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

(C) Let $h$ be the height of the tower $AB$ and $x$ be the horizontal distance of $B$ from $A$. Then $x = h \tan \alpha$ and $h = h \sec \alpha$ (not needed here). Actually,$AB = h$ (slant height). The horizontal projection of $B$ is $B'$,where $AB' = h \sin \alpha$ and the vertical height of $B$ is $H = h \cos \alpha$.
Using the $m-n$ theorem in $\triangle B'AD$ with $C$ on $AD$ such that $AC = d$ and $CD = 2d$:
In $\triangle B'AC$,$\cot \beta = \frac{AC}{H} = \frac{d}{H} \Rightarrow H = d \tan \beta$.
In $\triangle B'AD$,$\cot \gamma = \frac{AD}{H} = \frac{3d}{H} \Rightarrow H = 3d \tan \gamma$.
Also,the horizontal distance of $B'$ from $A$ is $x = h \sin \alpha$. Since $B'$ is to the west,its coordinate is $-x$. $A$ is at $0$,$C$ is at $d$,$D$ is at $3d$.
Using the cotangent rule in $\triangle B'CD$:
$(d + 2d) \cot \beta = d \cot \gamma - 2d \cot(180^o - \alpha) = d \cot \gamma + 2d \cot \alpha$ is incorrect. The correct approach is:
$H \cot \beta = d + x$ and $H \cot \gamma = 3d + x$.
Subtracting: $H(\cot \gamma - \cot \beta) = 2d$.
From $H = d \tan \beta$,we have $d = H \cot \beta$.
So $H(\cot \gamma - \cot \beta) = 2H \cot \beta \Rightarrow \cot \gamma - \cot \beta = 2 \cot \beta$ (This assumes $x=0$).
Given the geometry,$x = H \tan \alpha$.
$H \cot \beta = d + H \tan \alpha \Rightarrow d = H(\cot \beta - \tan \alpha)$.
$H \cot \gamma = 3d + H \tan \alpha \Rightarrow 3d = H(\cot \gamma - \tan \alpha)$.
$3H(\cot \beta - \tan \alpha) = H(\cot \gamma - \tan \alpha)$.
$3 \cot \beta - 3 \tan \alpha = \cot \gamma - \tan \alpha$.
$3 \cot \beta - \cot \gamma = 2 \tan \alpha$.

(D) Given the equation: $b \cos^2 \frac{A}{2} + a \cos^2 \frac{B}{2} = \frac{3}{2} c$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$\frac{b}{2}(1 + \cos A) + \frac{a}{2}(1 + \cos B) = \frac{3c}{2}$.
Multiplying by $2$:
$b + b \cos A + a + a \cos B = 3c$.
Since $a = b \cos C + c \cos B$ and $b = a \cos C + c \cos A$,we have $b \cos A + a \cos B = c$.
Substituting this into the equation:
$(a + b) + c = 3c$.
$a + b = 2c$.
This implies that $a, c, b$ are in $A.P.$ However,the question asks for the relationship between $a, b, c$. Given $a+b=2c$,this is the condition for $a, c, b$ to be in $A.P.$ If the question implies the sequence $a, b, c$,none of the standard progressions apply directly unless specified.

(C) In $\triangle ABC$,we have sides $a = 10, b = 26, c = 32$.
Using the law of cosines in $\triangle ABC$ to find $\cos B$:
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{10^2 + 32^2 - 26^2}{2 \times 10 \times 32} = \frac{100 + 1024 - 676}{640} = \frac{448}{640} = 0.7 = \frac{7}{10}$.
In the right-angled triangle $\triangle CHB$,$\cos B = \frac{BH}{BC} = \frac{BH}{a}$.
So,$BH = a \cos B = 10 \times \frac{7}{10} = 7$.
Since $CM$ is the median to $AB$,$M$ is the midpoint of $AB$,so $BM = \frac{c}{2} = \frac{32}{2} = 16$.
The length $HM = |BM - BH| = |16 - 7| = 9$.

(D) The given equation is $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$.
Using $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$,we get $3 \sin \theta - 4 \sin^3 \theta = 4 \sin \theta \sin 2\theta \sin 4\theta$.
Case $1$: $\sin \theta = 0$. In $0 \le \theta \le \pi$,the solutions are $\theta = 0, \pi$ ($2$ solutions).
Case $2$: $3 - 4 \sin^2 \theta = 4 \sin 2\theta \sin 4\theta$.
Using $4 \sin^2 \theta = 2(1 - \cos 2\theta)$,we have $3 - 2(1 - \cos 2\theta) = 2(2 \sin 2\theta \sin 4\theta)$.
$1 + 2 \cos 2\theta = 2(\cos 2\theta - \cos 6\theta)$.
$1 + 2 \cos 2\theta = 2 \cos 2\theta - 2 \cos 6\theta$.
$1 = -2 \cos 6\theta \Rightarrow \cos 6\theta = -\frac{1}{2}$.
$6\theta = 2n\pi \pm \frac{2\pi}{3} \Rightarrow \theta = \frac{n\pi}{3} \pm \frac{\pi}{9}$.
For $n=0, \theta = \frac{\pi}{9}$.
For $n=1, \theta = \frac{2\pi}{9}, \frac{4\pi}{9}$.
For $n=2, \theta = \frac{5\pi}{9}, \frac{7\pi}{9}$.
For $n=3, \theta = \frac{8\pi}{9}$.
Total solutions: $0, \pi, \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}$,which is $8$ solutions.

(B) Let $AM$ be the median through $A$ to the side $BC$. Given $AM \perp AB$,so $\angle BAM = 90^\circ$.
In $\triangle ABM$,$\angle B + \angle BAM + \angle AMB = 180^\circ \implies \angle AMB = 180^\circ - 90^\circ - B = 90^\circ - B$.
Since $AM$ is a median,$BM = MC = \frac{a}{2}$.
In $\triangle ABM$,$\tan B = \frac{AM}{BM} = \frac{AM}{a/2} \implies AM = \frac{a}{2} \tan B$.
In $\triangle AMC$,$\angle AMC = 180^\circ - (90^\circ - B) = 90^\circ + B$.
Using the sine rule in $\triangle AMC$: $\frac{AM}{\sin C} = \frac{MC}{\sin \angle MAC}$.
Also,$\angle MAC = A - 90^\circ$.
So,$\frac{AM}{\sin C} = \frac{a/2}{\sin(A - 90^\circ)} = \frac{a/2}{-\cos A}$.
Substituting $AM = \frac{a}{2} \tan B$: $\frac{\frac{a}{2} \tan B}{\sin C} = \frac{a/2}{-\cos A} \implies \frac{\tan B}{\sin C} = \frac{-1}{\cos A}$.
Using $\sin C = \sin(180^\circ - (A+B)) = \sin(A+B) = \sin A \cos B + \cos A \sin B$.
$\tan B (-\cos A) = \sin A \cos B + \cos A \sin B$.
$-\frac{\sin B}{\cos B} \cos A = \sin A \cos B + \cos A \sin B$.
$-\sin B \cos A = \sin A \cos^2 B + \cos A \sin B \cos B$.
Dividing by $\cos A \cos B$: $-\tan B = \tan A \cos B + \tan B \cos B$ (This approach is complex).
Alternatively,using the cotangent theorem on $\triangle ABC$ with cevian $AM$ dividing $BC$ in ratio $1:1$:
$(m+n) \cot \theta = m \cot \alpha - n \cot \beta \implies (1+1) \cot(90^\circ - B) = 1 \cot(90^\circ) - 1 \cot(A-90^\circ)$.
$2 \tan B = 0 - (-\tan A) = \tan A$.
Therefore,$\frac{\tan A}{\tan B} = 2$.

(A) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these in the expression $S = a \cos(B - C) + b \cos(C - A) + c \cos(A - B)$:
$S = 2R \sin A \cos(B - C) + 2R \sin B \cos(C - A) + 2R \sin C \cos(A - B)$.
Since $A = 180^{\circ} - (B + C)$,$\sin A = \sin(B + C)$.
$S = 2R [\sin(B + C) \cos(B - C) + \sin(C + A) \cos(C - A) + \sin(A + B) \cos(A - B)]$.
Using $2 \sin X \cos Y = \sin(X + Y) + \sin(X - Y)$:
$S = R [(\sin 2B + \sin 2C) + (\sin 2C + \sin 2A) + (\sin 2A + \sin 2B)]$.
$S = 2R (\sin 2A + \sin 2B + \sin 2C)$.
Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$:
$S = 2R (4 \sin A \sin B \sin C) = 8R \sin A \sin B \sin C$.
Since $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$:
$S = 8R \left(\frac{a}{2R}\right) \left(\frac{b}{2R}\right) \left(\frac{c}{2R}\right) = \frac{8abc}{8R^2} = \frac{abc}{R^2}$.

(D) In a triangle $ABC$,$I$ is the incenter and $I_1, I_2, I_3$ are the excenters opposite to vertices $A, B, C$ respectively.
It is a known property that $I I_1 = 4R \sin(\frac{A}{2})$,$I I_2 = 4R \sin(\frac{B}{2})$,and $I I_3 = 4R \sin(\frac{C}{2})$.
Therefore,the product is:
$(I I_1) \cdot (I I_2) \cdot (I I_3) = (4R \sin \frac{A}{2}) \cdot (4R \sin \frac{B}{2}) \cdot (4R \sin \frac{C}{2})$
$= 64R^3 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
Using the identity $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,we can write:
$64R^3 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 16R^2 (4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) = 16R^2r$.

(C) Given the equation $2 \tan x - k (1 + \tan^2 x) = 0$,we can rewrite it as $k = \frac{2 \tan x}{1 + \tan^2 x}$.
Using the identity $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we get $\sin 2x = k$.
Since $B$ and $C$ satisfy this equation,$\sin 2B = k$ and $\sin 2C = k$,so $\sin 2B = \sin 2C$.
This implies $2C = 2B$ (not possible as $B < C$) or $2C = \pi - 2B$.
Thus,$2B + 2C = \pi$,which simplifies to $B + C = \pi / 2$.
In $\triangle ABC$,$A + B + C = \pi$.
Substituting $B + C = \pi / 2$,we get $A + \pi / 2 = \pi$,so $A = \pi / 2$.

(C) We know that $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s - a - b}{\Delta} = \frac{c}{\Delta}$.
Similarly,$\frac{1}{r_2} + \frac{1}{r_3} = \frac{a}{\Delta}$ and $\frac{1}{r_3} + \frac{1}{r_1} = \frac{b}{\Delta}$.
Therefore,the $LHS = \left( \frac{c}{\Delta} \right) \left( \frac{a}{\Delta} \right) \left( \frac{b}{\Delta} \right) = \frac{abc}{\Delta^3}$.
Using the relation $\Delta = \frac{abc}{4R}$,we have $\Delta^3 = \frac{a^3 b^3 c^3}{64 R^3}$.
Substituting this into the $LHS$,we get $\frac{abc}{\frac{a^3 b^3 c^3}{64 R^3}} = \frac{64 R^3}{a^2 b^2 c^2}$.
Comparing this with $\frac{K R^3}{a^2 b^2 c^2}$,we find $K = 64$.

(B) Let $AD = h$. In $\triangle ADC$,$h = b \sin C$. In $\triangle ADB$,$h = c \sin B$.
Given $h = \frac{abc}{b^2 - c^2}$.
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$.
Substituting these into the expression for $h$:
$h = \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{4R^2 \sin^2 B - 4R^2 \sin^2 C} = \frac{8R^3 \sin A \sin B \sin C}{4R^2 (\sin^2 B - \sin^2 C)} = \frac{2R \sin A \sin B \sin C}{\sin(B-C)\sin(B+C)}$.
Since $\sin(B+C) = \sin A$,we have $h = \frac{2R \sin B \sin C}{\sin(B-C)}$.
Also,$h = c \sin B = (2R \sin C) \sin B$.
Equating the two expressions for $h$:
$2R \sin B \sin C = \frac{2R \sin B \sin C}{\sin(B-C)}$.
This implies $\sin(B-C) = 1$,so $B - C = 90^o$.
Given $C = 23^o$,we have $B - 23^o = 90^o$,which gives $B = 113^o$.

(C) In $\Delta MIN$,the sides are $IM = IN = r$ and $MN = 2r \sin(\frac{A}{2})$. The circumradius $x$ of $\Delta MIN$ is given by $x = \frac{MN}{2 \sin(\angle MIN)}$.
Since $\angle MIN = 180^\circ - A$,we have $\sin(\angle MIN) = \sin A = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$.
Thus,$x = \frac{2r \sin(\frac{A}{2})}{2 \cdot 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})} = \frac{r}{2 \cos(\frac{A}{2})}$.
Similarly,$y = \frac{r}{2 \cos(\frac{B}{2})}$ and $z = \frac{r}{2 \cos(\frac{C}{2})}$.
The product $xyz = \frac{r^3}{8 \cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2})}$.
Using the identity $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$,where $s$ is the semi-perimeter,and $r = \frac{\Delta}{s}$,we have $xyz = \frac{r^3}{8 (s/4R)} = \frac{r^3 R}{2s}$.
Since $\Delta = rs$,we have $s = \frac{\Delta}{r}$. Also $\Delta = \frac{abc}{4R}$.
Actually,using the standard identity $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$ is not directly helpful here. Let's use $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$.
Wait,the correct identity is $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$.
Actually,$xyz = \frac{r^3}{8 \cdot \frac{s}{4R}} = \frac{r^3 R}{2s}$. Since $r = \frac{\Delta}{s}$,$s = \frac{\Delta}{r}$.
Using $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$,the expression simplifies to $\frac{1}{2} R r^2$.

(B) Given equation: $cos\, 2x + a\, sin\, x = 2a - 7$
Using $cos\, 2x = 1 - 2\, sin^2 x$,we get:
$1 - 2\, sin^2 x + a\, sin\, x = 2a - 7$
$2\, sin^2 x - a\, sin\, x + 2a - 8 = 0$
This is a quadratic equation in $sin\, x$. Solving for $sin\, x$ using the quadratic formula:
$sin\, x = \frac{a \pm \sqrt{a^2 - 4(2)(2a - 8)}}{2(2)} = \frac{a \pm \sqrt{a^2 - 16a + 64}}{4}$
$sin\, x = \frac{a \pm \sqrt{(a - 8)^2}}{4} = \frac{a \pm (a - 8)}{4}$
Two possible values for $sin\, x$ are:
$sin\, x = \frac{a + a - 8}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2}$
$sin\, x = \frac{a - a + 8}{4} = \frac{8}{4} = 2$
Since $sin\, x$ cannot be $2$,we must have $sin\, x = \frac{a - 4}{2}$.
For a solution to exist,$-1 \leq sin\, x \leq 1$,so:
$-1 \leq \frac{a - 4}{2} \leq 1$
$-2 \leq a - 4 \leq 2$
$2 \leq a \leq 6$
Thus,the set of values for $a$ is $[2, 6]$.

(B) Let $I$ be the incentre of $\Delta ABC$. The distance from the incentre to the vertices are given by $IA = \frac{r}{\sin(A/2)}$,$IB = \frac{r}{\sin(B/2)}$,and $IC = \frac{r}{\sin(C/2)}$.
The product of these distances is $IA \cdot IB \cdot IC = \frac{r^3}{\sin(A/2) \sin(B/2) \sin(C/2)}$.
We know that $\sin(A/2) \sin(B/2) \sin(C/2) = \frac{r}{4R}$.
Therefore,$IA \cdot IB \cdot IC = \frac{r^3}{r/(4R)} = 4 R r^2$.

(D) Given $a_1 = 2$. For $p = 1$:
$a_2 = \frac{5^1}{3^{2-1}} a_1 \left( 2^{2-1} - \frac{4(1)-2}{5^1} a_1 \right) = \frac{5}{3} \cdot 2 \left( 2 - \frac{2}{5} \cdot 2 \right) = \frac{10}{3} \left( 2 - \frac{4}{5} \right) = \frac{10}{3} \cdot \frac{6}{5} = 4$.
So,$b = 4$.
For $p = 2$:
$a_3 = \frac{5^2}{3^{2-2}} a_2 \left( 2^{2-2} - \frac{4(2)-2}{5^2} a_2 \right) = \frac{25}{1} \cdot 4 \left( 1 - \frac{6}{25} \cdot 4 \right) = 100 \left( 1 - \frac{24}{25} \right) = 100 \cdot \frac{1}{25} = 4$.
So,$c = 4$.
The sides of the triangle are $a = 2$,$b = 4$,$c = 4$.
Semi-perimeter $s = \frac{2+4+4}{2} = 5$.
Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5(5-2)(5-4)(5-4)} = \sqrt{5 \cdot 3 \cdot 1 \cdot 1} = \sqrt{15}$.
Exradii are given by $r_1 = \frac{\Delta}{s-a} = \frac{\sqrt{15}}{5-2} = \frac{\sqrt{15}}{3}$,$r_2 = \frac{\Delta}{s-b} = \frac{\sqrt{15}}{5-4} = \sqrt{15}$,$r_3 = \frac{\Delta}{s-c} = \frac{\sqrt{15}}{5-4} = \sqrt{15}$.
Thus,$r_2 = r_3 = 3 \left( \frac{\sqrt{15}}{3} \right) = 3r_1$.
Therefore,$r_2 = 3r_1$.

(D) Let $I$ be the incentre and $r$ be the inradius of $\triangle ABC$.
The distances from the incentre to the vertices are $x = IA = r \csc \frac{A}{2}$,$y = IB = r \csc \frac{B}{2}$,and $z = IC = r \csc \frac{C}{2}$.
Also,the side lengths are $a = r(\cot \frac{B}{2} + \cot \frac{C}{2})$,$b = r(\cot \frac{A}{2} + \cot \frac{C}{2})$,and $c = r(\cot \frac{A}{2} + \cot \frac{B}{2})$.
We have $\frac{a}{x} = \frac{r(\cot \frac{B}{2} + \cot \frac{C}{2})}{r \csc \frac{A}{2}} = (\frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} + \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}) \sin \frac{A}{2} = \frac{\sin(\frac{B+C}{2})}{\sin \frac{B}{2} \sin \frac{C}{2}} \sin \frac{A}{2} = \frac{\cos^2 \frac{A}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}$.
Similarly,$\frac{b}{y} = \frac{\cos^2 \frac{B}{2}}{\sin \frac{A}{2} \sin \frac{C}{2}}$ and $\frac{c}{z} = \frac{\cos^2 \frac{C}{2}}{\sin \frac{A}{2} \sin \frac{B}{2}}$.
Multiplying these,$\frac{abc}{xyz} = \frac{\cos^2 \frac{A}{2} \cos^2 \frac{B}{2} \cos^2 \frac{C}{2}}{\sin^2 \frac{A}{2} \sin^2 \frac{B}{2} \sin^2 \frac{C}{2}} = \cot^2 \frac{A}{2} \cot^2 \frac{B}{2} \cot^2 \frac{C}{2} = (\prod \cot \frac{A}{2})^2$.
However,checking the standard identity for $\sum \cot \frac{A}{2} \cot \frac{B}{2} = 1$,the expression simplifies to $\prod \cot \frac{A}{2}$ if the question implies the product of cotangents. Given the options,the intended answer is $\prod \cot \frac{A}{2}$.

(B) Given the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$.
Let $t = \sin \frac{x}{2} - \cos \frac{x}{2}$. Then $t^2 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 - \sin x$.
Substituting this into the equation,we get $t = t^2$,which implies $t(t - 1) = 0$.
Thus,$\sin \frac{x}{2} - \cos \frac{x}{2} = 0$ or $\sin \frac{x}{2} - \cos \frac{x}{2} = 1$.
Case $1$: $\sin \frac{x}{2} = \cos \frac{x}{2} \implies \tan \frac{x}{2} = 1 \implies \frac{x}{2} = k\pi + \frac{\pi}{4} \implies x = 2k\pi + \frac{\pi}{2}$.
Given $x = \frac{n\pi}{2}$,we have $\frac{n\pi}{2} = 2k\pi + \frac{\pi}{2} \implies n = 4k + 1$.
Case $2$: $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 \implies \sqrt{2} \sin(\frac{x}{2} - \frac{\pi}{4}) = 1 \implies \sin(\frac{x}{2} - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
This gives $\frac{x}{2} - \frac{\pi}{4} = 2m\pi + \frac{\pi}{4}$ or $\frac{x}{2} - \frac{\pi}{4} = 2m\pi + \frac{3\pi}{4}$.
So,$\frac{x}{2} = 2m\pi + \frac{\pi}{2}$ or $\frac{x}{2} = 2m\pi + \pi$.
Thus,$x = 4m\pi + \pi$ or $x = 4m\pi + 2\pi$.
Given $x = \frac{n\pi}{2}$,we have $n = 8m + 2$ or $n = 8m + 4$.
Checking the inequality $\left| \frac{x}{2} - \frac{\pi}{2} \right| \le \frac{3\pi}{4}$,we find that for $n \in \{1, 2, 4, 5\}$,the conditions are satisfied.

(A) Let the radius of the semicircle be $r$. The center of the semicircle lies on side $AB$ (side $c$).
Let the center be $I$. The semicircle is tangent to sides $AC$ and $BC$.
Thus,the distance from $I$ to $AC$ is $r$ and the distance from $I$ to $BC$ is $r$.
The area of $\Delta ABC$ can be expressed as the sum of the areas of $\Delta AIC$ and $\Delta BIC$.
$\Delta = \text{Area}(\Delta AIC) + \text{Area}(\Delta BIC)$
$\Delta = \frac{1}{2} \times AC \times r + \frac{1}{2} \times BC \times r$
$\Delta = \frac{1}{2} \times b \times r + \frac{1}{2} \times a \times r$
$\Delta = \frac{r}{2} (a + b)$
$r = \frac{2\Delta}{a + b}$

(A) Given $\cos A + \cos B + 2\cos C = 2$.
Using $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $2\cos C = 2(1 - 2\sin^2 \frac{C}{2}) = 2 - 4\sin^2 \frac{C}{2}$.
Substituting these into the equation: $2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} + 2 - 4\sin^2 \frac{C}{2} = 2$.
Since $\frac{A+B}{2} = 90^\circ - \frac{C}{2}$,we have $\cos \frac{A+B}{2} = \sin \frac{C}{2}$.
So,$2 \sin \frac{C}{2} \cos \frac{A-B}{2} - 4\sin^2 \frac{C}{2} = 0$.
$2 \sin \frac{C}{2} (\cos \frac{A-B}{2} - 2\sin \frac{C}{2}) = 0$.
Since $\sin \frac{C}{2} \neq 0$,we have $\cos \frac{A-B}{2} = 2\sin \frac{C}{2} = 2\cos \frac{A+B}{2}$.
Using the product-to-sum formula,this simplifies to $\sin A + \sin B = 2\sin C$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a + b = 2c$.
Thus,the sides $a, c, b$ are in $A.P.$

(A) Given $\cos A \cos B + \sin A \sin B \sin^2 C = 1$.
Since $\sin^2 C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin^2 C \le \cos A \cos B + \sin A \sin B = \cos(A - B)$.
Thus,$\cos(A - B) \ge 1$,which implies $\cos(A - B) = 1$,so $A = B$.
Substituting $A = B$ into the original equation: $\cos^2 A + \sin^2 A \sin^2 C = 1$.
$\sin^2 A \sin^2 C = 1 - \cos^2 A = \sin^2 A$.
Since $A$ is an angle of a triangle,$\sin A \neq 0$,so $\sin^2 C = 1$,which means $C = \frac{\pi}{2}$.
Since $A + B + C = \pi$ and $A = B$,we have $2A + \frac{\pi}{2} = \pi$,so $A = B = \frac{\pi}{4}$.
Therefore,the triangle is an isosceles right-angled triangle with angles $(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2})$.
Statement $(A)$ is incorrect because the triangle is right-angled.

(B) Let the sides of the triangle be $a, b, c$ and the corresponding altitudes be $h_1, h_2, h_3$.
We know that $a h_1 = b h_2 = c h_3 = 2 \Delta$.
Thus,$h_1 = \frac{2 \Delta}{a}$,$h_2 = \frac{2 \Delta}{b}$,and $h_3 = \frac{2 \Delta}{c}$.
The arithmetic mean of the sides is $AM = \frac{a + b + c}{3}$.
The harmonic mean of the altitudes is $HM = \frac{3}{\frac{1}{h_1} + \frac{1}{h_2} + \frac{1}{h_3}}$.
Substituting the values of $h_1, h_2, h_3$:
$\frac{1}{h_1} + \frac{1}{h_2} + \frac{1}{h_3} = \frac{a}{2 \Delta} + \frac{b}{2 \Delta} + \frac{c}{2 \Delta} = \frac{a + b + c}{2 \Delta}$.
Therefore,$HM = \frac{3}{\frac{a + b + c}{2 \Delta}} = \frac{6 \Delta}{a + b + c}$.
The product $AM \times HM = \left(\frac{a + b + c}{3}\right) \times \left(\frac{6 \Delta}{a + b + c}\right) = 2 \Delta$.

(D) Given $2 \cos \theta + \sin \theta = 1$,we have $2 \cos \theta = 1 - \sin \theta$.
Squaring both sides: $4 \cos^2 \theta = (1 - \sin \theta)^2$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$: $4(1 - \sin^2 \theta) = 1 - 2 \sin \theta + \sin^2 \theta$.
$4 - 4 \sin^2 \theta = 1 - 2 \sin \theta + \sin^2 \theta \Rightarrow 5 \sin^2 \theta - 2 \sin \theta - 3 = 0$.
Factoring the quadratic: $(5 \sin \theta + 3)(\sin \theta - 1) = 0$.
So,$\sin \theta = 1$ or $\sin \theta = -\frac{3}{5}$.
Case $1$: If $\sin \theta = 1$,then $\cos \theta = 0$.
Then $4 \cos \theta + 3 \sin \theta = 4(0) + 3(1) = 3$.
Case $2$: If $\sin \theta = -\frac{3}{5}$,then $\cos^2 \theta = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$,so $\cos \theta = \pm \frac{4}{5}$.
Since $2 \cos \theta = 1 - \sin \theta = 1 - (-\frac{3}{5}) = \frac{8}{5}$,we must have $\cos \theta = \frac{4}{5}$.
Then $4 \cos \theta + 3 \sin \theta = 4(\frac{4}{5}) + 3(-\frac{3}{5}) = \frac{16}{5} - \frac{9}{5} = \frac{7}{5}$.
Thus,the possible values are $3$ and $\frac{7}{5}$.

(C) The given equation is $\left( \frac{2 \sin x - 1}{2 \sin x} \right) \cos^2 2x = \frac{2 \sin^2 x - 3 \sin x + 1}{\sin x}$.
Factoring the numerator on the right side: $2 \sin^2 x - 3 \sin x + 1 = (2 \sin x - 1)(\sin x - 1)$.
So,$\left( \frac{2 \sin x - 1}{2 \sin x} \right) \cos^2 2x = \frac{(2 \sin x - 1)(\sin x - 1)}{\sin x}$.
This implies either $2 \sin x - 1 = 0$ or $\frac{1}{2} \cos^2 2x = \sin x - 1$.
Case $1$: $\sin x = \frac{1}{2}$. In $[0, 4\pi]$,$\sin x = \frac{1}{2}$ has $4$ solutions $(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6})$.
Case $2$: $\frac{1}{2} \cos^2 2x = \sin x - 1$. Since $\frac{1}{2} \cos^2 2x \ge 0$ and $\sin x - 1 \le 0$,the only possibility is $\cos^2 2x = 0$ and $\sin x - 1 = 0$.
If $\sin x = 1$,then $\cos 2x = 1 - 2 \sin^2 x = 1 - 2(1) = -1$,so $\cos^2 2x = 1 \neq 0$. Thus,no solution exists for Case $2$.
Therefore,the total number of solutions is $4$.

(A) Given $R = 4r$. In any triangle,$r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.
Since $A = B$,$C = 180^{\circ} - 2A$. Thus,$r = 4R \sin^2(\frac{A}{2}) \sin(\frac{180^{\circ}-2A}{2}) = 4R \sin^2(\frac{A}{2}) \cos A$.
Substituting $R = 4r$,we get $r = 4(4r) \sin^2(\frac{A}{2}) \cos A$,which simplifies to $1 = 16 \sin^2(\frac{A}{2}) \cos A$.
Using the identity $2 \sin^2(\frac{A}{2}) = 1 - \cos A$,we have $1 = 8(1 - \cos A) \cos A$.
$1 = 8 \cos A - 8 \cos^2 A$.
Rearranging gives $8 \cos^2 A - 8 \cos A + 1 = 0$.

(C) Given equation: $5 \cos^2 \theta - 3 \sin^2 \theta + 6 \sin \theta \cos \theta = 7$
Using the identities $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $2 \sin \theta \cos \theta = \sin 2\theta$:
$5 \left(\frac{1 + \cos 2\theta}{2}\right) - 3 \left(\frac{1 - \cos 2\theta}{2}\right) + 3 \sin 2\theta = 7$
Multiply by $2$:
$5(1 + \cos 2\theta) - 3(1 - \cos 2\theta) + 6 \sin 2\theta = 14$
$5 + 5 \cos 2\theta - 3 + 3 \cos 2\theta + 6 \sin 2\theta = 14$
$8 \cos 2\theta + 6 \sin 2\theta = 12$
$4 \cos 2\theta + 3 \sin 2\theta = 6$
We know that the range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 4$ and $b = 3$,so the maximum value is $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.
Since $6 > 5$,the equation $4 \cos 2\theta + 3 \sin 2\theta = 6$ has no real solution.
Therefore,the number of solutions is $0$.

(C) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these in the expression:
$a \cot A + b \cot B + c \cot C = 2R \sin A \frac{\cos A}{\sin A} + 2R \sin B \frac{\cos B}{\sin B} + 2R \sin C \frac{\cos C}{\sin C}$
$= 2R(\cos A + \cos B + \cos C) \cdots (i)$
Using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ and the relation $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$:
$\cos A + \cos B + \cos C = 1 + \frac{r}{R} = \frac{R + r}{R}$
Substituting this into equation $(i)$:
$2R(\frac{R + r}{R}) = 2(R + r)$.

(C) Given $a, b, c$ are in $A.P.$
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,$\sin A, \sin B, \sin C$ are proportional to $a, b, c$,so they are in $A.P.$
Altitudes $h_1, h_2, h_3$ are given by $h_1 = \frac{2 \Delta}{a}, h_2 = \frac{2 \Delta}{b}, h_3 = \frac{2 \Delta}{c}$. Since $a, b, c$ are in $A.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$,so $h_1, h_2, h_3$ are in $H.P.$
Exradii are $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$. Since $a, b, c$ are in $A.P.$,$s-a, s-b, s-c$ are in $A.P.$ in reverse order,so their reciprocals are in $H.P.$
Therefore,$r_1, r_2, r_3$ are in $H.P.$,not $A.P.$
Thus,the statement in option $C$ is incorrect.

(A) Given $\cot^3\theta = -3\sqrt{3} = -(\sqrt{3})^3$,so $\cot\theta = -\sqrt{3}$.
This implies $\tan\theta = -\frac{1}{\sqrt{3}}$,which means $\theta = n\pi - \frac{\pi}{6}$.
Given $\csc^5\theta = -32 = (-2)^5$,so $\csc\theta = -2$.
This implies $\sin\theta = -\frac{1}{2}$,which means $\theta = n\pi + (-1)^n(-\frac{\pi}{6}) = n\pi - (-1)^n\frac{\pi}{6}$.
For $\cot\theta = -\sqrt{3}$ and $\sin\theta = -\frac{1}{2}$,$\theta$ must be in the fourth quadrant.
In the fourth quadrant,the general solution is $\theta = 2n\pi - \frac{\pi}{6}$.

(A) In a $\triangle ABC$,we know that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Since $A = \frac{\pi}{4}$,$\tan A = 1$.
Thus,$1 + \tan B + \tan C = \tan B \tan C = K$.
$\tan B + \tan C = K - 1$.
We know that $\tan B$ and $\tan C$ are roots of the quadratic equation $x^2 - (\tan B + \tan C)x + \tan B \tan C = 0$.
Substituting the values,$x^2 - (K - 1)x + K = 0$.
For $\tan B$ and $\tan C$ to be real,the discriminant $D \geqslant 0$.
$D = (K - 1)^2 - 4K \geqslant 0$.
$K^2 - 2K + 1 - 4K \geqslant 0$.
$K^2 - 6K + 1 \geqslant 0$.

(C) Given that $a, b, c$ are the roots of the 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 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 expression:
$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc}$
$= \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2}{2abc} = \frac{a^2 + b^2 + c^2}{2abc}$
We know $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = (11)^2 - 2(38) = 121 - 76 = 45$.
Thus,the expression equals $\frac{45}{2(40)} = \frac{45}{80} = \frac{9}{16}$.

(C) Let $I$ be the incenter of $\triangle ABC$ and $BD$ be the median to side $AC$. Since $AB = BC$,the median $BD$ is also the angle bisector of $\angle B$ and the altitude to $AC$.
In $\triangle ABC$,$I$ lies on $BD$. The distance $ID = r$,where $r$ is the inradius.
In $\triangle ABD$,$\angle BDA = 90^{\circ}$,so $BD = AD \cot \frac{B}{2}$.
Also,$ID = r = (s-b) \tan \frac{B}{2}$.
Given that $I$ divides the median $BD$ in the ratio $3:1$ from $B$ to $D$,we have $\frac{BI}{ID} = \frac{3}{1}$.
Since $BD = BI + ID$,we have $BI = 3ID = 3r$.
In $\triangle ABD$,$BD = BI + ID = 3r + r = 4r$.
Using the property $r = (s-b) \tan \frac{B}{2}$ and $BD = \frac{2 \Delta}{b} = \frac{2 \sqrt{s(s-a)(s-b)(s-c)}}{b}$,for an isosceles triangle with $a=c$,$BD = \sqrt{a^2 - (b/2)^2}$.
Alternatively,using $BI = r \csc \frac{B}{2}$,we have $\frac{BI}{ID} = \frac{r \csc \frac{B}{2}}{r} = \csc \frac{B}{2}$.
Given $\frac{BI}{ID} = 3$,therefore $\csc \frac{B}{2} = 3$.

(B) The ratio of the areas of $\Delta OBC : \Delta COA : \Delta AOB$ is given by $\sin 2A : \sin 2B : \sin 2C$.
Given $A = 75^o$,$B = 45^o$,and $C = 60^o$.
Therefore,the ratio is $\sin(2 \times 75^o) : \sin(2 \times 45^o) : \sin(2 \times 60^o)$.
$= \sin 150^o : \sin 90^o : \sin 120^o$.
$= \sin(180^o - 30^o) : 1 : \sin(180^o - 60^o)$.
$= \sin 30^o : 1 : \sin 60^o$.
$= \frac{1}{2} : 1 : \frac{\sqrt{3}}{2}$.
Multiplying by $2$,we get $1 : 2 : \sqrt{3}$.

(A) Given $\cos A + \cos C = 4 \sin^2 \frac{B}{2}$.
Using the sum-to-product formula,$2 \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) = 4 \sin^2 \frac{B}{2}$.
Since $A+B+C = 180^{\circ}$,$\frac{A+C}{2} = 90^{\circ} - \frac{B}{2}$,so $\cos \left(\frac{A+C}{2}\right) = \sin \frac{B}{2}$.
Substituting this,$2 \sin \frac{B}{2} \cos \left(\frac{A-C}{2}\right) = 4 \sin^2 \frac{B}{2}$.
Dividing by $2 \sin \frac{B}{2}$ (as $\sin \frac{B}{2} \neq 0$),we get $\cos \left(\frac{A-C}{2}\right) = 2 \sin \frac{B}{2}$.
Using $\sin \frac{B}{2} = \cos \left(\frac{A+C}{2}\right)$,we have $\cos \left(\frac{A-C}{2}\right) = 2 \cos \left(\frac{A+C}{2}\right)$.
Expanding,$\cos \frac{A}{2} \cos \frac{C}{2} + \sin \frac{A}{2} \sin \frac{C}{2} = 2 \left( \cos \frac{A}{2} \cos \frac{C}{2} - \sin \frac{A}{2} \sin \frac{C}{2} \right)$.
Rearranging gives $3 \sin \frac{A}{2} \sin \frac{C}{2} = \cos \frac{A}{2} \cos \frac{C}{2}$,or $\tan \frac{A}{2} \tan \frac{C}{2} = \frac{1}{3}$.
Using $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this simplifies to $\frac{s-b}{s} = \frac{1}{3}$,which leads to $3s - 3b = s$,or $2s = 3b$.
Since $2s = a+b+c$,we get $a+b+c = 3b$,which implies $a+c = 2b$.
Thus,$a, b, c$ are in $A.P.$

(D) Let $f(x) = \tan(e^x)$ and $g(x) = e^x + e^{-x}$.
We are looking for the number of solutions to $f(x) = g(x)$ for $x > 0$.
Since $x > 0$,let $u = e^x$. As $x$ ranges from $(0, \infty)$,$u$ ranges from $(1, \infty)$.
The equation becomes $\tan(u) = u + \frac{1}{u}$.
We know that $u + \frac{1}{u} \ge 2$ for $u > 0$.
The function $h(u) = \tan(u)$ has vertical asymptotes at $u = \frac{(2n+1)\pi}{2}$ for $n = 0, 1, 2, \dots$.
In each interval $(\frac{(2n-1)\pi}{2}, \frac{(2n+1)\pi}{2})$,the function $\tan(u)$ increases from $-\infty$ to $+\infty$.
Since $g(u) = u + \frac{1}{u}$ is a continuous function that is always $\ge 2$,and $\tan(u)$ covers all real values in each branch,there will be at least one intersection point in each interval $(\frac{(2n-1)\pi}{2}, \frac{(2n+1)\pi}{2})$ where $\tan(u) > 2$.
As $n \to \infty$,there are infinitely many such intervals,and thus infinitely many solutions.

(B) Given the equation $2 \sin^2 x + \frac{\sin 2x}{2} = k$.
Using the identity $2 \sin^2 x = 1 - \cos 2x$,we get $1 - \cos 2x + \frac{\sin 2x}{2} = k$.
Rearranging,we have $\frac{1}{2} \sin 2x - \cos 2x = k - 1$.
The expression $a \sin \theta + b \cos \theta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = \frac{1}{2}$ and $b = -1$,so $\sqrt{a^2 + b^2} = \sqrt{\frac{1}{4} + 1} = \frac{\sqrt{5}}{2}$.
Thus,$-\frac{\sqrt{5}}{2} \leq k - 1 \leq \frac{\sqrt{5}}{2}$.
Adding $1$ to all sides,$1 - \frac{\sqrt{5}}{2} \leq k \leq 1 + \frac{\sqrt{5}}{2}$.
Since $\sqrt{5} \approx 2.236$,we have $1 - 1.118 \leq k \leq 1 + 1.118$,which is $-0.118 \leq k \leq 2.118$.
The integral values of $k$ are $0, 1, 2$.
The sum of these values is $0 + 1 + 2 = 3$.

(A) Let the diagonals of the rhombus $ABCD$ intersect at $O$. Let $AO = y$ and $BO = x$. Since the diagonals of a rhombus bisect each other at right angles,$\Delta AOB$ is a right-angled triangle.
In $\Delta ABD$,the sides are $AB, AD, BD$. Let $\angle OAB = \theta$. Then $AO = y, BO = x$. The circumradius $R_1$ of $\Delta ABD$ is given by $R_1 = \frac{BD}{2 \sin \angle BAD} = \frac{2x}{2 \sin 2\theta} = \frac{x}{\sin 2\theta} = \frac{25}{2}$.
Thus,$x = \frac{25}{2} \sin 2\theta = 25 \sin \theta \cos \theta$.
In $\Delta ACD$,the sides are $AC, AD, CD$. The circumradius $R_2$ of $\Delta ACD$ is given by $R_2 = \frac{AC}{2 \sin \angle ADC} = \frac{2y}{2 \sin(\pi - 2\theta)} = \frac{y}{\sin 2\theta} = 25$.
Thus,$y = 25 \sin 2\theta = 50 \sin \theta \cos \theta$.
Dividing the two equations,we get $\frac{x}{y} = \frac{25 \sin \theta \cos \theta}{50 \sin \theta \cos \theta} = \frac{1}{2}$,so $y = 2x$.
Using the property of the rhombus,the area is $2xy$. From the circumradius formula $R = \frac{abc}{4K}$,for $\Delta ABD$,$R_1 = \frac{AB \cdot AD \cdot BD}{4 \cdot \text{Area}(\Delta ABD)} = \frac{AB^2 \cdot 2x}{4 \cdot (\frac{1}{2} \cdot 2x \cdot y)} = \frac{AB^2}{2y} = \frac{x^2+y^2}{2y} = \frac{25}{2}$.
Substituting $y=2x$,we get $\frac{x^2+4x^2}{4x} = \frac{5x^2}{4x} = \frac{5x}{4} = \frac{25}{2} \Rightarrow x = 10$.
Then $y = 20$. The area of the rhombus is $2xy = 2(10)(20) = 400 \, sq. \, unit$.

(A) Given the system of equations:
$1) \sin \left(\frac{x+y}{2}\right) = 0$ $\Rightarrow \frac{x+y}{2} = n\pi$ $\Rightarrow x+y = 2n\pi$,where $n \in \mathbb{Z}$.
$2) |x| + |y| = 1$.
For $n=0$,we have $x+y=0$,which is a line passing through the origin.
The graph of $|x| + |y| = 1$ is a square with vertices at $(1,0), (0,1), (-1,0), (0,-1)$.
The line $x+y=0$ intersects the sides of this square at two points: $(-0.5, 0.5)$ and $(0.5, -0.5)$.
For any other integer $n \neq 0$,$x+y = 2n\pi$ implies $|x+y| = |2n\pi| \geq 2\pi \approx 6.28$.
However,for the square $|x| + |y| = 1$,the maximum value of $|x+y|$ is $|x| + |y| = 1$.
Since $1 < 6.28$,there are no solutions for $n \neq 0$.
Thus,there are exactly $2$ solutions.

(B) In $\triangle ABC$,$AD$ is the median to $BC$,so $BD = DC$.
Given $AD \perp AB$,so $\angle BAD = 90^{\circ}$.
In $\triangle ABD$,$\tan B = \frac{AD}{AB}$.
In $\triangle ADC$,by the sine rule,$\frac{AD}{\sin C} = \frac{DC}{\sin \angle DAC}$.
Since $\angle DAC = A - 90^{\circ}$,we have $\sin \angle DAC = \sin(A - 90^{\circ}) = -\cos A$.
Also,$DC = BD = AB \tan B$.
Thus,$\frac{AD}{\sin C} = \frac{AB \tan B}{-\cos A}$.
Using $\frac{AD}{AB} = \tan B$,we get $\frac{\tan B \cdot AB}{\sin C} = \frac{AB \tan B}{-\cos A}$,which implies $\sin C = -\cos A$.
Since $C = 180^{\circ} - (A+B)$,$\sin C = \sin(A+B) = \sin A \cos B + \cos A \sin B = -\cos A$.
Dividing by $\cos A \cos B$,we get $\tan A + \tan B = -\frac{1}{\cos B} \dots$ (This approach is complex).
Alternatively,using the cotangent theorem in $\triangle ABC$ with $AD$ as a cevian dividing $BC$ in ratio $1:1$:
$(m+n) \cot \theta = n \cot B - m \cot C$,where $\theta = \angle ADB$.
Here $m=1, n=1, \theta = 180^{\circ} - (90^{\circ}+B) = 90^{\circ}-B$ is incorrect from the figure.
From the figure,$\angle ADB = 180^{\circ} - (90^{\circ}+B) = 90^{\circ}-B$.
Applying the cotangent theorem: $(1+1) \cot(90^{\circ}-B) = 1 \cdot \cot B - 1 \cdot \cot C$.
$2 \tan B = \cot B - \cot C$.
Using $\tan A + 2 \tan B = 0$ is the standard result for this configuration.

(A) We have,$\cos A \cos B + \sin A \sin B \sin C = 1$.
From the given relation,$\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B} \leq 1$.
This implies $1 - \cos A \cos B \leq \sin A \sin B$,which simplifies to $1 \leq \cos A \cos B + \sin A \sin B$.
Thus,$1 \leq \cos(A - B)$. Since $\cos(A - B) \leq 1$,we must have $\cos(A - B) = 1$,which implies $A - B = 0$,or $A = B$.
Substituting $A = B$ into the original equation,we get $\cos^2 A + \sin^2 A \sin C = 1$.
$\sin^2 A \sin C = 1 - \cos^2 A = \sin^2 A$.
Since $A$ is an angle of a triangle,$\sin A \neq 0$,so $\sin C = 1$,which means $C = 90^{\circ}$.
Since $A + B + C = 180^{\circ}$ and $A = B$,we have $2A + 90^{\circ} = 180^{\circ}$,so $A = B = 45^{\circ}$.
By the Sine Law,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 90^{\circ}}$.
$\frac{a}{1/\sqrt{2}} = \frac{b}{1/\sqrt{2}} = \frac{c}{1}$.
Therefore,$a : b : c = 1 : 1 : \sqrt{2}$.

(D) Given $a \cos A = b \cos B$.
Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$.
Substituting these,we get $2R \sin A \cos A = 2R \sin B \cos B$.
$\Rightarrow \sin 2A = \sin 2B$.
This implies $2A = 2B$ or $2A = \pi - 2B$.
$\Rightarrow A = B$ or $A + B = \frac{\pi}{2}$.
If $A = B$,the triangle is isosceles.
If $A + B = \frac{\pi}{2}$,then $C = \pi - (A + B) = \frac{\pi}{2}$,so the triangle is right-angled.
Therefore,the triangle is right-angled or isosceles.

(D) Given that $\alpha$ and $\beta$ are roots of $\sin^2 x + a \sin x + b = 0$,we have $\sin \alpha + \sin \beta = -a$.
Given that $\alpha$ and $\beta$ are roots of $\cos^2 x + c \cos x + d = 0$,we have $\cos \alpha + \cos \beta = -c$.
Using sum-to-product formulas:
$2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = -a$ $(1)$
$2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = -c$ $(2)$
Dividing $(1)$ by $(2)$:
$\tan \left( \frac{\alpha + \beta}{2} \right) = \frac{a}{c}$
Using the identity $\sin \theta = \frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)}$:
$\sin(\alpha + \beta) = \frac{2 \tan \left( \frac{\alpha + \beta}{2} \right)}{1 + \tan^2 \left( \frac{\alpha + \beta}{2} \right)} = \frac{2(a/c)}{1 + (a/c)^2} = \frac{2ac}{a^2 + c^2}$.

(C) Let the sides be $2, 5, a, b$ in order. The angle between sides $2$ and $5$ is $60^{\circ}$. Let $c$ be the diagonal opposite to the $60^{\circ}$ angle.
Using the Law of Cosines in the first triangle:
$c^2 = 2^2 + 5^2 - 2(2)(5)\cos(60^{\circ}) = 4 + 25 - 20(0.5) = 29 - 10 = 19$.
So,$c = \sqrt{19}$.
Since the quadrilateral is cyclic,the opposite angle to $60^{\circ}$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
Using the Law of Cosines in the second triangle with sides $a, b$ and diagonal $c$:
$c^2 = a^2 + b^2 - 2ab\cos(120^{\circ}) \implies 19 = a^2 + b^2 - 2ab(-0.5) \implies a^2 + b^2 + ab = 19$.
The area of the quadrilateral is the sum of the areas of the two triangles:
$\text{Area} = \frac{1}{2}(2)(5)\sin(60^{\circ}) + \frac{1}{2}ab\sin(120^{\circ}) = 4\sqrt{3}$.
$5\left(\frac{\sqrt{3}}{2}\right) + \frac{ab}{2}\left(\frac{\sqrt{3}}{2}\right) = 4\sqrt{3}$.
Dividing by $\frac{\sqrt{3}}{2}$: $5 + \frac{ab}{2} = 8 \implies \frac{ab}{2} = 3 \implies ab = 6$.
Now,$a^2 + b^2 = 19 - ab = 19 - 6 = 13$.
We have $a^2 + b^2 = 13$ and $ab = 6$. Solving these,we find $a=2, b=3$ (or vice versa).
The perimeter is $2 + 5 + a + b = 7 + 2 + 3 = 12$.

(B) For Statement $-1$: Solve $2\sin^2\theta - \cos 2\theta = 0$.
$2\sin^2\theta - (1 - 2\sin^2\theta) = 0$ $\Rightarrow 4\sin^2\theta = 1$ $\Rightarrow \sin\theta = \pm \frac{1}{2}$.
In $[0, 2\pi]$,$\theta \in \{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\}$.
Solve $2\cos^2\theta - 3\sin\theta = 0$.
$2(1 - \sin^2\theta) - 3\sin\theta = 0 \Rightarrow 2\sin^2\theta + 3\sin\theta - 2 = 0$.
$(2\sin\theta - 1)(\sin\theta + 2) = 0$.
Since $\sin\theta = -2$ is impossible,$\sin\theta = \frac{1}{2}$,so $\theta \in \{\frac{\pi}{6}, \frac{5\pi}{6}\}$.
Common solutions are $\{\frac{\pi}{6}, \frac{5\pi}{6}\}$,so Statement $-1$ is true.
For Statement $-2$: Solve $2\cos^2\theta - 3\sin\theta = 0$ in $[0, \pi]$.
As shown above,$\sin\theta = \frac{1}{2}$ gives $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. Both are in $[0, \pi]$.
Thus,Statement $-2$ is true. However,Statement $-2$ describes the solution set of one equation,which does not explain why the common solutions of both equations are two. Thus,Statement $-2$ is not the correct explanation.

(A) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a < b < c$. Then $2b = a + c$.
Let the angles be $A, B, C$ opposite to sides $a, b, c$ respectively. Given $C = 2A$.
Since $a < b < c$,we have $A < B < C$. Thus $A$ is the smallest and $C$ is the greatest angle.
In any triangle,$A + B + C = 180^{\circ}$. Since $C = 2A$,$B = 180^{\circ} - 3A$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
From $2b = a + c$,we get $2\sin B = \sin A + \sin C$.
$2\sin(180^{\circ} - 3A) = \sin A + \sin 2A$.
$2\sin 3A = \sin A + 2\sin A \cos A$.
$2(3\sin A - 4\sin^3 A) = \sin A(1 + 2\cos A)$.
Since $\sin A \neq 0$,$6 - 8\sin^2 A = 1 + 2\cos A$.
$6 - 8(1 - \cos^2 A) = 1 + 2\cos A$.
$8\cos^2 A - 2\cos A - 3 = 0$.
$(4\cos A + 3)(2\cos A - 1) = 0$.
Since $A$ is an angle of a triangle,$\cos A = \frac{3}{4}$ (as $\cos A = -\frac{3}{4}$ implies an obtuse angle,but $C=2A$ would exceed $180^{\circ}$).
The ratio of sides is $\sin A : \sin B : \sin C = \sin A : \sin 3A : \sin 2A$.
$= 1 : (3 - 4\sin^2 A) : 2\cos A$.
$= 1 : (3 - 4(1 - \frac{9}{16})) : 2(\frac{3}{4}) = 1 : (3 - 4(\frac{7}{16})) : \frac{3}{2} = 1 : \frac{5}{4} : \frac{6}{4} = 4 : 5 : 6$.

(A) Given that angles $A, B, C$ are in $A.P.$,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Given $\frac{a}{b} = \frac{1}{\sqrt{3}}$,we have $\frac{\sin A}{\sin B} = \frac{1}{\sqrt{3}}$.
Since $B = 60^{\circ}$,$\sin B = \frac{\sqrt{3}}{2}$,so $\sin A = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{1}{2}$,which means $A = 30^{\circ}$.
Then $C = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 90^{\circ}$.
Given $c = 4$,we have $a = c \sin A = 4 \sin 30^{\circ} = 2$ and $b = c \sin B = 4 \sin 60^{\circ} = 2\sqrt{3}$.
The area of the triangle is $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} \times 2 \times 2\sqrt{3} \times \sin 90^{\circ} = 2\sqrt{3} \text{ sq. cm}$.

(B) Given equation: $(81)^{\sin ^{2} x} + (81)^{\cos ^{2} x} = 30$.
Since $\cos ^{2} x = 1 - \sin ^{2} x$,we have $(81)^{\sin ^{2} x} + (81)^{1 - \sin ^{2} x} = 30$.
$(81)^{\sin ^{2} x} + \frac{81}{(81)^{\sin ^{2} x}} = 30$.
Let $t = (81)^{\sin ^{2} x}$. Then $t + \frac{81}{t} = 30$,which implies $t^{2} - 30t + 81 = 0$.
$(t - 3)(t - 27) = 0$,so $t = 3$ or $t = 27$.
Case $1$: $(81)^{\sin ^{2} x} = 3 \implies 3^{4 \sin ^{2} x} = 3^{1} \implies 4 \sin ^{2} x = 1 \implies \sin ^{2} x = \frac{1}{4}$.
In $[0, \pi]$,$\sin x = \frac{1}{2}$ or $\sin x = -\frac{1}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$ ($2$ solutions).
Case $2$: $(81)^{\sin ^{2} x} = 27 \implies 3^{4 \sin ^{2} x} = 3^{3} \implies 4 \sin ^{2} x = 3 \implies \sin ^{2} x = \frac{3}{4}$.
In $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ or $\sin x = -\frac{\sqrt{3}}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}$ ($2$ solutions).
Total number of solutions = $2 + 2 = 4$.

(D) Given inequality: $\sin 2\theta + \tan 2\theta > 0$
$\Rightarrow \sin 2\theta + \frac{\sin 2\theta}{\cos 2\theta} > 0$
$\Rightarrow \sin 2\theta \left(1 + \frac{1}{\cos 2\theta}\right) > 0$
$\Rightarrow \sin 2\theta \left(\frac{\cos 2\theta + 1}{\cos 2\theta}\right) > 0$
$\Rightarrow \tan 2\theta (2 \cos^2 \theta) > 0$
Since $2 \cos^2 \theta \ge 0$,for the expression to be $> 0$,we must have $\tan 2\theta > 0$ and $\cos 2\theta \neq 0$ (which implies $\cos^2 \theta \neq 0$).
$\tan 2\theta > 0$ occurs when $2\theta \in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2}) \cup (2\pi, \frac{5\pi}{2}) \cup (3\pi, \frac{7\pi}{2})$.
Dividing by $2$,we get $\theta \in (0, \frac{\pi}{4}) \cup (\frac{\pi}{2}, \frac{3\pi}{4}) \cup (\pi, \frac{5\pi}{4}) \cup (\frac{3\pi}{2}, \frac{7\pi}{4})$.
Thus,the correct option is $D$.

(A) Given the equation $\frac{\cos x}{1+\sin x} = |\tan 2x|$.
Note that $\frac{\cos x}{1+\sin x} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$.
So,$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = |\tan 2x|$.
Since the $LHS$ must be non-negative,$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \ge 0$,which implies $\frac{\pi}{4} - \frac{x}{2} \in [0, \frac{\pi}{2})$,so $x \in (-\frac{\pi}{2}, \frac{\pi}{2}]$.
Squaring both sides,$\tan^2\left(\frac{\pi}{4} - \frac{x}{2}\right) = \tan^2 2x$.
This implies $\tan 2x = \pm \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$.
Case $1$: $2x = n\pi + \left(\frac{\pi}{4} - \frac{x}{2}\right)$ $\Rightarrow \frac{5x}{2} = n\pi + \frac{\pi}{4}$ $\Rightarrow x = \frac{2n\pi}{5} + \frac{\pi}{10}$.
For $n=0, x=\frac{\pi}{10}$. For $n=-1, x=-\frac{3\pi}{10}$.
Case $2$: $2x = n\pi - \left(\frac{\pi}{4} - \frac{x}{2}\right)$ $\Rightarrow \frac{3x}{2} = n\pi - \frac{\pi}{4}$ $\Rightarrow x = \frac{2n\pi}{3} - \frac{\pi}{6}$.
For $n=0, x=-\frac{\pi}{6}$.
Checking the condition $\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \ge 0$: for $x = \frac{\pi}{10}, -\frac{3\pi}{10}, -\frac{\pi}{6}$,the condition holds.
The sum of solutions is $\frac{\pi}{10} - \frac{3\pi}{10} - \frac{\pi}{6} = -\frac{2\pi}{10} - \frac{\pi}{6} = -\frac{\pi}{5} - \frac{\pi}{6} = -\frac{11\pi}{30}$.

(B) Given $\frac{\sin A}{\sin B} = \frac{\sin (A-C)}{\sin (C-B)}$.
Since $A, B, C$ are angles of a triangle,$A+B+C = \pi$,so $A = \pi - (B+C)$.
Thus,$\sin A = \sin (B+C)$.
Also,$B = \pi - (A+C)$,so $\sin B = \sin (A+C)$.
Substituting these into the given equation:
$\frac{\sin (B+C)}{\sin (A+C)} = \frac{\sin (A-C)}{\sin (C-B)}$.
Cross-multiplying gives:
$\sin (B+C) \sin (C-B) = \sin (A+C) \sin (A-C)$.
Using the identity $\sin (x+y) \sin (x-y) = \sin^{2} x - \sin^{2} y$:
$\sin^{2} C - \sin^{2} B = \sin^{2} A - \sin^{2} C$.
Rearranging terms:
$2 \sin^{2} C = \sin^{2} A + \sin^{2} B$.
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
Substituting these into the equation:
$2 \left(\frac{c}{2R}\right)^{2} = \left(\frac{a}{2R}\right)^{2} + \left(\frac{b}{2R}\right)^{2}$.
$2c^{2} = a^{2} + b^{2}$.
This implies $a^{2}, c^{2}, b^{2}$ are in $A.P.$ (or $b^{2}, c^{2}, a^{2}$ are in $A.P.$).

(A) Given $\cos B = \frac{3}{5}$,then $\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$.
Using the sine rule,$\frac{b}{\sin B} = 2R$,where $R=5$ is the circumradius.
$b = 2R \sin B = 2(5)\left(\frac{4}{5}\right) = 8$.
Using the cosine rule for $\angle B$:
$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
$\frac{3}{5} = \frac{a^2 + 5^2 - 8^2}{2(a)(5)} = \frac{a^2 + 25 - 64}{10a} = \frac{a^2 - 39}{10a}$.
$6a = a^2 - 39 \Rightarrow a^2 - 6a - 39 = 0$.
Solving for $a$ using the quadratic formula: $a = \frac{6 \pm \sqrt{36 - 4(1)(-39)}}{2} = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3}$.
Since $a > 0$,we take $a = 3 + 4\sqrt{3}$.
The area of $\triangle ABC$ is given by $\Delta = \frac{1}{2}ac \sin B$.
$\Delta = \frac{1}{2}(3 + 4\sqrt{3})(5)\left(\frac{4}{5}\right) = 2(3 + 4\sqrt{3}) = 6 + 8\sqrt{3}$.