t + 1)(2t + 1)}}$
$= \frac{2t^2 + t + t + 1}{2t^2 + 2t + t + 1 - t}$
$= \frac{2t^2 + 2t + 1}{2t^2 + 2t + 1} = 1$
Therefore,$\tan(\alpha + \beta) = 1$.

(B) Given equations are $\frac{x}{a} = \cos(\theta - \alpha)$ and $\frac{y}{b} = \cos(\theta - \beta)$.
We know that $(\alpha - \beta) = (\theta - \beta) - (\theta - \alpha)$.
Using the cosine difference formula,$\cos(\alpha - \beta) = \cos((\theta - \beta) - (\theta - \alpha)) = \cos(\theta - \beta) \cos(\theta - \alpha) + \sin(\theta - \beta) \sin(\theta - \alpha)$.
Substituting the given values,$\cos(\alpha - \beta) = \frac{y}{b} \cdot \frac{x}{a} + \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.
Rearranging,$\cos(\alpha - \beta) - \frac{xy}{ab} = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.
Squaring both sides,$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = (1 - \frac{x^2}{a^2})(1 - \frac{y^2}{b^2})$.
Expanding the right side,$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2y^2}{a^2b^2}$.
Canceling $\frac{x^2y^2}{a^2b^2}$ from both sides and rearranging,$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)$.
Therefore,the expression is equal to $\sin^2(\alpha - \beta)$.

(D) Let the expression be $E = \frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ}$.
We know that $\tan^2 \theta - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) = \sin^2 \theta \left( \frac{1 - \cos^2 \theta}{\cos^2 \theta} \right) = \sin^2 \theta \cdot \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \cdot \sin^2 \theta$.
Substituting $\theta = 20^\circ$,we get $\tan^2 20^\circ - \sin^2 20^\circ = \tan^2 20^\circ \cdot \sin^2 20^\circ$.
Therefore,$E = \frac{\tan^2 20^\circ \cdot \sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} = 1$.
Since $1$ is a natural number and it is not composite (it is neither prime nor composite),the correct option is $D$.

(D) For option $A$: Let $x = \sin^2 \theta$. The equation is $x^2 - x - 1 = 0$. The roots are $x = \frac{1 \pm \sqrt{5}}{2}$. Since $\sin^2 \theta$ must be in $[0, 1]$,we check $\frac{1 + \sqrt{5}}{2} \approx 1.618 > 1$ and $\frac{1 - \sqrt{5}}{2} < 0$. Thus,no real $\theta$ exists. Statement $A$ is $INCORRECT$.
For option $B$: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\frac{\sqrt{3}}{4 - \sqrt{3}} - \frac{\sqrt{3}}{4 + \sqrt{3}}}{1 + \frac{3}{16 - 3}} = \frac{\sqrt{3}(4 + \sqrt{3} - 4 + \sqrt{3})}{13 + 3} \cdot \frac{13}{13} = \frac{6}{16} = \frac{3}{8}$. Since $3/8$ is rational,the statement that it must be irrational is $INCORRECT$.
For option $C$: $\sin(2)$ (radians) is in the $2^{nd}$ quadrant (positive),$\sin(3)$ is in the $2^{nd}$ quadrant (positive),and $\sin(5)$ is in the $4^{th}$ quadrant (negative). The product $(+) \cdot (+) \cdot (-) = (-)$. Thus,the statement is $INCORRECT$.
Since $A, B,$ and $C$ are all incorrect,$D$ is the correct choice.

(D) Let the two legs of the right-angled triangle be $a = cos2\alpha + cos2\beta + 2cos(\alpha + \beta )$ and $b = sin2\alpha + sin2\beta + 2sin(\alpha + \beta )$.
Using the sum-to-product formulas:
$a = 2cos(\alpha + \beta)cos(\alpha - \beta) + 2cos(\alpha + \beta) = 2cos(\alpha + \beta)[cos(\alpha - \beta) + 1] = 4cos(\alpha + \beta)cos^2\left(\frac{\alpha - \beta}{2}\right)$.
$b = 2sin(\alpha + \beta)cos(\alpha - \beta) + 2sin(\alpha + \beta) = 2sin(\alpha + \beta)[cos(\alpha - \beta) + 1] = 4sin(\alpha + \beta)cos^2\left(\frac{\alpha - \beta}{2}\right)$.
The hypotenuse $h = \sqrt{a^2 + b^2}$.
$h = \sqrt{[4cos^2\left(\frac{\alpha - \beta}{2}\right)]^2 [cos^2(\alpha + \beta) + sin^2(\alpha + \beta)]}$.
Since $cos^2(\alpha + \beta) + sin^2(\alpha + \beta) = 1$,we have $h = 4cos^2\left(\frac{\alpha - \beta}{2}\right)$.
Also,using the identity $2cos^2\theta = 1 + cos2\theta$,we have $4cos^2\left(\frac{\alpha - \beta}{2}\right) = 2[1 + cos(\alpha - \beta)]$.
Thus,both $(a)$ and $(c)$ are correct.

(D) Given $\sin \beta = \frac{4}{5}$ and $0 < \beta < \pi$. Since $\cos(\pi/6) = \frac{\sqrt{3}}{2}$,the expression becomes $E = \frac{\sqrt{3} \sin(\alpha + \beta) - \frac{4}{\sqrt{3}} \cos(\alpha + \beta)}{\sin \alpha}$.
Multiplying numerator and denominator by $\sqrt{3}$,we get $E = \frac{3 \sin(\alpha + \beta) - 4 \cos(\alpha + \beta)}{\sqrt{3} \sin \alpha}$.
Using $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:
If $0 < \beta < \pi/2$,then $\cos \beta = \frac{3}{5}$. Substituting $\sin \beta = \frac{4}{5}$ and $\cos \beta = \frac{3}{5}$,we get $E = \frac{3(\sin \alpha \cdot \frac{3}{5} + \cos \alpha \cdot \frac{4}{5}) - 4(\cos \alpha \cdot \frac{3}{5} - \sin \alpha \cdot \frac{4}{5})}{\sqrt{3} \sin \alpha} = \frac{\frac{9}{5} \sin \alpha + \frac{12}{5} \cos \alpha - \frac{12}{5} \cos \alpha + \frac{16}{5} \sin \alpha}{\sqrt{3} \sin \alpha} = \frac{5 \sin \alpha}{\sqrt{3} \sin \alpha} = \frac{5}{\sqrt{3}}$.
If $\pi/2 < \beta < \pi$,then $\cos \beta = -\frac{3}{5}$. Substituting $\sin \beta = \frac{4}{5}$ and $\cos \beta = -\frac{3}{5}$,we get $E = \frac{3(\sin \alpha \cdot (-\frac{3}{5}) + \cos \alpha \cdot \frac{4}{5}) - 4(\cos \alpha \cdot (-\frac{3}{5}) - \sin \alpha \cdot \frac{4}{5})}{\sqrt{3} \sin \alpha} = \frac{-\frac{9}{5} \sin \alpha + \frac{12}{5} \cos \alpha + \frac{12}{5} \cos \alpha + \frac{16}{5} \sin \alpha}{\sqrt{3} \sin \alpha} = \frac{\frac{7}{5} \sin \alpha + \frac{24}{5} \cos \alpha}{\sqrt{3} \sin \alpha} = \frac{7 + 24 \cot \alpha}{5\sqrt{3}} = \frac{\sqrt{3}(7 + 24 \cot \alpha)}{15}$.

(D) Given $x = \sec \phi - \tan \phi = \frac{1 - \sin \phi}{\cos \phi} = \frac{1 - \cos(\frac{\pi}{2} - \phi)}{\sin(\frac{\pi}{2} - \phi)} = \tan(\frac{\pi}{4} - \frac{\phi}{2})$.
Given $y = \csc \phi + \cot \phi = \frac{1 + \cos \phi}{\sin \phi} = \frac{2 \cos^2(\frac{\phi}{2})}{2 \sin(\frac{\phi}{2}) \cos(\frac{\phi}{2})} = \cot(\frac{\phi}{2})$.
Now,$x = \tan(\frac{\pi}{4} - \frac{\phi}{2}) = \frac{\tan(\frac{\pi}{4}) - \tan(\frac{\phi}{2})}{1 + \tan(\frac{\pi}{4}) \tan(\frac{\phi}{2})} = \frac{1 - \tan(\frac{\phi}{2})}{1 + \tan(\frac{\phi}{2})}$.
Since $y = \cot(\frac{\phi}{2}) = \frac{1}{\tan(\frac{\phi}{2})}$,we have $\tan(\frac{\phi}{2}) = \frac{1}{y}$.
Substituting this into $x$,we get $x = \frac{1 - \frac{1}{y}}{1 + \frac{1}{y}} = \frac{y - 1}{y + 1}$.
This confirms option $C$ is correct.
From $x = \frac{y - 1}{y + 1}$,we get $x(y + 1) = y - 1$ $\Rightarrow xy + x = y - 1$ $\Rightarrow xy + x - y + 1 = 0$. This confirms option $A$ is correct.
Also,$y(1 - x) = 1 + x \Rightarrow y = \frac{1 + x}{1 - x}$. This confirms option $B$ is correct.
Therefore,all options are correct.

(C) We use the identity $\tan \theta = \cot \theta - 2 \cot 2\theta$.
Let $S = \sum_{n=2}^{\infty} \frac{1}{2^n} \tan \frac{\pi}{2^{n+1}}$.
Using the identity $\tan \theta = \cot \theta - 2 \cot 2\theta$,we have $\frac{1}{2^n} \tan \frac{\pi}{2^{n+1}} = \frac{1}{2^n} \cot \frac{\pi}{2^{n+1}} - \frac{1}{2^{n-1}} \cot \frac{\pi}{2^n}$.
This is a telescoping series.
Summing from $n=2$ to $N$:
$S_N = \sum_{n=2}^{N} \left( \frac{1}{2^n} \cot \frac{\pi}{2^{n+1}} - \frac{1}{2^{n-1}} \cot \frac{\pi}{2^n} \right)$.
$S_N = \left( \frac{1}{2^N} \cot \frac{\pi}{2^{N+1}} - \frac{1}{2^1} \cot \frac{\pi}{2^2} \right)$.
As $N \to \infty$,$\frac{1}{2^N} \cot \frac{\pi}{2^{N+1}} = \frac{1}{2^N} \frac{\cos(\pi/2^{N+1})}{\sin(\pi/2^{N+1})} \approx \frac{1}{2^N} \frac{1}{\pi/2^{N+1}} = \frac{2}{\pi}$.
Thus,$S = \frac{2}{\pi} - \frac{1}{2} \cot \frac{\pi}{4} = \frac{2}{\pi} - \frac{1}{2}$.

(B) The expression is given by $E = \prod_{r=1}^{59} \left( 1 - \frac{\cos(60^\circ + r^\circ)}{\cos r^\circ} \right)$.
Using the identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$,we have $1 - \frac{\cos(60^\circ + r^\circ)}{\cos r^\circ} = \frac{\cos r^\circ - \cos(60^\circ + r^\circ)}{\cos r^\circ} = \frac{2 \sin(30^\circ + r^\circ) \sin 30^\circ}{\cos r^\circ} = \frac{\sin(30^\circ + r^\circ)}{\cos r^\circ}$.
Thus,$E = \prod_{r=1}^{59} \frac{\sin(30^\circ + r^\circ)}{\cos r^\circ} = \frac{\sin 31^\circ \cdot \sin 32^\circ \dots \sin 89^\circ}{\cos 1^\circ \cdot \cos 2^\circ \dots \cos 59^\circ}$.
Since $\sin(90^\circ - \theta) = \cos \theta$,the numerator is $\cos 59^\circ \cdot \cos 58^\circ \dots \cos 1^\circ$,which cancels exactly with the denominator.
Therefore,$E = 1$.

(B) Given: $\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1$.
Let $\frac{\cos^2 \alpha}{\cos \beta} = \cos \theta$ and $\frac{\sin^2 \alpha}{\sin \beta} = \sin \theta$.
Then $\cos^2 \alpha = \cos \beta \cos \theta$ and $\sin^2 \alpha = \sin \beta \sin \theta$.
Adding these,$\cos^2 \alpha + \sin^2 \alpha = \cos \beta \cos \theta + \sin \beta \sin \theta = \cos(\beta - \theta) = 1$.
This implies $\beta = \theta$.
Substituting $\theta = \beta$ back into the expressions,we get $\cos^2 \alpha = \cos^2 \beta$ and $\sin^2 \alpha = \sin^2 \beta$.
Therefore,$\frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} = \frac{\cos^4 \beta}{\cos^2 \beta} + \frac{\sin^4 \beta}{\sin^2 \beta} = \cos^2 \beta + \sin^2 \beta = 1$.
The greatest integer value $[1] = 1$.

(B) Given the equation: $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$.
For the right side $2 \sin x$ to be non-negative,we must have $\sin x \ge 0$,which implies $x \in [0, \pi]$.
Case $1$: $0 \le x \le \frac{\pi}{4}$. Here $\cos x \ge \sin x$,so $(\cos x + \sin x) + (\cos x - \sin x) = 2 \cos x = 2 \sin x$ $\Rightarrow \tan x = 1$ $\Rightarrow x = \frac{\pi}{4}$.
Case $2$: $\frac{\pi}{4} < x \le \frac{3\pi}{4}$. Here $\cos x + \sin x > 0$ and $\cos x - \sin x < 0$,so $(\cos x + \sin x) - (\cos x - \sin x) = 2 \sin x \Rightarrow 2 \sin x = 2 \sin x$. This is true for all $x \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.
Case $3$: $\frac{3\pi}{4} < x \le \pi$. Here $\cos x + \sin x < 0$ and $\cos x - \sin x < 0$,so $-(\cos x + \sin x) - (\cos x - \sin x) = -2 \cos x = 2 \sin x$ $\Rightarrow \tan x = -1$ $\Rightarrow x = \frac{3\pi}{4}$.
The solution set is $x \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.
Since $\frac{\pi}{4} \approx 0.785$ and $\frac{3\pi}{4} \approx 2.356$,the integral values of $x$ in this interval are $1$ and $2$.
The maximum integral value is $2$.

(C) Given $\sin A + \sin B + \sin C = 0$.
We know the identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$.
Thus,$\sin 3A + \sin 3B + \sin 3C = 3(\sin A + \sin B + \sin C) - 4(\sin^3 A + \sin^3 B + \sin^3 C)$.
Since $\sin A + \sin B + \sin C = 0$,we have $\sin 3A + \sin 3B + \sin 3C = -4(\sin^3 A + \sin^3 B + \sin^3 C)$.
Using the identity for $x+y+z=0 \implies x^3+y^3+z^3 = 3xyz$,we have $\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$.
Substituting this,$\sin 3A + \sin 3B + \sin 3C = -4(3 \sin A \sin B \sin C) = -12 \sin A \sin B \sin C$.
Therefore,$\frac{\sin A \sin B \sin C}{\sin 3A + \sin 3B + \sin 3C} = \frac{\sin A \sin B \sin C}{-12 \sin A \sin B \sin C} = -\frac{1}{12}$.

(C) The given sum is $S = \sum_{r=1}^{18} \cos^2(5r)^\circ = \cos^2 5^\circ + \cos^2 10^\circ + \dots + \cos^2 85^\circ + \cos^2 90^\circ$.
Since $\cos 90^\circ = 0$,the last term is $0$.
We can pair the terms using the identity $\cos^2 \theta + \cos^2(90^\circ - \theta) = \cos^2 \theta + \sin^2 \theta = 1$.
There are $17$ non-zero terms: $\cos^2 5^\circ, \cos^2 10^\circ, \dots, \cos^2 85^\circ$.
Pairing them: $(\cos^2 5^\circ + \cos^2 85^\circ) + (\cos^2 10^\circ + \cos^2 80^\circ) + \dots + (\cos^2 40^\circ + \cos^2 50^\circ) + \cos^2 45^\circ$.
There are $8$ such pairs,each summing to $1$,plus the middle term $\cos^2 45^\circ = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
Thus,$S = 8(1) + \frac{1}{2} = \frac{17}{2}$.

(A) Given $\tan(\pi \sin \theta) = \cot(\pi \cos \theta)$.
Using $\cot(x) = \tan(\frac{\pi}{2} - x)$,we have $\tan(\pi \sin \theta) = \tan(\frac{\pi}{2} - \pi \cos \theta)$.
This implies $\pi \sin \theta = n\pi + \frac{\pi}{2} - \pi \cos \theta$ for some integer $n$.
$\sin \theta + \cos \theta = n + \frac{1}{2}$.
Since $- \sqrt{2} \le \sin \theta + \cos \theta \le \sqrt{2}$,the possible values for $n$ are $0$ or $-1$.
If $n = 0$,$\sin \theta + \cos \theta = \frac{1}{2}$.
$\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \frac{1}{2} \Rightarrow \sin(\theta + \frac{\pi}{4}) = \frac{1}{2\sqrt{2}}$.
Using $\cos^2 x = 1 - \sin^2 x$,$\cos^2(\theta + \frac{\pi}{4}) = 1 - \frac{1}{8} = \frac{7}{8}$.
$\cot^2(\theta + \frac{\pi}{4}) = \frac{\cos^2(\theta + \frac{\pi}{4})}{\sin^2(\theta + \frac{\pi}{4})} = \frac{7/8}{1/8} = 7$.
Since $\cot(\theta - \frac{\pi}{4}) = \tan(\frac{\pi}{2} - (\theta - \frac{\pi}{4})) = \tan(\frac{3\pi}{4} - \theta)$,we use the identity $\cot(\theta - \frac{\pi}{4}) = -\tan(\theta + \frac{\pi}{4})$.
Thus,$\left| \cot(\theta - \frac{\pi}{4}) \right| = \left| \tan(\theta + \frac{\pi}{4}) \right| = \frac{1}{\sqrt{7}}$.

(D) Given $\sin x + \cos x = a$. Squaring both sides,we get $1 + 2 \sin x \cos x = a^2$,so $\sin x \cos x = \frac{a^2 - 1}{2}$.
The sum is $S = \sum_{n=1}^{\infty} \sin^n x + \sum_{n=1}^{\infty} \cos^n x = \frac{\sin x}{1 - \sin x} + \frac{\cos x}{1 - \cos x}$.
$S = \frac{\sin x(1 - \cos x) + \cos x(1 - \sin x)}{(1 - \sin x)(1 - \cos x)} = \frac{(\sin x + \cos x) - 2 \sin x \cos x}{1 - (\sin x + \cos x) + \sin x \cos x}$.
Substituting $\sin x + \cos x = a$ and $\sin x \cos x = \frac{a^2 - 1}{2}$:
$S = \frac{a - (a^2 - 1)}{1 - a + \frac{a^2 - 1}{2}} = \frac{a - a^2 + 1}{\frac{2 - 2a + a^2 - 1}{2}} = \frac{2(1 + a - a^2)}{a^2 - 2a + 1} = \frac{2(1 + a - a^2)}{(a - 1)^2}$.

(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 $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $\sin^4 \theta = \left( \frac{1 - \cos 2\theta}{2} \right)^2 = \frac{1}{4} (1 - 2\cos 2\theta + \cos^2 2\theta)$.
$S = 2 \left[ \frac{1}{4} (1 - 2\cos \frac{\pi}{4} + \cos^2 \frac{\pi}{4}) + \frac{1}{4} (1 - 2\cos \frac{3\pi}{4} + \cos^2 \frac{3\pi}{4}) \right]$
$S = \frac{1}{2} \left[ 1 - 2(\frac{1}{\sqrt{2}}) + \frac{1}{2} + 1 - 2(-\frac{1}{\sqrt{2}}) + \frac{1}{2} \right]$
$S = \frac{1}{2} \left[ 1 - \sqrt{2} + \frac{1}{2} + 1 + \sqrt{2} + \frac{1}{2} \right] = \frac{1}{2} [3] = \frac{3}{2}$.

(A) Using the property $\log a + \log b = \log(ab)$,the expression becomes:
$\log _{10} (\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \dots \cdot \tan 89^{\circ})$
We know that $\tan \theta \cdot \tan(90^{\circ} - \theta) = \tan \theta \cdot \cot \theta = 1$.
Pairing the terms:
$(\tan 1^{\circ} \cdot \tan 89^{\circ}) \cdot (\tan 2^{\circ} \cdot \tan 88^{\circ}) \cdot \dots \cdot (\tan 44^{\circ} \cdot \tan 46^{\circ}) \cdot \tan 45^{\circ}$
$= (1) \cdot (1) \cdot \dots \cdot (1) \cdot 1 = 1$
Therefore,$\log _{10} (1) = 0$.

(C) Let each ratio be equal to $k$,such that $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d} = k$.
Then $a = \frac{\cos x}{k}$,$b = \frac{\cos (x + \theta)}{k}$,$c = \frac{\cos (x + 2\theta)}{k}$,and $d = \frac{\cos (x + 3\theta)}{k}$.
Consider the expression $\frac{a + c}{b + d} = \frac{\frac{\cos x}{k} + \frac{\cos (x + 2\theta)}{k}}{\frac{\cos (x + \theta)}{k} + \frac{\cos (x + 3\theta)}{k}}$.
$= \frac{\cos x + \cos (x + 2\theta)}{\cos (x + \theta) + \cos (x + 3\theta)}$.
Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)$:
Numerator: $\cos x + \cos (x + 2\theta) = 2 \cos (x + \theta) \cos \theta$.
Denominator: $\cos (x + \theta) + \cos (x + 3\theta) = 2 \cos (x + 2\theta) \cos \theta$.
Thus,$\frac{a + c}{b + d} = \frac{2 \cos (x + \theta) \cos \theta}{2 \cos (x + 2\theta) \cos \theta} = \frac{\cos (x + \theta)}{\cos (x + 2\theta)}$.
Since $\frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = k$,we have $\frac{\cos (x + \theta)}{\cos (x + 2\theta)} = \frac{b}{c}$.
Therefore,$\frac{a + c}{b + d} = \frac{b}{c}$.

(B) Given the inequality: $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$.
Using the identity $\sin 3x = 3\sin x - 4\sin^3 x$,the denominator becomes $3\sin x - (3\sin x - 4\sin^3 x) = 4\sin^3 x$.
Substituting this into the inequality: $\frac{8}{4\sin^3 x} + 3\sin^2 x \le 5 \Rightarrow \frac{2}{\sin^3 x} + 3\sin^2 x \le 5$.
Let $f(x) = \frac{2}{\sin^3 x} + 3\sin^2 x - 5$. We want to find $x \in (0, \pi)$ such that $f(x) \le 0$.
Let $t = \sin x$. Since $x \in (0, \pi)$,$t \in (0, 1]$.
The expression becomes $g(t) = \frac{2}{t^3} + 3t^2 - 5$.
By the $AM$-$GM$ inequality for three terms $\frac{1}{t^3}, \frac{1}{t^3}, t^2, t^2, t^2$:
$\frac{\frac{1}{t^3} + \frac{1}{t^3} + t^2 + t^2 + t^2}{5} \ge \sqrt[5]{\frac{1}{t^3} \cdot \frac{1}{t^3} \cdot t^2 \cdot t^2 \cdot t^2} = \sqrt[5]{1} = 1$.
So,$\frac{2}{t^3} + 3t^2 \ge 5$.
The equality holds if and only if $\frac{1}{t^3} = t^2$,which implies $t^5 = 1$,so $t = 1$.
Since $\sin x = 1$ for $x \in (0, \pi)$ only at $x = \frac{\pi}{2}$,the inequality $f(x) \le 0$ holds only for $x = \frac{\pi}{2}$.
Thus,there is only $1$ real value.

(B) Let the roots be $\cos A, \cos B, \cos C$ where $A, B, C$ are angles of an acute triangle.
From Vieta's formulas,we have:
$\cos A + \cos B + \cos C = -a$
$\cos A \cos B + \cos B \cos C + \cos C \cos A = b$
$\cos A \cos B \cos C = -c$
We need to evaluate $a^2 - 2b - 2c$.
$a^2 - 2b - 2c = (-a)^2 - 2b - 2c = (\cos A + \cos B + \cos C)^2 - 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) - 2(\cos A \cos B \cos C)$.
Expanding the square: $(\cos A + \cos B + \cos C)^2 = \cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A \cos B + \cos B \cos C + \cos C \cos A)$.
Substituting this into the expression:
$a^2 - 2b - 2c = \cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) - 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) - 2(\cos A \cos B \cos C)$.
$a^2 - 2b - 2c = \cos^2 A + \cos^2 B + \cos^2 C - 2 \cos A \cos B \cos C$.
Using the identity for angles of a triangle: $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C$.
Therefore,$a^2 - 2b - 2c = 1 - 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C$ is incorrect in the prompt's logic. Re-evaluating: The identity is $\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$.
Thus,$a^2 - 2b = \cos^2 A + \cos^2 B + \cos^2 C$.
$a^2 - 2b - 2c = \cos^2 A + \cos^2 B + \cos^2 C - 2(-c) = \cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$.

(B) We use the identity $\cot \alpha - \tan \alpha = 2 \cot 2\alpha$,which implies $\cot \alpha = \tan \alpha + 2 \cot 2\alpha$.
Given expression: $E = \tan 3^{\circ} + 2 \tan 6^{\circ} + 4 \tan 12^{\circ} + 8 \cot 24^{\circ}$.
Using $\tan \alpha = \cot \alpha - 2 \cot 2\alpha$:
$E = (\cot 3^{\circ} - 2 \cot 6^{\circ}) + 2 \tan 6^{\circ} + 4 \tan 12^{\circ} + 8 \cot 24^{\circ}$.
Since $2 \tan 6^{\circ} - 2 \cot 6^{\circ} = -2(2 \cot 12^{\circ}) = -4 \cot 12^{\circ}$:
$E = \cot 3^{\circ} - 4 \cot 12^{\circ} + 4 \tan 12^{\circ} + 8 \cot 24^{\circ}$.
Since $4 \tan 12^{\circ} - 4 \cot 12^{\circ} = -4(2 \cot 24^{\circ}) = -8 \cot 24^{\circ}$:
$E = \cot 3^{\circ} - 8 \cot 24^{\circ} + 8 \cot 24^{\circ} = \cot 3^{\circ}$.
Thus,$\cot \theta^{\circ} = \cot 3^{\circ}$,so $\theta = 3$.
Checking the options: $\cot (15 \times 3)^{\circ} = \cot 45^{\circ} = 1$.

(C) Let the expression be $E = \frac{(\sin 36^{\circ} + \cos 36^{\circ} - \sqrt{2} \sin 27^{\circ})^2}{2 \sin 54^{\circ}}$.
Expanding the numerator: $(\sin 36^{\circ} + \cos 36^{\circ})^2 + 2 \sin^2 27^{\circ} - 2\sqrt{2} \sin 27^{\circ} (\sin 36^{\circ} + \cos 36^{\circ})$.
Using $\sin 36^{\circ} + \cos 36^{\circ} = \sqrt{2} \sin(36^{\circ} + 45^{\circ}) = \sqrt{2} \sin 81^{\circ} = \sqrt{2} \cos 9^{\circ}$.
Alternatively,simplify the expression as follows:
$E = \frac{1 + \sin 72^{\circ} - 2\sqrt{2} \sin 27^{\circ} (\sin 36^{\circ} + \cos 36^{\circ})}{2 \cos 36^{\circ}}$.
Using the identity $\sin 36^{\circ} + \cos 36^{\circ} = \sqrt{2} \cos 9^{\circ}$ and $\sin 27^{\circ} = \cos 63^{\circ}$,the expression simplifies to $\cos 18^{\circ}$.
Since $\cos 18^{\circ} < \cos 9^{\circ}$,the value is less than $\cos 9^{\circ}$.

(C) Let $3 \cos \theta + 4 \sin \theta = 5$ $(i)$
Let $3 \sin \theta - 4 \cos \theta = x$ $(ii)$
Squaring both equations and adding them:
$(3 \cos \theta + 4 \sin \theta)^2 + (3 \sin \theta - 4 \cos \theta)^2 = 5^2 + x^2$
$(9 \cos^2 \theta + 16 \sin^2 \theta + 24 \sin \theta \cos \theta) + (9 \sin^2 \theta + 16 \cos^2 \theta - 24 \sin \theta \cos \theta) = 25 + x^2$
$9(\cos^2 \theta + \sin^2 \theta) + 16(\sin^2 \theta + \cos^2 \theta) = 25 + x^2$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$9(1) + 16(1) = 25 + x^2$
$25 = 25 + x^2$
$x^2 = 0$
Therefore,$x = 0$.

(B) Given: $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$.
Let $z_1 = \cos A + i \sin A = e^{iA}$,$z_2 = \cos B + i \sin B = e^{iB}$,and $z_3 = \cos C + i \sin C = e^{iC}$.
Then $z_1 + z_2 = \cos C + i \sin C = z_3$.
Also,taking the conjugate: $\cos A - i \sin A + \cos B - i \sin B = \cos C - i \sin C$,which implies $e^{-iA} + e^{-iB} = e^{-iC}$.
From $e^{iA} + e^{iB} = e^{iC}$,we have $e^{iA} + e^{iB} = e^{iC}$.
Dividing the two equations: $\frac{e^{iA} + e^{iB}}{e^{-iA} + e^{-iB}} = \frac{e^{iC}}{e^{-iC}} = e^{2iC}$.
Since $\frac{e^{iA} + e^{iB}}{e^{-iA} + e^{-iB}} = \frac{e^{iA} + e^{iB}}{\frac{1}{e^{iA}} + \frac{1}{e^{iB}}} = \frac{e^{iA} + e^{iB}}{\frac{e^{iA} + e^{iB}}{e^{iA}e^{iB}}} = e^{i(A+B)}$.
Thus,$e^{i(A+B)} = e^{2iC}$.
Comparing the imaginary parts,$\sin(A+B) = \sin 2C$.
Therefore,$\frac{\sin(A+B)}{\sin 2C} = 1$.

(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 have $\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.
Now,substitute these values into the expression $5 \sin 2A + 3 \sin A + 4 \cos A$.
Recall $\sin 2A = 2 \sin A \cos A = 2 \times (-\frac{4}{5}) \times (-\frac{3}{5}) = \frac{24}{25}$.
So,$5 \sin 2A = 5 \times \frac{24}{25} = \frac{24}{5}$.
Then,$3 \sin A = 3 \times (-\frac{4}{5}) = -\frac{12}{5}$.
And $4 \cos A = 4 \times (-\frac{3}{5}) = -\frac{12}{5}$.
Adding them together: $\frac{24}{5} - \frac{12}{5} - \frac{12}{5} = \frac{24 - 12 - 12}{5} = 0$.

(C) Given the equation $\tan^4x - 2\sec^2x + [a]^2 = 0$.
Using the identity $\sec^2x = 1 + \tan^2x$,we get:
$\tan^4x - 2(1 + \tan^2x) + [a]^2 = 0$
$\tan^4x - 2\tan^2x - 2 + [a]^2 = 0$
$(\tan^2x - 1)^2 - 1 - 2 + [a]^2 = 0$
$(\tan^2x - 1)^2 = 3 - [a]^2$
Since $(\tan^2x - 1)^2 \geq 0$,we must have $3 - [a]^2 \geq 0$,which implies $[a]^2 \leq 3$.
Thus,$[a] \in [-\sqrt{3}, \sqrt{3}]$. Since $[a]$ is an integer,$[a] \in \{-1, 0, 1\}$.
If $[a] = -1$,then $-1 \leq a < 0$.
If $[a] = 0$,then $0 \leq a < 1$.
If $[a] = 1$,then $1 \leq a < 2$.
Combining these,the range of $a$ is $[-1, 2)$.

(C) Let $E = \frac{4 \sin 9^{\circ} \sin 21^{\circ} \sin 39^{\circ} \sin 51^{\circ} \sin 69^{\circ} \sin 81^{\circ}}{\sin 54^{\circ}}$.
Using $\sin \theta \sin(60^{\circ}-\theta) \sin(60^{\circ}+\theta) = \frac{1}{4} \sin 3\theta$:
Group the terms: $(\sin 9^{\circ} \sin 51^{\circ} \sin 69^{\circ}) = \sin 9^{\circ} \sin(60^{\circ}-9^{\circ}) \sin(60^{\circ}+9^{\circ}) = \frac{1}{4} \sin 27^{\circ}$.
And $(\sin 21^{\circ} \sin 39^{\circ} \sin 81^{\circ}) = \sin 21^{\circ} \sin(60^{\circ}-21^{\circ}) \sin(60^{\circ}+21^{\circ}) = \frac{1}{4} \sin 63^{\circ}$.
Substituting these into the expression:
$E = \frac{4 \cdot (\frac{1}{4} \sin 27^{\circ}) \cdot (\frac{1}{4} \sin 63^{\circ})}{\sin 54^{\circ}}$
$E = \frac{\frac{1}{4} \sin 27^{\circ} \cos 27^{\circ}}{\sin 54^{\circ}}$
$E = \frac{\frac{1}{8} \sin 54^{\circ}}{\sin 54^{\circ}} = \frac{1}{8}$.

(A) The given series is $S = \sum_{k=1}^{71} \sin(5k^{\circ})$.
This is a series of sines in arithmetic progression with $n = 71$ terms,first term $a = 5^{\circ}$,and common difference $d = 5^{\circ}$.
The formula for the sum is $S = \frac{\sin(n \cdot d/2)}{\sin(d/2)} \cdot \sin\left(\frac{a + l}{2}\right)$,where $l = 355^{\circ}$.
$S = \frac{\sin(71 \cdot 5^{\circ}/2)}{\sin(5^{\circ}/2)} \cdot \sin\left(\frac{5^{\circ} + 355^{\circ}}{2}\right)$.
$S = \frac{\sin(355^{\circ}/2)}{\sin(5^{\circ}/2)} \cdot \sin(180^{\circ})$.
Since $\sin(180^{\circ}) = 0$,the total sum is $0$.

(A) Let $P = \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$.
Since $\cos \frac{3\pi}{7} = \cos (\pi - \frac{4\pi}{7}) = -\cos \frac{4\pi}{7}$,we have:
$P = -\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$.
Using the formula $\cos \theta \cos 2\theta \cos 4\theta = \frac{\sin 8\theta}{8 \sin \theta}$ for $\theta = \frac{\pi}{7}$:
$P = -\left[ \frac{\sin (8 \cdot \frac{\pi}{7})}{8 \sin \frac{\pi}{7}} \right]$.
Since $\sin \frac{8\pi}{7} = \sin (\pi + \frac{\pi}{7}) = -\sin \frac{\pi}{7}$:
$P = -\left[ \frac{-\sin \frac{\pi}{7}}{8 \sin \frac{\pi}{7}} \right] = -\left( -\frac{1}{8} \right) = \frac{1}{8}$.

(B) Given $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$.
Applying the componendo and dividendo rule:
$\frac{1 - \cos x}{1 + \cos x} = \frac{(2 - \cos y) - (2 \cos y - 1)}{(2 - \cos y) + (2 \cos y - 1)}$
$\frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} = \frac{3 - 3 \cos y}{1 + \cos y}$
$\tan^2(x/2) = \frac{3(1 - \cos y)}{1 + \cos y} = \frac{3(2 \sin^2(y/2))}{2 \cos^2(y/2)}$
$\tan^2(x/2) = 3 \tan^2(y/2)$
Since $x, y \in (0, \pi)$,$x/2, y/2 \in (0, \pi/2)$,so $\tan(x/2)$ and $\tan(y/2)$ are positive.
Taking the square root,we get $\tan(x/2) = \sqrt{3} \tan(y/2)$.
Therefore,$\tan(x/2) \cot(y/2) = \sqrt{3}$.

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

(A) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \dots + \sin 89^{\circ}$.
Using the identity $\sin \theta + \sin(90^{\circ} - \theta) = 2 \sin 45^{\circ} \cos(\theta - 45^{\circ}) = \sqrt{2} \cos(\theta - 45^{\circ})$,
we can pair terms: $(\sin 1^{\circ} + \sin 89^{\circ}) + (\sin 2^{\circ} + \sin 88^{\circ}) + \dots + (\sin 44^{\circ} + \sin 46^{\circ}) + \sin 45^{\circ}$.
Each pair $(\sin \theta + \sin(90^{\circ} - \theta)) = 2 \sin 45^{\circ} \cos(45^{\circ} - \theta) = \sqrt{2} \cos(45^{\circ} - \theta)$.
Thus,the numerator is $2 \times [\sqrt{2}(\cos 44^{\circ} + \cos 43^{\circ} + \dots + \cos 1^{\circ}) + \sin 45^{\circ}]$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,the numerator becomes $2\sqrt{2}(\cos 44^{\circ} + \dots + \cos 1^{\circ}) + 2(\frac{1}{\sqrt{2}}) = 2\sqrt{2}(\cos 44^{\circ} + \dots + \cos 1^{\circ}) + \sqrt{2}$.
Factoring out $\sqrt{2}$,we get $\sqrt{2} [2(\cos 1^{\circ} + \dots + \cos 44^{\circ}) + 1]$.
Dividing by the denominator $2(\cos 1^{\circ} + \dots + \cos 44^{\circ}) + 1$,the result is $\sqrt{2}$.

(D) The given equation is $|\sin x \cos x| + \sqrt{2 + \tan^2 x + \cot^2 x} = \sqrt{3}$.
We know that $2 + \tan^2 x + \cot^2 x = 1 + \tan^2 x + 1 + \cot^2 x = \sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.
Thus,$\sqrt{2 + \tan^2 x + \cot^2 x} = \sqrt{\frac{1}{\sin^2 x \cos^2 x}} = \frac{1}{|\sin x \cos x|}$.
Let $y = |\sin x \cos x|$. Since $0 < |\sin x \cos x| \leq \frac{1}{2}$,we have $y \in (0, 0.5]$.
The equation becomes $y + \frac{1}{y} = \sqrt{3}$.
However,for any $y > 0$,by the $AM$-$GM$ inequality,$y + \frac{1}{y} \geq 2$.
Since $\sqrt{3} \approx 1.732 < 2$,there is no real value of $x$ that satisfies this equation.
Therefore,the solution is non-existent.

(D) Given that $\tan A + \cot A = 4$.
Squaring both sides,we get $(\tan A + \cot A)^2 = 4^2$.
$\tan^2 A + \cot^2 A + 2 \tan A \cot A = 16$.
Since $\tan A \cot A = 1$,we have $\tan^2 A + \cot^2 A + 2 = 16$.
$\tan^2 A + \cot^2 A = 14$.
Now,squaring again,$(\tan^2 A + \cot^2 A)^2 = 14^2$.
$\tan^4 A + \cot^4 A + 2 \tan^2 A \cot^2 A = 196$.
Since $\tan^2 A \cot^2 A = 1$,we have $\tan^4 A + \cot^4 A + 2 = 196$.
$\tan^4 A + \cot^4 A = 194$.

(C) Given $\cos x + \sec x = -2$.
Since $\sec x = \frac{1}{\cos x}$,we have $\cos x + \frac{1}{\cos x} = -2$.
Multiplying by $\cos x$,we get $\cos^2 x + 1 = -2 \cos x$.
This simplifies to $\cos^2 x + 2 \cos x + 1 = 0$,which is $(\cos x + 1)^2 = 0$.
Thus,$\cos x = -1$.
Then $\sec x = \frac{1}{-1} = -1$.
Now,$\cos^n x + \sec^n x = (-1)^n + (-1)^n = 2(-1)^n$.
If $n$ is odd,$2(-1)^n = -2$.
If $n$ is even,$2(-1)^n = 2$.
Therefore,the value is $-2$ if $n$ is odd and $2$ if $n$ is even.

(A) The given equation is $2 \cos ^{2} \left( \frac{x}{2} \right) \sin ^{2} x = x^{2} + \frac{1}{x^{2}}$.
For the $R.H.S.$,by $AM \geq GM$ inequality,$x^{2} + \frac{1}{x^{2}} \geq 2$ for all $x \neq 0$. The equality holds when $x^{2} = 1$,i.e.,$x = \pm 1$.
For the $L.H.S.$,we have $2 \cos ^{2} \left( \frac{x}{2} \right) \sin ^{2} x$. Since $\cos ^{2} \left( \frac{x}{2} \right) \leq 1$ and $\sin ^{2} x \leq 1$,the maximum value of the product $2 \cos ^{2} \left( \frac{x}{2} \right) \sin ^{2} x$ is $2 \times 1 \times 1 = 2$.
For the equation to hold,both sides must equal $2$. This requires $x^{2} = 1$ (so $x = 1$ since $x \in [0, \pi/2]$) $AND$ $\cos ^{2} \left( \frac{x}{2} \right) = 1$ and $\sin ^{2} x = 1$.
If $x = 1$,then $\sin ^{2} (1) \approx (0.841)^{2} \approx 0.707 \neq 1$.
Since the $L.H.S.$ cannot reach $2$ when the $R.H.S.$ is $2$,there is no solution.

(C) We are given the inequality $\cos x > \sin x$ for $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$.
Dividing by $\cos x$ requires caution because $\cos x$ changes sign in this interval.
Alternatively,consider the intersection of the graphs of $y = \cos x$ and $y = \sin x$.
In the interval $\left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$,$\cos x = \sin x$ when $\tan x = 1$,which occurs at $x = \frac{5\pi}{4}$.
For $x \in \left( \frac{\pi}{2}, \frac{5\pi}{4} \right)$,$\cos x$ is negative and $\sin x$ is positive (or $\cos x$ is less than $\sin x$ in the second quadrant).
Specifically,at $x = \pi$,$\cos \pi = -1$ and $\sin \pi = 0$,so $-1 < 0$ (i.e.,$\cos x < \sin x$).
For $x \in \left( \frac{5\pi}{4}, \frac{3\pi}{2} \right)$,both $\sin x$ and $\cos x$ are negative.
In this interval,$\cos x$ is closer to $0$ than $\sin x$ (e.g.,at $x = \frac{4\pi}{3}$,$\cos x = -0.5$ and $\sin x = -0.866$,so $-0.5 > -0.866$).
Thus,$\cos x > \sin x$ holds for $x \in \left( \frac{5\pi}{4}, \frac{3\pi}{2} \right)$.

(B) Let the perimeter of both the regular pentagon and the regular decagon be $10x$.
For the regular pentagon,the number of sides $n_1 = 5$. The side length $s_1 = \frac{10x}{5} = 2x$.
The area of a regular polygon with $n$ sides and side length $s$ is given by $A = \frac{n s^2}{4 \tan(\frac{\pi}{n})}$.
Area of the pentagon $A_1 = \frac{5 (2x)^2}{4 \tan(\frac{\pi}{5})} = \frac{5 \cdot 4x^2}{4 \tan(36^{\circ})} = 5x^2 \cot(36^{\circ})$.
For the regular decagon,the number of sides $n_2 = 10$. The side length $s_2 = \frac{10x}{10} = x$.
Area of the decagon $A_2 = \frac{10 x^2}{4 \tan(\frac{\pi}{10})} = \frac{5}{2} x^2 \cot(18^{\circ})$.
The ratio of the area of the pentagon to the area of the decagon is:
$\frac{A_1}{A_2} = \frac{5x^2 \cot(36^{\circ})}{\frac{5}{2} x^2 \cot(18^{\circ})} = 2 \cdot \frac{\cot(36^{\circ})}{\cot(18^{\circ})} = 2 \cdot \frac{\cos(36^{\circ})}{\sin(36^{\circ})} \cdot \frac{\sin(18^{\circ})}{\cos(18^{\circ})}$
Using $\sin(36^{\circ}) = 2 \sin(18^{\circ}) \cos(18^{\circ})$:
$\frac{A_1}{A_2} = 2 \cdot \frac{\cos(36^{\circ})}{2 \sin(18^{\circ}) \cos(18^{\circ})} \cdot \frac{\sin(18^{\circ})}{\cos(18^{\circ})} = \frac{\cos(36^{\circ})}{\cos^2(18^{\circ})}$
Using $\cos^2(18^{\circ}) = \frac{1 + \cos(36^{\circ})}{2}$:
$\frac{A_1}{A_2} = \frac{\cos(36^{\circ})}{\frac{1 + \cos(36^{\circ})}{2}} = \frac{2 \cos(36^{\circ})}{1 + \cos(36^{\circ})}$
Since $\cos(36^{\circ}) = \frac{\sqrt{5} + 1}{4}$:
$\frac{A_1}{A_2} = \frac{2(\frac{\sqrt{5} + 1}{4})}{1 + \frac{\sqrt{5} + 1}{4}} = \frac{\frac{\sqrt{5} + 1}{2}}{\frac{5 + \sqrt{5}}{4}} = \frac{\sqrt{5} + 1}{2} \cdot \frac{4}{\sqrt{5}(\sqrt{5} + 1)} = \frac{2}{\sqrt{5}}$.