Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance Questions in English

(B) Let the sides of the right-angled triangle be $a, b, c$ where $c$ is the hypotenuse. Thus,$c^2 = a^2 + b^2$.
Given that the triangle formed by the reciprocals of the sides is also a right-angled triangle,the largest side among $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must be the hypotenuse. Since $c$ is the largest side,$\frac{1}{c}$ is the smallest,so $\frac{1}{a}$ is the largest.
Thus,$(\frac{1}{a})^2 = (\frac{1}{b})^2 + (\frac{1}{c})^2$.
Substituting $a = c \sin \theta$ and $b = c \cos \theta$:
$\frac{1}{c^2 \sin^2 \theta} = \frac{1}{c^2 \cos^2 \theta} + \frac{1}{c^2}$
$\frac{1}{\sin^2 \theta} = \frac{1}{\cos^2 \theta} + 1$
$1 = \frac{\sin^2 \theta}{\cos^2 \theta} + \sin^2 \theta$
$1 = \tan^2 \theta + \sin^2 \theta$
$1 = \frac{\sin^2 \theta}{1 - \sin^2 \theta} + \sin^2 \theta$
Let $x = \sin^2 \theta$. Then $1 = \frac{x}{1-x} + x \implies 1-x = x + x(1-x) \implies 1-x = x + x - x^2 \implies x^2 - 3x + 1 = 0$.
Using the quadratic formula,$x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.
Since $\sin^2 \theta < 1$,we take $x = \frac{3 - \sqrt{5}}{2}$.
$\sin \theta = \sqrt{\frac{3 - \sqrt{5}}{2}} = \sqrt{\frac{6 - 2\sqrt{5}}{4}} = \frac{\sqrt{5}-1}{2}$.

(B) Given equation: $\sin \theta \tan \theta + \tan \theta = \sin 2 \theta$
$\tan \theta (\sin \theta + 1) = \frac{2 \tan \theta}{1 + \tan^2 \theta} = 2 \sin \theta \cos \theta$
Case $1$: $\tan \theta = 0 \implies \theta = 0, \pi, -\pi$ (since $\theta \in [-\pi, \pi]$).
Case $2$: $\sin \theta + 1 = 2 \cos^2 \theta = 2(1 - \sin^2 \theta) = 2(1 - \sin \theta)(1 + \sin \theta)$.
If $\sin \theta = -1$,then $\theta = -\frac{\pi}{2}$,which is excluded.
If $\sin \theta \neq -1$,then $1 = 2(1 - \sin \theta) \implies 1 = 2 - 2 \sin \theta \implies \sin \theta = \frac{1}{2}$.
Thus,$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$.
The set $S = \{0, \pi, -\pi, \frac{\pi}{6}, \frac{5\pi}{6} \}$,so $n(S) = 5$.
$T = \sum_{\theta \in S} \cos 2 \theta = \cos(0) + \cos(2\pi) + \cos(-2\pi) + \cos(\frac{\pi}{3}) + \cos(\frac{5\pi}{3})$
$T = 1 + 1 + 1 + \frac{1}{2} + \frac{1}{2} = 4$.
Therefore,$T + n(S) = 4 + 5 = 9$.

(D) Given the equation: $7 \cos^2 \theta - 3 \sin^2 \theta - 2 \cos^2 2\theta = 2$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we get $7 \cos^2 \theta - 3(1 - \cos^2 \theta) - 2 \cos^2 2\theta = 2$,which simplifies to $10 \cos^2 \theta - 3 - 2 \cos^2 2\theta = 2$.
Since $2 \cos^2 \theta = 1 + \cos 2\theta$,we have $5(1 + \cos 2\theta) - 3 - 2 \cos^2 2\theta = 2$,leading to $5 + 5 \cos 2\theta - 3 - 2 \cos^2 2\theta = 2$.
This simplifies to $2 \cos^2 2\theta - 5 \cos 2\theta = 0$,so $\cos 2\theta(2 \cos 2\theta - 5) = 0$.
Since $\cos 2\theta$ cannot be $2.5$,we have $\cos 2\theta = 0$,which gives $2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.
Thus,$S = \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}$.
For any $\theta \in S$,$\tan^2 \theta = 1$ and $\cot^2 \theta = 1$,so $\tan^2 \theta + \cot^2 \theta = 2$.
Also,$\sin^2 \theta = \frac{1}{2}$.
The equation becomes $x^2 - 2(2)x + 6(\frac{1}{2}) = 0$,which is $x^2 - 4x + 3 = 0$.
The sum of roots for each equation is $4$.
Since there are $4$ such equations,the total sum of roots is $4 \times 4 = 16$.

(A) Let $\angle CAD = \theta_2$ and $\angle BAD = \theta_1$. We are given $\cot \theta_2 : \cot \theta_1 = 2 : 1$,so $\cot \theta_2 = 2 \cot \theta_1$.
Using the cotangent rule in $\triangle ADC$ and $\triangle ADB$ with $AD$ as a common side and $BD = DC = a/2$:
$\cot \theta_2 = \frac{AD^2 + b^2 - (a/2)^2}{2 \cdot AD \cdot (a/2) \cdot \sin \theta_2} \cdot \sin \theta_2 = \frac{AD^2 + b^2 - a^2/4}{2 \cdot \text{Area}(\triangle ADC)}$
Since $\text{Area}(\triangle ADC) = \text{Area}(\triangle ADB)$,the condition $\cot \theta_2 = 2 \cot \theta_1$ implies:
$AD^2 + b^2 - a^2/4 = 2(AD^2 + c^2 - a^2/4)$
$AD^2 = b^2 - 2c^2 + a^2/4$
Using Apollonius theorem,$AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$.
Equating the two expressions for $AD^2$:
$\frac{2b^2 + 2c^2 - a^2}{4} = b^2 - 2c^2 + \frac{a^2}{4}$
$2b^2 + 2c^2 - a^2 = 4b^2 - 8c^2 + a^2$
$10c^2 = 2b^2 + 2a^2 \Rightarrow b^2 + a^2 = 5c^2$.
In $\triangle BGA$,by the Law of Cosines:
$\cos(\angle BGA) = \frac{AG^2 + BG^2 - AB^2}{2 \cdot AG \cdot BG}$.
Using $AG = \frac{2}{3}m_a$ and $BG = \frac{2}{3}m_b$,where $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$ and $m_b^2 = \frac{2a^2+2c^2-b^2}{4}$,and substituting $a^2+b^2=5c^2$,we find $\cos(\angle BGA) = 0$,so $\angle BGA = 90^{\circ}$.

(D) The given equation is $\sin(\pi \sin^2 \theta) + \sin(\pi \cos^2 \theta) = 2 \cos(\frac{\pi}{2} \cos \theta)$.
Using the identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:
$2 \sin(\frac{\pi(\sin^2 \theta + \cos^2 \theta)}{2}) \cos(\frac{\pi(\sin^2 \theta - \cos^2 \theta)}{2}) = 2 \cos(\frac{\pi}{2} \cos \theta)$.
Since $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin^2 \theta - \cos^2 \theta = -\cos 2\theta$,we have:
$2 \sin(\frac{\pi}{2}) \cos(-\frac{\pi}{2} \cos 2\theta) = 2 \cos(\frac{\pi}{2} \cos \theta)$.
Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(-x) = \cos x$,this simplifies to:
$\cos(\frac{\pi}{2} \cos 2\theta) = \cos(\frac{\pi}{2} \cos \theta)$.
This implies $\frac{\pi}{2} \cos 2\theta = 2n\pi \pm \frac{\pi}{2} \cos \theta$,so $\cos 2\theta = 4n \pm \cos \theta$ for $n \in Z$.
Case $I$: $\cos 2\theta - \cos \theta = 4n$. For $n=0$,$2\cos^2 \theta - 1 - \cos \theta = 0$,so $(2\cos \theta + 1)(\cos \theta - 1) = 0$. Thus $\cos \theta = 1$ or $\cos \theta = -1/2$. Solutions in $[0, 2\pi]$ are $\theta = 0, 2\pi, 2\pi/3, 4\pi/3$.
Case $II$: $\cos 2\theta + \cos \theta = 4n$. For $n=0$,$2\cos^2 \theta + \cos \theta - 1 = 0$,so $(2\cos \theta - 1)(\cos \theta + 1) = 0$. Thus $\cos \theta = 1/2$ or $\cos \theta = -1$. Solutions in $[0, 2\pi]$ are $\theta = \pi, \pi/3, 5\pi/3$.
Total solutions are ${0, 2\pi, 2\pi/3, 4\pi/3, \pi, \pi/3, 5\pi/3}$,which is $7$ solutions.

(D) Given that $\alpha, \beta, \gamma$ are the angles of a triangle,we have $\alpha + \beta + \gamma = 180^{\circ}$.
We are given the equations:
$(i) \quad 2 \sin \alpha + 3 \cos \beta = 3 \sqrt{2}$
$(ii) \quad 3 \sin \beta + 2 \cos \alpha = 1$
Squaring both equations:
$(2 \sin \alpha + 3 \cos \beta)^2 = (3 \sqrt{2})^2 = 18$
$4 \sin^2 \alpha + 9 \cos^2 \beta + 12 \sin \alpha \cos \beta = 18$
$(3 \sin \beta + 2 \cos \alpha)^2 = 1^2 = 1$
$9 \sin^2 \beta + 4 \cos^2 \alpha + 12 \sin \beta \cos \alpha = 1$
Adding these two equations:
$4(\sin^2 \alpha + \cos^2 \alpha) + 9(\sin^2 \beta + \cos^2 \beta) + 12(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 18 + 1$
$4(1) + 9(1) + 12 \sin(\alpha + \beta) = 19$
$13 + 12 \sin(\alpha + \beta) = 19$
$12 \sin(\alpha + \beta) = 6$
$\sin(\alpha + \beta) = \frac{1}{2}$
Since $\alpha + \beta + \gamma = 180^{\circ}$,we have $\alpha + \beta = 180^{\circ} - \gamma$. Thus,$\sin(180^{\circ} - \gamma) = \sin \gamma = \frac{1}{2}$.
Therefore,$\gamma = 30^{\circ}$ or $\gamma = 150^{\circ}$.
Checking the original equations,if $\gamma = 150^{\circ}$,then $\alpha + \beta = 30^{\circ}$,which is not possible for positive angles $\alpha, \beta$ satisfying the given equations. Thus,$\gamma = 30^{\circ}$.

(A) Given equations are:
$3 \sin A + 4 \cos B = 6$ $(i)$
$4 \sin B + 3 \cos A = 1$ $(ii)$
Squaring and adding equations $(i)$ and $(ii)$:
$(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$
$9 \sin^2 A + 16 \cos^2 B + 24 \sin A \cos B + 16 \sin^2 B + 9 \cos^2 A + 24 \sin B \cos A = 37$
$9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$
$9(1) + 16(1) + 24 \sin(A + B) = 37$
$25 + 24 \sin(A + B) = 37$
$24 \sin(A + B) = 12$
$\sin(A + B) = \frac{1}{2}$
Since $A + B + C = 180^{\circ}$,we have $A + B = 180^{\circ} - C$.
$\sin(180^{\circ} - C) = \frac{1}{2}$
$\sin C = \frac{1}{2}$
Thus,$C = 30^{\circ}$ or $C = 150^{\circ}$.
Checking the original equations with $C = 150^{\circ}$ leads to a contradiction,so $C = 30^{\circ}$.

(C) To find the number of solutions to $\sin x = \frac{6}{x}$ in the interval $0 \leq x \leq 12 \pi$,we look for the points of intersection between the graphs of $y = \sin x$ and $y = \frac{6}{x}$.
$1$. The function $y = \sin x$ oscillates between $-1$ and $1$ with a period of $2 \pi$.
$2$. The function $y = \frac{6}{x}$ is a hyperbola that decreases as $x$ increases.
$3$. For $x > 0$,the intersection points occur where $\sin x = \frac{6}{x}$. Since $|\sin x| \leq 1$,we must have $|\frac{6}{x}| \leq 1$,which implies $x \geq 6$.
$4$. In the interval $[0, 12 \pi]$,the function $y = \frac{6}{x}$ intersects the peaks of the sine wave. By observing the graph,the curve $y = \frac{6}{x}$ intersects the sine wave twice in each period of $2 \pi$ starting from the first positive cycle where the value of $\sin x$ can reach $\frac{6}{x}$.
$5$. Counting the intersection points from the graph,there are $10$ such points.
Thus,the correct option is $C$.

(B) Given relations for the sides of the triangle are:
$c^2 = 2ab$ $(i)$
$a^2 + c^2 = 3b^2$ $(ii)$
Substituting $(i)$ into $(ii)$:
$a^2 + 2ab = 3b^2$
$a^2 + 2ab - 3b^2 = 0$
$(a + 3b)(a - b) = 0$
Since $a$ and $b$ are side lengths,$a, b > 0$,so $a = b$.
Substituting $a = b$ into $(i)$:
$c^2 = 2(b)(b) = 2b^2$
$c = \sqrt{2}b$
Now,we have sides $a = b$,$b = b$,and $c = \sqrt{2}b$.
Since $a^2 + b^2 = b^2 + b^2 = 2b^2 = c^2$,the triangle is a right-angled triangle with the right angle at $C$ (i.e.,$\angle C = 90^{\circ}$).
Since $a = b$,the triangle is an isosceles right-angled triangle.
Therefore,$\angle A = \angle B = 45^{\circ}$.
The measure of $\angle BAC$ is $45^{\circ}$.

(D) Let $O$ be the center of the circumcircle of the regular nonagon $A_1 A_2 \ldots A_9$. The side length is $s = 2$.
The central angle subtended by each side is $\theta = \frac{2\pi}{9}$.
Let $r$ be the circumradius. In $\triangle O A_1 A_2$,by the law of cosines:
$s^2 = r^2 + r^2 - 2r^2 \cos(\frac{2\pi}{9}) = 2r^2(1 - \cos(\frac{2\pi}{9})) = 4r^2 \sin^2(\frac{\pi}{9})$.
Since $s = 2$,we have $4 = 4r^2 \sin^2(\frac{\pi}{9})$,so $r = \frac{1}{\sin(\frac{\pi}{9})}$.
The length of a chord $A_i A_j$ subtending a central angle $\phi$ is $2r \sin(\frac{\phi}{2})$.
For $A_1 A_5$,the central angle is $4 \times \frac{2\pi}{9} = \frac{8\pi}{9}$,so $A_1 A_5 = 2r \sin(\frac{4\pi}{9})$.
For $A_2 A_4$,the central angle is $2 \times \frac{2\pi}{9} = \frac{4\pi}{9}$,so $A_2 A_4 = 2r \sin(\frac{2\pi}{9})$.
The difference is $A_1 A_5 - A_2 A_4 = 2r(\sin(\frac{4\pi}{9}) - \sin(\frac{2\pi}{9}))$.
Using the identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$:
$A_1 A_5 - A_2 A_4 = 2r \cdot 2 \sin(\frac{\pi}{9}) \cos(\frac{3\pi}{9}) = 4r \sin(\frac{\pi}{9}) \cos(\frac{\pi}{3})$.
Substituting $r = \frac{1}{\sin(\frac{\pi}{9})}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$:
$A_1 A_5 - A_2 A_4 = 4 \cdot \frac{1}{\sin(\frac{\pi}{9})} \cdot \sin(\frac{\pi}{9}) \cdot \frac{1}{2} = 2$.

(B) Given,in $\triangle ABC$,the angle bisectors $BD$ and $CE$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively.
$\frac{BI}{ID} = \frac{3}{2}$ and $\frac{CI}{IE} = \frac{2}{1}$.
We know that for an incentre $I$ of $\triangle ABC$ with sides $a, b, c$ opposite to vertices $A, B, C$ respectively,the ratio in which $I$ divides the angle bisectors are:
$\frac{AI}{IF} = \frac{b+c}{a}$,$\frac{BI}{ID} = \frac{a+c}{b}$,and $\frac{CI}{IE} = \frac{a+b}{c}$.
Given $\frac{a+c}{b} = \frac{3}{2} \Rightarrow 2a + 2c = 3b \quad \dots(i)$
Given $\frac{a+b}{c} = \frac{2}{1} \Rightarrow a + b = 2c \quad \dots(ii)$
From $(ii)$,$c = \frac{a+b}{2}$. Substituting this into $(i)$:
$2a + 2(\frac{a+b}{2}) = 3b$
$2a + a + b = 3b$
$3a = 2b \Rightarrow b = \frac{3}{2}a$
Now,substitute $b = \frac{3}{2}a$ into $(ii)$:
$a + \frac{3}{2}a = 2c$
$\frac{5}{2}a = 2c \Rightarrow c = \frac{5}{4}a$
Finally,the ratio $\frac{AI}{IF} = \frac{b+c}{a} = \frac{\frac{3}{2}a + \frac{5}{4}a}{a} = \frac{6+5}{4} = \frac{11}{4}$.
Thus,the ratio is $11:4$.

(C) Given the expression $E = \frac{\sin A \cot C + \cos A}{\sin B \cot C + \cos B}$.
Substituting $\cot C = \frac{\cos C}{\sin C}$,we get $E = \frac{\frac{\sin A \cos C + \cos A \sin C}{\sin C}}{\frac{\sin B \cos C + \cos B \sin C}{\sin C}} = \frac{\sin(A+C)}{\sin(B+C)}$.
Since $A+B+C = \pi$,we have $A+C = \pi - B$ and $B+C = \pi - A$.
Thus,$E = \frac{\sin(\pi - B)}{\sin(\pi - A)} = \frac{\sin B}{\sin A} = \frac{b}{a}$.
Given $b^2 = ac$,the sides $a, b, c$ are in a Geometric Progression $(G.P.)$. Let $b = ar$ and $c = ar^2$.
Then $E = \frac{b}{a} = \frac{ar}{a} = r$.
For a triangle to exist,the sum of any two sides must be greater than the third side:
$1) a+b > c$ $\Rightarrow a+ar > ar^2$ $\Rightarrow 1+r > r^2$ $\Rightarrow r^2-r-1 < 0$. Solving this gives $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Since $r > 0$,$r \in \left(0, \frac{1+\sqrt{5}}{2}\right)$.
$2) a+c > b$ $\Rightarrow a+ar^2 > ar$ $\Rightarrow r^2-r+1 > 0$. This is true for all $r$.
$3) b+c > a$ $\Rightarrow ar+ar^2 > a$ $\Rightarrow r^2+r-1 > 0$. Solving this gives $r > \frac{-1+\sqrt{5}}{2}$ (since $r > 0$).
Combining these,$r \in \left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$.

(C) Let the sides of the right-angled triangle be $b$,$c$ (legs) and $x$ (hypotenuse).
Given $b + c + x = 42$ --- $(1)$
In a right-angled triangle,the median to the hypotenuse is half the hypotenuse,so $M = \frac{x}{2}$.
The altitude $h$ to the hypotenuse is given by $h = \frac{bc}{x}$.
Given $M - h = 2$,so $\frac{x}{2} - \frac{bc}{x} = 2 \Rightarrow x^2 - 2bc = 4x$ --- $(2)$
From $(1)$,$b + c = 42 - x$. Squaring both sides: $b^2 + c^2 + 2bc = (42 - x)^2$.
Since $b^2 + c^2 = x^2$,we have $x^2 + 2bc = (42 - x)^2$ --- $(3)$
Adding $(2)$ and $(3)$: $2x^2 = (42 - x)^2 + 4x$.
$2x^2 = 1764 - 84x + x^2 + 4x \Rightarrow x^2 + 80x - 1764 = 0$.
Solving the quadratic equation: $(x + 98)(x - 18) = 0$. Since $x > 0$,$x = 18$.
Substitute $x = 18$ into $(2)$: $18^2 - 2bc = 4(18)$ $\Rightarrow 324 - 2bc = 72$ $\Rightarrow 2bc = 252$ $\Rightarrow bc = 126$.
The area of the triangle is $\frac{1}{2}bc = \frac{1}{2} \times 126 = 63$.

(D) Given $\tan (\pi \cos \theta) + \tan (\pi \sin \theta) = 0$.
This implies $\tan (\pi \cos \theta) = -\tan (\pi \sin \theta) = \tan (-\pi \sin \theta)$.
Thus,$\pi \cos \theta = n\pi - \pi \sin \theta$,where $n \in \mathbb{Z}$.
Dividing by $\pi$,we get $\sin \theta + \cos \theta = n$.
Since $-\sqrt{2} \leq \sin \theta + \cos \theta \leq \sqrt{2}$,the possible integer values for $n$ are $-1, 0, 1$.
Case $1$: $\sin \theta + \cos \theta = 0 \implies \tan \theta = -1$. In $[0, 2\pi)$,$\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$.
Case $2$: $\sin \theta + \cos \theta = 1 \implies \sin(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. In $[0, 2\pi)$,$\theta = 0, \frac{\pi}{2}$.
Case $3$: $\sin \theta + \cos \theta = -1 \implies \sin(\theta + \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$. In $[0, 2\pi)$,$\theta = \pi, \frac{3\pi}{2}$.
The set $S = \{0, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{3\pi}{2}, \frac{7\pi}{4}\}$.
We need to calculate $\sum_{\theta \in S} \sin^2(\theta + \frac{\pi}{4})$.
For $\theta \in \{0, \frac{\pi}{2}\}$,$\sin^2(\theta + \frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
For $\theta \in \{\pi, \frac{3\pi}{2}\}$,$\sin^2(\theta + \frac{\pi}{4}) = (-\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
For $\theta \in \{\frac{3\pi}{4}, \frac{7\pi}{4}\}$,$\sin^2(\theta + \frac{\pi}{4}) = (0)^2 = 0$.
Sum $= 4 \times \frac{1}{2} + 2 \times 0 = 2$.

(D) The expression $\cos 2A + \cos 2B + \cos 2C$ is minimized when $\triangle ABC$ is an equilateral triangle,i.e.,$A = B = C = 60^{\circ}$.
Given inradius $r = 3$. For an equilateral triangle,$r = \frac{a}{2\sqrt{3}}$,so $3 = \frac{a}{2\sqrt{3}} \implies a = 6\sqrt{3}$.
Perimeter $= 3a = 18\sqrt{3}$. (Option $A$ is correct).
Area $= \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} (108) = 27\sqrt{3}$. (Option $D$ is incorrect as it states $\frac{27\sqrt{3}}{2}$).
For $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 3\sin 60^{\circ} = \frac{3\sqrt{3}}{2}$. Since $4\sin^3 60^{\circ} = 4(\frac{\sqrt{3}}{2})^3 = \frac{3\sqrt{3}}{2}$,Option $B$ is correct.
For $\overrightarrow{MA} \cdot \overrightarrow{MB} = |MA||MB| \cos(120^{\circ}) = (2r)(2r)(-1/2) = -2r^2 = -2(3^2) = -18$. (Option $C$ is correct).

(C) Given $\cos A + \cos C = 2(1 - \cos B)$.
Using the sum-to-product formula,$2 \cos \frac{A+C}{2} \cos \frac{A-C}{2} = 4 \sin^2 \frac{B}{2}$.
Since $\cos \frac{A+C}{2} = \sin \frac{B}{2}$,we have $2 \sin \frac{B}{2} \cos \frac{A-C}{2} = 4 \sin^2 \frac{B}{2}$,which simplifies to $\cos \frac{A-C}{2} = 2 \sin \frac{B}{2}$.
Using the sine rule,$\sin A + \sin C = 2 \sin B$,which implies $a + c = 2b$.
Given $a = 3$ and $c = 7$,we get $3 + 7 = 2b$,so $b = 5$.
Now,$\cos A - \cos C = \frac{b^2 + c^2 - a^2}{2bc} - \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos A - \cos C = \frac{25 + 49 - 9}{2(5)(7)} - \frac{9 + 25 - 49}{2(3)(5)}$.
$= \frac{65}{70} - \frac{-15}{30} = \frac{13}{14} + \frac{1}{2} = \frac{13 + 7}{14} = \frac{20}{14} = \frac{10}{7}$.

(D) Given $2 \tan ^2 \theta - 5 \sec \theta = 1$.
Using $\tan ^2 \theta = \sec ^2 \theta - 1$,we get $2(\sec ^2 \theta - 1) - 5 \sec \theta - 1 = 0$.
$2 \sec ^2 \theta - 5 \sec \theta - 3 = 0$.
$(2 \sec \theta + 1)(\sec \theta - 3) = 0$.
Since $\sec \theta = -\frac{1}{2}$ is impossible,we have $\sec \theta = 3$,which means $\cos \theta = \frac{1}{3}$.
In the interval $[0, 2\pi]$,there are $2$ solutions for $\cos \theta = \frac{1}{3}$.
For $7$ solutions,we need $3$ full periods of $2\pi$ plus one more solution in the next interval,so $n = 13$.
We need to calculate $S = \sum_{k=1}^{13} \frac{k}{2^k}$.
$S = \frac{1}{2} + \frac{2}{2^2} + \dots + \frac{13}{2^{13}}$.
$\frac{1}{2}S = \frac{1}{2^2} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}$.
Subtracting gives $\frac{S}{2} = \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{13}} - \frac{13}{2^{14}} = \frac{1/2(1 - 1/2^{13})}{1 - 1/2} - \frac{13}{2^{14}} = 1 - \frac{1}{2^{13}} - \frac{13}{2^{14}} = 1 - \frac{15}{2^{14}}$.
$S = 2 - \frac{15}{2^{13}} = \frac{2^{14} - 15}{2^{13}}$.

(C) Given equation: $\sin ^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$
Rewrite the middle term: $2 + 2x - x^2 = 3 - (x^2 - 2x + 1) = 3 - (x - 1)^2$
Let $u = \sin x$ and $v = (x - 1)^2$. The equation becomes $u^2 + (3 - v)u - 3v = 0$.
Factoring the quadratic: $u^2 + 3u - vu - 3v = 0 \implies u(u + 3) - v(u + 3) = 0 \implies (u - v)(u + 3) = 0$.
This gives two cases: $\sin x = -3$ (which is impossible as $-1 \leq \sin x \leq 1$) or $\sin x = (x - 1)^2$.
We need to find the number of solutions for $\sin x = (x - 1)^2$ in the interval $[-\pi, \pi]$.
Let $f(x) = \sin x$ and $g(x) = (x - 1)^2$.
$g(x) = 0$ at $x = 1$. Since $1 < \pi \approx 3.14$,$x = 1$ is in the interval.
At $x = 1$,$f(1) = \sin(1) \approx 0.84$ and $g(1) = 0$. Since $f(1) > g(1)$,and $g(x)$ is a parabola opening upwards with vertex at $(1, 0)$,we check intersection points.
For $x > 1$,$g(x)$ increases rapidly. At $x = 1 + \sqrt{1} = 2$,$g(2) = 1$ and $\sin(2) < 1$. At $x = 1 + \sqrt{\sin x}$,there are two intersection points between $x=1$ and $x=1+\pi/2$.
Graphically,the curve $y = \sin x$ and $y = (x - 1)^2$ intersect at $2$ points in the given interval $[-\pi, \pi]$.

(B) Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Given $a=7$,$b=8$,$c=\alpha$,and $\cos A = \frac{2}{3}$:
$\frac{2}{3} = \frac{8^2+\alpha^2-7^2}{2 \times 8 \times \alpha} = \frac{64+\alpha^2-49}{16\alpha} = \frac{15+\alpha^2}{16\alpha}$.
$32\alpha = 45 + 3\alpha^2 \implies 3\alpha^2 - 32\alpha + 45 = 0$.
$(3\alpha - 5)(\alpha - 9) = 0$. Since $\alpha \in N$,we have $\alpha = 9$.
Now,find $\cos C$ using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{7^2+8^2-9^2}{2 \times 7 \times 8} = \frac{49+64-81}{112} = \frac{32}{112} = \frac{2}{7}$.
We need to evaluate $49 \cos(3C) + 42$.
Using $\cos(3C) = 4\cos^3 C - 3\cos C$:
$49(4(\frac{2}{7})^3 - 3(\frac{2}{7})) + 42 = 49(4 \times \frac{8}{343} - \frac{6}{7}) + 42 = 49(\frac{32}{343} - \frac{6}{7}) + 42 = 49(\frac{32 - 294}{343}) + 42 = \frac{32 - 294}{7} + 42 = \frac{-262}{7} + \frac{294}{7} = \frac{32}{7}$.
Thus,$\frac{m}{n} = \frac{32}{7}$,so $m=32$ and $n=7$. Since $\operatorname{gcd}(32, 7)=1$,$m+n = 32+7 = 39$.

(C) In $\triangle ABC$,we have $A + B + C = 180^{\circ}$,so $\frac{B+C}{2} = 90^{\circ} - \frac{A}{2}$.
Given $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$.
Using the sum-to-product formula: $2 \cos \frac{B+C}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$.
Since $\cos \frac{B+C}{2} = \sin \frac{A}{2}$,we have $2 \sin \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$.
Dividing by $2 \sin \frac{A}{2}$ (assuming $A \neq 0$),we get $\cos \frac{B-C}{2} = 2 \sin \frac{A}{2}$.
Multiplying both sides by $2 \cos \frac{A}{2}$,we get $2 \cos \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin \frac{A}{2} \cos \frac{A}{2}$.
Using $2 \cos \frac{A}{2} = 2 \sin \frac{B+C}{2}$,we get $2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} = 2 \sin A$.
This simplifies to $\sin B + \sin C = 2 \sin A$.
By the Sine Rule,$\frac{b}{2R} + \frac{c}{2R} = 2 \frac{a}{2R}$,which implies $b + c = 2a$.
Since $b + c = AB + AC = 2BC$,and $BC$ is a fixed base,the sum of the distances of $A$ from two fixed points $B$ and $C$ is constant $(2BC > BC)$.
Therefore,the locus of point $A$ is an ellipse with foci at $B$ and $C$.

(A) Let the angles of the triangle be $A, C, B$ such that $A < C < B$. Given $B - A = \frac{\pi}{2}$.
Let the sides be $a_1 = n-d, a_2 = n, a_3 = n+d$ in arithmetic progression.
Using the sine rule with circumradius $R=1$,the sides are $2R \sin A, 2R \sin C, 2R \sin B$.
Thus,$n-d = 2 \sin A$,$n = 2 \sin C$,$n+d = 2 \sin B$.
Since $A+B+C = \pi$ and $B = A + \frac{\pi}{2}$,we have $C = \pi - (A + B) = \pi - (2A + \frac{\pi}{2}) = \frac{\pi}{2} - 2A$.
Since $A+C+B = \pi$,$A + (\frac{\pi}{2} - 2A) + (A + \frac{\pi}{2}) = \pi$,which is consistent.
From $2n = (n-d) + (n+d) = 2 \sin A + 2 \sin B$,we get $n = \sin A + \sin B = \sin A + \cos A$.
Also $n = 2 \sin C = 2 \sin(\frac{\pi}{2} - 2A) = 2 \cos 2A$.
Equating $n$: $\sin A + \cos A = 2 \cos 2A = 2(\cos^2 A - \sin^2 A) = 2(\cos A - \sin A)(\cos A + \sin A)$.
Since $\sin A + \cos A \neq 0$,we have $1 = 2(\cos A - \sin A) \Rightarrow \cos A - \sin A = \frac{1}{2}$.
Squaring both sides: $1 - 2 \sin A \cos A = \frac{1}{4} \Rightarrow \sin 2A = \frac{3}{4}$.
Area $a = \frac{1}{2} (n-d)(n+d) \sin C = \frac{1}{2} (2 \sin A)(2 \sin B) \sin C = 2 \sin A \cos A \sin C = \sin 2A \cos 2A$.
Since $\sin 2A = \frac{3}{4}$,$\cos 2A = \sqrt{1 - (3/4)^2} = \frac{\sqrt{7}}{4}$.
$a = \frac{3}{4} \times \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{16}$.
$(64a)^2 = (64 \times \frac{3\sqrt{7}}{16})^2 = (4 \times 3\sqrt{7})^2 = 144 \times 7 = 1008$.
Inradius $r = \frac{a}{s} = \frac{a}{(3n/2)} = \frac{2a}{3n} = \frac{2 \sin 2A \cos 2A}{3(2 \cos 2A)} = \frac{\sin 2A}{3} = \frac{3/4}{3} = \frac{1}{4} = 0.25$.

(A) Given $\tan \theta = \cot 5 \theta = \tan\left(\frac{\pi}{2} - 5\theta\right)$.
This implies $\theta = n\pi + \frac{\pi}{2} - 5\theta$,so $6\theta = n\pi + \frac{\pi}{2}$,or $\theta = \frac{(2n+1)\pi}{12}$.
Also,$\sin 2\theta = \cos 4\theta = 1 - 2\sin^2 2\theta$.
Let $x = \sin 2\theta$,then $2x^2 + x - 1 = 0$,which gives $(2x-1)(x+1) = 0$.
So $\sin 2\theta = \frac{1}{2}$ or $\sin 2\theta = -1$.
If $\sin 2\theta = -1$,then $2\theta = -\frac{\pi}{2} \Rightarrow \theta = -\frac{\pi}{4}$.
If $\sin 2\theta = \frac{1}{2}$,then $2\theta = \frac{\pi}{6}$ or $\frac{5\pi}{6} \Rightarrow \theta = \frac{\pi}{12}$ or $\frac{5\pi}{12}$.
Checking the condition $\theta \neq \frac{n\pi}{5}$ for $n=0, \pm 1, \pm 2$:
For $\theta = -\frac{\pi}{4}$,$\theta = \frac{\pi}{12}$,and $\theta = \frac{5\pi}{12}$,none of these are equal to $0, \pm \frac{\pi}{5}, \pm \frac{2\pi}{5}$.
Thus,there are $3$ such values.

(C) Using the sine rule, $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c} = 2(1) = 2$.
Given $p=\sqrt{3}, q=1$, we have $\sin P = \frac{\sqrt{3}}{2}$ and $\sin Q = \frac{1}{2}$.
Since $p > q$, $P > Q$. Possible values for $P$ are $60^{\circ}$ or $120^{\circ}$, and for $Q$ are $30^{\circ}$ or $150^{\circ}$.
If $P=60^{\circ}, Q=30^{\circ}$, then $R=90^{\circ}$ (rejected as it is non-right-angled).
If $P=120^{\circ}, Q=30^{\circ}$, then $R=30^{\circ}$. Thus, $\triangle PQR$ is isosceles with $q=r=1$.
Area $\Delta = \frac{1}{2}qr \sin P = \frac{1}{2}(1)(1)\sin 120^{\circ} = \frac{\sqrt{3}}{4}$.
Semi-perimeter $s = \frac{\sqrt{3}+1+1}{2} = \frac{\sqrt{3}+2}{2}$.
Inradius $r_{in} = \frac{\Delta}{s} = \frac{\sqrt{3}/4}{(\sqrt{3}+2)/2} = \frac{\sqrt{3}}{2(2+\sqrt{3})} = \frac{\sqrt{3}}{2}(2-\sqrt{3})$. (Option $2$ is correct).
Length of median $RS = \frac{1}{2}\sqrt{2p^2+2q^2-r^2} = \frac{1}{2}\sqrt{2(3)+2(1)-1} = \frac{\sqrt{7}}{2}$. (Option $3$ is correct).
$PE$ is the altitude to $QR$. $PE = q \sin R = 1 \sin 30^{\circ} = \frac{1}{2}$.
$O$ is the centroid of $\triangle PQR$ (since $S$ is midpoint of $PQ$ and $PE$ is altitude, $O$ is intersection of median $RS$ and altitude $PE$ is not generally the centroid, but here $PQR$ is isosceles with $q=r$, so altitude $PE$ is also the median). Thus $O$ is the centroid, $OE = \frac{1}{3}PE = \frac{1}{6}$. (Option $4$ is correct).
Area $\triangle SOE = \frac{1}{6} \Delta = \frac{1}{6} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{24}$. (Option $1$ is incorrect).

(B) Given $\tan \frac{X}{2} + \tan \frac{Z}{2} = \frac{2y}{x+y+z}$.
Using the formula $\tan \frac{X}{2} = \frac{\Delta}{S(S-x)}$ where $S = \frac{x+y+z}{2}$ and $\Delta$ is the area of the triangle,we have:
$\frac{\Delta}{S(S-x)} + \frac{\Delta}{S(S-z)} = \frac{2y}{2S} = \frac{y}{S}$
$\frac{\Delta}{S} \left( \frac{S-z + S-x}{(S-x)(S-z)} \right) = \frac{y}{S}$
$\Delta \left( \frac{2S - (x+z)}{(S-x)(S-z)} \right) = y$
Since $2S = x+y+z$,$2S - (x+z) = y$.
$\Delta \left( \frac{y}{(S-x)(S-z)} \right) = y \implies \Delta = (S-x)(S-z)$
Squaring both sides: $\Delta^2 = (S-x)^2(S-z)^2$
$S(S-x)(S-y)(S-z) = (S-x)^2(S-z)^2$
$S(S-y) = (S-x)(S-z)$
Substituting $S = \frac{x+y+z}{2}$,we get $y^2 = x^2 + z^2$.
This implies $\angle Y = 90^\circ$.
Since $X+Y+Z = 180^\circ$,$X+Z = 90^\circ$,so $Y = X+Z$. Thus $(B)$ is true.
Also,for a right-angled triangle at $Y$,$\tan \frac{X}{2} = \sqrt{\frac{(S-y)(S-z)}{S(S-x)}} = \frac{x}{y+z}$. Thus $(C)$ is true.
Therefore,$(B)$ and $(C)$ are true.

(D) By the Law of Cosines,$\cos P = \frac{q^2+r^2-p^2}{2qr} = \frac{q^2+r^2}{2qr} - \frac{p^2}{2qr}$. Since $q^2+r^2 \geq 2qr$ (by $AM \geq GM$),we have $\frac{q^2+r^2}{2qr} \geq 1$. Thus,$\cos P \geq 1 - \frac{p^2}{2qr}$. So,$(A)$ is correct.
$(B)$ The inequality $(p+q) \cos R \geq (q-r) \cos P + (p-r) \cos Q$ can be rewritten as $(p \cos R + r \cos P) + (q \cos R + r \cos Q) \geq q \cos P + p \cos Q$. Using the projection formula,this simplifies to $q + p \geq r$,which is true by the triangle inequality. So,$(B)$ is correct.
$(C)$ By the Law of Sines,$\frac{q+r}{p} = \frac{\sin Q + \sin R}{\sin P}$. Since $\sin Q + \sin R \geq 2 \sqrt{\sin Q \sin R}$,we have $\frac{q+r}{p} \geq 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$. Thus,$(C)$ is incorrect.
$(D)$ The condition $\cos Q > \frac{p}{r}$ implies $\sin R \cos Q > \sin P$,which simplifies to $\sin P + \sin(R-Q) > 2 \sin P$,or $\sin(R-Q) > \sin P$. This is not necessarily true for all triangles where $p < q$ and $p < r$. Thus,$(D)$ is incorrect.
Hence,the correct statements are $(A)$ and $(B)$.

(A) Given $a=2, b=\frac{7}{2}, c=\frac{5}{2}$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$.
The expression is $\frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P} = \frac{2 \sin P - 2 \sin P \cos P}{2 \sin P + 2 \sin P \cos P} = \frac{2 \sin P(1 - \cos P)}{2 \sin P(1 + \cos P)} = \frac{1 - \cos P}{1 + \cos P}$.
Using half-angle formulas,$\frac{1 - \cos P}{1 + \cos P} = \frac{2 \sin^2(P/2)}{2 \cos^2(P/2)} = \tan^2(P/2)$.
We know $\tan^2(P/2) = \frac{(s-b)(s-c)}{s(s-a)}$.
Substituting the values: $s-a = 4-2 = 2$,$s-b = 4-3.5 = 0.5 = 1/2$,$s-c = 4-2.5 = 1.5 = 3/2$.
$\tan^2(P/2) = \frac{(1/2)(3/2)}{4(2)} = \frac{3/4}{8} = \frac{3}{32}$.
Also,$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{4 \times 2 \times 0.5 \times 1.5} = \sqrt{6}$.
Thus,$\left(\frac{3}{4 \Delta}\right)^2 = \frac{9}{16 \Delta^2} = \frac{9}{16 \times 6} = \frac{9}{96} = \frac{3}{32}$.
Therefore,the expression equals $\left(\frac{3}{4 \Delta}\right)^2$.

(B) Let the lengths of the tangents from the vertices be $x, y, z$. Given $PN=x, QL=y, RM=z$ are consecutive even integers. Let $x=2k, y=2k+2, z=2k+4$.
Since $P$ is the largest angle,the side opposite to $P$ $(QR = y+z)$ must be the largest side.
$QR = (2k+2) + (2k+4) = 4k+6$
$PQ = x+y = 2k + 2k+2 = 4k+2$
$PR = x+z = 2k + 2k+4 = 4k+4$
Using the Law of Cosines: $\cos P = \frac{PQ^2 + PR^2 - QR^2}{2(PQ)(PR)} = \frac{1}{3}$
$\frac{(4k+2)^2 + (4k+4)^2 - (4k+6)^2}{2(4k+2)(4k+4)} = \frac{1}{3}$
$\frac{16k^2+16k+4 + 16k^2+32k+16 - (16k^2+48k+36)}{2(16k^2+24k+8)} = \frac{1}{3}$
$\frac{16k^2}{32k^2+48k+16} = \frac{1}{3} \Rightarrow \frac{k^2}{2k^2+3k+1} = \frac{1}{3}$
$3k^2 = 2k^2+3k+1 \Rightarrow k^2-3k-1 = 0$. This gives no integer solution for $k$.
Re-evaluating the consecutive even integers as $2n, 2n+2, 2n+4$ where $PN=2n, QL=2n+2, RM=2n+4$:
$PQ = 4n+2, PR = 4n+4, QR = 4n+6$. $\cos P = \frac{(4n+2)^2+(4n+4)^2-(4n+6)^2}{2(4n+2)(4n+4)} = \frac{1}{3}$
$\frac{16n^2+16n+4+16n^2+32n+16-16n^2-48n-36}{2(16n^2+24n+8)} = \frac{1}{3}$ $\Rightarrow \frac{16n^2-16}{32n^2+48n+16} = \frac{1}{3}$
$\frac{n^2-1}{2n^2+3n+1} = \frac{1}{3}$ $\Rightarrow 3n^2-3 = 2n^2+3n+1$ $\Rightarrow n^2-3n-4 = 0$
$(n-4)(n+1) = 0 \Rightarrow n=4$.
Sides are $PQ = 18, PR = 20, QR = 22$. The possible lengths are $18$ and $22$.

(B) Let the two sides be $a$ and $b$. Given $a+b=x$ and $ab=y$.
Given $x^2-c^2=y$,substituting $x=a+b$ and $y=ab$,we get $(a+b)^2-c^2=ab$.
$a^2+b^2+2ab-c^2=ab \implies a^2+b^2-c^2=-ab$.
Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{-ab}{2ab} = -\frac{1}{2}$.
Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$.
The ratio of in-radius $r$ to circum-radius $R$ is given by $\frac{r}{R} = 4 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}$.
Alternatively,$\frac{r}{R} = \frac{\Delta/s}{abc/4\Delta} = \frac{4\Delta^2}{sabc} = \frac{4 \cdot (\frac{1}{2}ab \sin C)^2}{(\frac{a+b+c}{2})abc} = \frac{a^2b^2 \sin^2(120^\circ)}{(\frac{x+c}{2})abc} = \frac{y^2 \cdot (\frac{\sqrt{3}}{2})^2}{(\frac{x+c}{2})abc} = \frac{3y^2}{2(x+c)abc}$.
Since $ab=y$,this simplifies to $\frac{3y}{2c(x+c)}$.
Wait,re-evaluating: $\frac{r}{R} = \frac{4\Delta^2}{sabc} = \frac{a^2b^2 \sin^2 C}{sabc} = \frac{y^2 \cdot (3/4)}{(\frac{x+c}{2})abc} = \frac{3y^2}{2(x+c)abc} = \frac{3y^2}{2(x+c)yc} = \frac{3y}{2c(x+c)}$.

(D) Given the equation: $\sin x + 2 \sin 2x - \sin 3x = 3$.
Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we substitute:
$\sin x + 2(2 \sin x \cos x) - (3 \sin x - 4 \sin^3 x) = 3$.
$-2 \sin x + 4 \sin x \cos x + 4 \sin^3 x = 3$.
Since $\sin x \in (0, 1]$ for $x \in (0, \pi)$,we analyze the maximum value of the expression $f(x) = \sin x + 2 \sin 2x - \sin 3x$.
We know that $\sin x \leq 1$,$\sin 2x \leq 1$,and $-\sin 3x \leq 1$.
However,these cannot all be $1$ simultaneously for the same $x$.
Alternatively,consider the range of $f(x) = \sin x + 4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 4 \sin x \cos x + 4 \sin^3 x - 2 \sin x$.
$f(x) = 2 \sin x (2 \cos x + 2 \sin^2 x - 1) = 2 \sin x (2 \cos x + 2(1 - \cos^2 x) - 1) = 2 \sin x (-2 \cos^2 x + 2 \cos x + 1)$.
Let $u = \cos x$,where $u \in (-1, 1)$. The function $g(u) = 2 \sqrt{1-u^2} (-2u^2 + 2u + 1)$.
By checking the maximum value,it is found that $f(x) < 3$ for all $x \in (0, \pi)$.
Thus,there is no solution.

(C) In $\triangle PQR$,$\angle PRQ = \angle QRS - \angle PRS = 70^{\circ} - 40^{\circ} = 30^{\circ}$.
In $\triangle QRS$,$\angle RQS = \angle PQR - \angle PQS = 70^{\circ} - 15^{\circ} = 55^{\circ}$.
Then $\angle QSR = 180^{\circ} - 70^{\circ} - 55^{\circ} = 55^{\circ}$.
Since $\angle RQS = \angle QSR = 55^{\circ}$,we have $QR = RS = 1$.
In $\triangle PQR$,$\angle QPR = 180^{\circ} - 70^{\circ} - 30^{\circ} = 80^{\circ}$.
Applying the sine rule in $\triangle PQR$:
$\frac{\alpha}{\sin 30^{\circ}} = \frac{1}{\sin 80^{\circ}} \Rightarrow \alpha = \frac{1}{2 \sin 80^{\circ}}$.
Applying the sine rule in $\triangle PRS$:
$\frac{\beta}{\sin 40^{\circ}} = \frac{1}{\sin \theta} \Rightarrow \beta \sin \theta = \sin 40^{\circ}$.
Now,$4 \alpha \beta \sin \theta = 4 \left( \frac{1}{2 \sin 80^{\circ}} \right) \sin 40^{\circ} = \frac{2 \sin 40^{\circ}}{\sin 80^{\circ}} = \frac{2 \sin 40^{\circ}}{2 \sin 40^{\circ} \cos 40^{\circ}} = \sec 40^{\circ}$.
Since $30^{\circ} < 40^{\circ} < 45^{\circ}$,we have $\sec 30^{\circ} < \sec 40^{\circ} < \sec 45^{\circ}$,which means $\frac{2}{\sqrt{3}} < \sec 40^{\circ} < \sqrt{2}$.
Since $\frac{2}{\sqrt{3}} \approx 1.15$ and $\sqrt{2} \approx 1.414$,the value $\sec 40^{\circ}$ lies in the intervals $(0, \sqrt{2})$ and $(1, 2)$.
Thus,the correct options are $(A)$ and $(B)$.

(D) Given equations are $2 \sin^2 \theta = \cos 2\theta$ and $2 \cos^2 \theta = 3 \sin \theta$.
From the first equation: $2 \sin^2 \theta = 1 - 2 \sin^2 \theta \implies 4 \sin^2 \theta = 1 \implies \sin^2 \theta = \frac{1}{4} \implies \sin \theta = \pm \frac{1}{2}$.
From the second equation: $2(1 - \sin^2 \theta) = 3 \sin \theta \implies 2 - 2 \sin^2 \theta = 3 \sin \theta \implies 2 \sin^2 \theta + 3 \sin \theta - 2 = 0$.
Factoring the quadratic: $(2 \sin \theta - 1)(\sin \theta + 2) = 0$.
Since $\sin \theta$ cannot be $-2$,we have $\sin \theta = \frac{1}{2}$.
Comparing both results,the common solution is $\sin \theta = \frac{1}{2}$.
For $\theta \in [0, 2\pi]$,the values are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
The sum of these values is $\frac{\pi}{6} + \frac{5\pi}{6} = \pi$.

(A) Given that $\tan A, \tan B, \tan C$ are in $H.P.$
$\frac{2}{\tan B} = \frac{1}{\tan A} + \frac{1}{\tan C}$
$\frac{2 \cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}$
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Also,$\cos B = \frac{a^{2}+c^{2}-b^{2}}{2ac}, \cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}, \cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$.
Substituting these values:
$2 \left( \frac{a^{2}+c^{2}-b^{2}}{2ac} \right) \cdot \frac{2R}{b} = \frac{b^{2}+c^{2}-a^{2}}{2bc} \cdot \frac{2R}{a} + \frac{a^{2}+b^{2}-c^{2}}{2ab} \cdot \frac{2R}{c}$
Multiplying both sides by $\frac{abc}{2R}$:
$2(a^{2}+c^{2}-b^{2}) = (b^{2}+c^{2}-a^{2}) + (a^{2}+b^{2}-c^{2})$
$2a^{2} + 2c^{2} - 2b^{2} = 2b^{2}$
$2a^{2} + 2c^{2} = 4b^{2}$
$a^{2} + c^{2} = 2b^{2}$
Thus,$a^{2}, b^{2}, c^{2}$ are in $A.P.$

(B) We know from the Sine Rule that $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$.
Given that $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
Dividing the Sine Rule equation by the given equation,we get $\tan A = \tan B = \tan C$.
Since $A, B, C$ are angles of a triangle,this implies $A = B = C = 60^{\circ}$,so the triangle is equilateral.
The area of an equilateral triangle with side $a$ is given by $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given $a = \sqrt{6}$,the area is $\frac{\sqrt{3}}{4} (\sqrt{6})^2 = \frac{\sqrt{3}}{4} \times 6 = \frac{3 \sqrt{3}}{2}$ sq. units.

(C) Given the cubic equation $x^3-11x^2+38x-40=0$,the roots $a, b, c$ represent the sides of the triangle.
By Vieta's formulas,we have:
$a+b+c = 11$
$ab+bc+ca = 38$
$abc = 40$
Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Substituting this into the expression $\frac{\cos A}{a} = \frac{b^2+c^2-a^2}{2abc}$.
Similarly,$\frac{\cos B}{b} = \frac{a^2+c^2-b^2}{2abc}$ and $\frac{\cos C}{c} = \frac{a^2+b^2-c^2}{2abc}$.
Summing these,we get $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2abc} = \frac{a^2+b^2+c^2}{2abc}$.
We know $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$.
$11^2 = a^2+b^2+c^2+2(38) \implies 121 = a^2+b^2+c^2+76 \implies a^2+b^2+c^2 = 45$.
Thus,the expression equals $\frac{45}{2(40)} = \frac{45}{80} = \frac{9}{16}$.

(B) Given $a \cos B = b \cos A$. Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$.
Substituting these,we get $2R \sin A \cos B = 2R \sin B \cos A$,which implies $\sin A \cos B - \cos A \sin B = 0$,so $\sin(A - B) = 0$.
Since $A$ and $B$ are angles of a triangle,$A = B$,meaning the triangle is isosceles with $a = b$.
However,the condition $a \cos C \neq c \cos A$ implies $\sin A \cos C \neq \sin C \cos A$,so $\sin(A - C) \neq 0$,meaning $A \neq C$.
Thus,the triangle is isosceles with $a = b$ and $a \neq c$.
The area of an isosceles triangle with sides $a, a, c$ is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
The base is $c$. The height $h$ is $\sqrt{a^2 - (c/2)^2} = \frac{1}{2} \sqrt{4a^2 - c^2}$.
Area $= \frac{1}{2} \times c \times \frac{1}{2} \sqrt{4a^2 - c^2} = \frac{c}{4} \sqrt{4a^2 - c^2}$.

(A) Let the sides of the triangle be $a-d$,$a$,and $a+d$,where $d > 0$. The greatest side is $a+d = 7$.
Since the angle opposite to the greatest side is $120^{\circ}$,we use the Law of Cosines:
$(a+d)^2 = (a-d)^2 + a^2 - 2a(a-d) \cos(120^{\circ})$.
Substituting $a+d=7$,we have $d = 7-a$.
$49 = (a-(7-a))^2 + a^2 - 2a(a-(7-a))(-1/2)$.
$49 = (2a-7)^2 + a^2 + a(2a-7)$.
$49 = 4a^2 - 28a + 49 + a^2 + 2a^2 - 7a$.
$7a^2 - 35a = 0$.
Since $a \neq 0$,$7a = 35$,so $a = 5$.
The sides are $5-(7-5) = 3$,$5$,and $7$.
The area of the triangle is $\frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(120^{\circ})$.
Area $= \frac{1}{2} \times 3 \times 5 \times \frac{\sqrt{3}}{2} = \frac{15 \sqrt{3}}{4} \ m^2$.

(A) The sides $a, b, c$ are the roots of the cubic equation $x^3-11x^2+38x-40=0$.
By Vieta's formulas,we have:
$a+b+c = 11$
$ab+bc+ca = 38$
$abc = 40$
Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Thus,$\frac{\cos A}{a} = \frac{b^2+c^2-a^2}{2abc}$.
Similarly,$\frac{\cos B}{b} = \frac{a^2+c^2-b^2}{2abc}$ and $\frac{\cos C}{c} = \frac{a^2+b^2-c^2}{2abc}$.
Adding these expressions:
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2abc} = \frac{a^2+b^2+c^2}{2abc}$.
We know that $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 11^2 - 2(38) = 121 - 76 = 45$.
Therefore,the sum is $\frac{45}{2(40)} = \frac{45}{80} = \frac{9}{16}$.

(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting these into the given expression:
$\frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc} = \frac{k+7}{30}$
$\frac{a^2+b^2+c^2}{2abc} = \frac{k+7}{30}$
Given $a=2, b=3, c=5$,we have $a^2+b^2+c^2 = 4+9+25 = 38$ and $2abc = 2 \times 2 \times 3 \times 5 = 60$.
So,$\frac{38}{60} = \frac{k+7}{30}$
$\frac{38}{60} = \frac{2(k+7)}{60}$
$38 = 2k + 14$
$2k = 24 \Rightarrow k = 12$.

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
The expression is $E = a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right)$.
Using the identity $2 \cos^2\theta = 1 + \cos(2\theta)$,we get:
$E = \frac{a}{2}(1 + \cos C) + \frac{c}{2}(1 + \cos A)$
$E = \frac{1}{2}(a + a \cos C + c + c \cos A)$
By the projection formula,$a \cos C + c \cos A = b$.
$E = \frac{1}{2}(a + c + b)$
Since $a + c = 2b$,we have:
$E = \frac{1}{2}(2b + b) = \frac{3b}{2}$.

(A) In $\triangle PQR$,$\angle R = \frac{\pi}{2}$,so $P + Q = \frac{\pi}{2}$.
Thus,$\frac{P+Q}{2} = \frac{\pi}{4}$.
We know that $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are roots of $ax^2 + bx + c = 0$.
Sum of roots: $\tan \frac{P}{2} + \tan \frac{Q}{2} = -\frac{b}{a}$.
Product of roots: $\tan \frac{P}{2} \tan \frac{Q}{2} = \frac{c}{a}$.
Using the identity $\tan(\frac{P}{2} + \frac{Q}{2}) = \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}}$.
Since $\frac{P+Q}{2} = \frac{\pi}{4}$,$\tan(\frac{\pi}{4}) = 1$.
So,$1 = \frac{-b/a}{1 - c/a} = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.
$a - c = -b$,which implies $a + b = c$.

(A) In $\triangle ABC$,we have $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$.
This gives:
$a+b=7k$ $(i)$
$b+c=8k$ $(ii)$
$c+a=9k$ $(iii)$
Adding these equations,we get $2(a+b+c)=24k$,so $a+b+c=12k$ $(iv)$.
Subtracting $(i), (ii), (iii)$ from $(iv)$ respectively:
$c = (a+b+c) - (a+b) = 12k - 7k = 5k$
$a = (a+b+c) - (b+c) = 12k - 8k = 4k$
$b = (a+b+c) - (c+a) = 12k - 9k = 3k$
Since $a^2+b^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2 = (5k)^2 = c^2$,the triangle is a right-angled triangle with hypotenuse $c$.
Thus,$\angle C = 90^{\circ}$.
Area of $\triangle ABC = \frac{1}{2} \times a \times b = \frac{1}{2} \times 4k \times 3k = 6k^2$.
Therefore,$\frac{(A(\triangle ABC))^2}{k^4} = \frac{(6k^2)^2}{k^4} = \frac{36k^4}{k^4} = 36$.

(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$,which implies $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$,so the triangle is equilateral.
Given $a = \frac{1}{\sqrt{6}}$,the area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2$.
Area $= \frac{\sqrt{3}}{4} \left( \frac{1}{\sqrt{6}} \right)^2 = \frac{\sqrt{3}}{4} \times \frac{1}{6} = \frac{\sqrt{3}}{24} = \frac{1}{8 \sqrt{3}}$ sq. units.

(D) Given that $A, B, C$ are in an Arithmetic Progression ($A$.$P$.).
Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
By the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$,so $\sin A = ak, \sin B = bk, \sin C = ck$.
The expression is $E = \frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$.
$E = \frac{a}{c} (2 \sin C \cos C) + \frac{c}{a} (2 \sin A \cos A)$.
Substituting $\sin C = ck$ and $\sin A = ak$:
$E = \frac{a}{c} (2 ck \cos C) + \frac{c}{a} (2 ak \cos A) = 2ak \cos C + 2ck \cos A$.
$E = 2k (a \cos C + c \cos A)$.
Using the projection formula $b = a \cos C + c \cos A$,we get:
$E = 2kb = 2 \sin B$.
Since $B = 60^{\circ}$,$E = 2 \sin 60^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.

(D) Given the expression $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.
Simplifying the terms inside the parentheses:
$= \left(\frac{c+a+b}{c}\right) \left(\frac{b+c-a}{b}\right)$
$= \frac{(b+c)+a}{c} \times \frac{(b+c)-a}{b}$
$= \frac{(b+c)^2 - a^2}{bc}$
$= \frac{b^2 + c^2 + 2bc - a^2}{bc}$
$= \frac{b^2 + c^2 - a^2}{bc} + 2$
Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,we have $\frac{b^2 + c^2 - a^2}{bc} = 2 \cos A$.
$= 2 \cos A + 2$
Given $\angle A = 60^{\circ}$,$\cos 60^{\circ} = 1/2$.
$= 2(1/2) + 2 = 1 + 2 = 3$.

(A) Given $b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=\frac{3 a}{2}$.
Using the identity $\cos ^{2} \theta = \frac{1+\cos 2\theta}{2}$,we have:
$b \left( \frac{1+\cos C}{2} \right) + c \left( \frac{1+\cos B}{2} \right) = \frac{3 a}{2}$.
Multiplying by $2$ on both sides:
$b(1+\cos C) + c(1+\cos B) = 3a$.
$b + b \cos C + c + c \cos B = 3a$.
Rearranging terms:
$(b \cos C + c \cos B) + b + c = 3a$.
By the projection rule,$b \cos C + c \cos B = a$,so:
$a + b + c = 3a$.
$b + c = 2a$.
This implies that $b, a, c$ are in $A$.$P$.

(C) Given $\sin^2 A + \sin^2 B = \sin^2 C$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation: $(\frac{a}{2R})^2 + (\frac{b}{2R})^2 = (\frac{c}{2R})^2 \Rightarrow a^2 + b^2 = c^2$.
This implies that $\triangle ABC$ is a right-angled triangle with the hypotenuse $c = l(AB) = 10$.
The area of the triangle is $Area = \frac{1}{2}ab$.
Since $c = 10$,we have $a = 10 \sin A$ and $b = 10 \cos A$.
$Area = \frac{1}{2}(10 \sin A)(10 \cos A) = 50 \sin A \cos A = 25 \sin(2A)$.
The maximum value of $\sin(2A)$ is $1$ (when $2A = 90^{\circ}$ or $A = 45^{\circ}$).
Therefore,the maximum area is $25 \times 1 = 25$.

(B) In a triangle $ABC$ right-angled at $A$,we have $B+C = 90^\circ$,so $\frac{B+C}{2} = 45^\circ$.
Since $\tan \frac{B}{2}$ and $\tan \frac{C}{2}$ are roots of $ax^2+bx+c=0$,the sum of roots is $\tan \frac{B}{2} + \tan \frac{C}{2} = -\frac{b}{a}$ and the product of roots is $\tan \frac{B}{2} \tan \frac{C}{2} = \frac{c}{a}$.
Using the identity $\tan(\frac{B}{2} + \frac{C}{2}) = \frac{\tan \frac{B}{2} + \tan \frac{C}{2}}{1 - \tan \frac{B}{2} \tan \frac{C}{2}}$,we substitute $\tan 45^\circ = 1$:
$1 = \frac{-b/a}{1 - c/a} = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.
This simplifies to $a-c = -b$,which rearranges to $a+b=c$.

(A) In $\triangle ABC$,since $\angle B = \frac{\pi}{2}$,we have $A + C = \frac{\pi}{2}$,which implies $\frac{A}{2} + \frac{C}{2} = \frac{\pi}{4}$.
Taking tangent on both sides,$\tan(\frac{A}{2} + \frac{C}{2}) = \tan(\frac{\pi}{4}) = 1$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get $\frac{\tan \frac{A}{2} + \tan \frac{C}{2}}{1 - \tan \frac{A}{2} \tan \frac{C}{2}} = 1$.
Let $\alpha = \tan \frac{A}{2}$ and $\beta = \tan \frac{C}{2}$ be the roots of $px^2 + qx + r = 0$.
From the properties of roots,$\alpha + \beta = -\frac{q}{p}$ and $\alpha \beta = \frac{r}{p}$.
Substituting these into the equation $\alpha + \beta = 1 - \alpha \beta$,we get $-\frac{q}{p} = 1 - \frac{r}{p}$.
Multiplying by $p$,we get $-q = p - r$,which simplifies to $r = p + q$ or $p + q = r$.

(A) In $\triangle PQR$,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Since $\angle R = \frac{\pi}{2} = 90^{\circ}$,we have $\angle P + \angle Q = 90^{\circ}$.
Dividing by $2$,we get $\frac{P}{2} + \frac{Q}{2} = 45^{\circ} = \frac{\pi}{4}$.
Given that $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are roots of $ax^2 + bx + c = 0$,by Vieta's formulas:
Sum of roots: $\tan \left(\frac{P}{2}\right) + \tan \left(\frac{Q}{2}\right) = -\frac{b}{a}$.
Product of roots: $\tan \left(\frac{P}{2}\right) \tan \left(\frac{Q}{2}\right) = \frac{c}{a}$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\tan \left(\frac{P}{2} + \frac{Q}{2}\right) = \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}}$.
Substituting the values: $\tan \left(\frac{\pi}{4}\right) = \frac{-b/a}{1 - c/a}$.
$1 = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.
$a - c = -b$,which implies $a + b = c$.