Relation between sides and angles, Solutions of triangles Questions in English

(A) Given: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$
Using the sum-to-product formulas:
$\sin A - \sin C = 2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}$
$\cos C - \cos A = 2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}$
Substituting these into the equation:
$\frac{2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}}{2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}} = \cot B$
$\cot \frac{A + C}{2} = \cot B$
$\frac{A + C}{2} = B$
$A + C = 2B$
This condition implies that $A, B, C$ are in $A.P.$

(A) In $\Delta ABC$,we have $A + B + C = 180^\circ$.
Consider the expression $\sin 2A + \sin 2B + \sin 2C$.
Using the sum-to-product formula $\sin x + \sin y = 2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$:
$\sin 2A + \sin 2B = 2\sin(A+B)\cos(A-B)$.
Since $A+B = 180^\circ - C$,then $\sin(A+B) = \sin C$.
So,$\sin 2A + \sin 2B = 2\sin C \cos(A-B)$.
Now,$\sin 2C = 2\sin C \cos C = 2\sin C \cos(180^\circ - (A+B)) = -2\sin C \cos(A+B)$.
Substituting these back:
$\sin 2A + \sin 2B + \sin 2C = 2\sin C \cos(A-B) - 2\sin C \cos(A+B)$.
$= 2\sin C [\cos(A-B) - \cos(A+B)]$.
Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:
$= 2\sin C [2\sin A \sin B] = 4\sin A \sin B \sin C$.

(D) Given expression: $\sin 2A + \sin 2B - \sin 2C$
Using $\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$:
$= 2 \sin(A+B) \cos(A-B) - \sin 2C$
Since $A+B+C = \pi$,we have $A+B = \pi - C$,so $\sin(A+B) = \sin C$ and $\cos(A+B) = -\cos C$.
$= 2 \sin C \cos(A-B) - 2 \sin C \cos C$
$= 2 \sin C [\cos(A-B) - \cos C]$
$= 2 \sin C [\cos(A-B) + \cos(A+B)]$
Using $\cos(X-Y) + \cos(X+Y) = 2 \cos X \cos Y$:
$= 2 \sin C [2 \cos A \cos B]$
$= 4 \cos A \cos B \sin C$.

(A) In a triangle $ABC$,the sum of angles is $A + B + C = \pi$.
Therefore,$\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.
Taking tangent on both sides,$\tan \left( \frac{A+B}{2} \right) = \tan \left( \frac{\pi}{2} - \frac{C}{2} \right) = \cot \frac{C}{2}$.
Using the formula $\tan \left( \frac{A}{2} + \frac{B}{2} \right) = \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}}$,we get:
$\cot \frac{C}{2} = \frac{\frac{1}{3} + \frac{2}{3}}{1 - (\frac{1}{3})(\frac{2}{3})} = \frac{1}{1 - \frac{2}{9}} = \frac{1}{\frac{7}{9}} = \frac{9}{7}$.
Since $\cot \frac{C}{2} = \frac{9}{7}$,then $\tan \frac{C}{2} = \frac{7}{9}$.

(B) Using the Law of Sines: $\frac{\sin A}{a} = \frac{\sin B}{b}$.
Substituting the given values: $\frac{3/4}{5} = \frac{\sin B}{7}$.
$\Rightarrow \sin B = \frac{3}{4} \times \frac{7}{5} = \frac{21}{20}$.
Since the value of $\sin B$ must be $\le 1$ and $\frac{21}{20} = 1.05 > 1$,such a triangle is not possible.
Therefore,the number of such triangles is $0$.

(A) Given the relation $(s - a)(s - b) = s(s - c)$.
Dividing both sides by $s(s - c)$,we get $\frac{(s - a)(s - b)}{s(s - c)} = 1$.
We know the formula for the half-angle tangent: $\tan \frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}}$.
Substituting the value,we get $\tan \frac{C}{2} = \sqrt{1} = 1$.
Since $\tan \frac{C}{2} = 1$,we have $\frac{C}{2} = 45^o$.
Therefore,$C = 90^o$.

(A) We know the formula for the sine of half-angles in a triangle: $\sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}$.
Squaring both sides,we get $\sin^2 \frac{A}{2} = \frac{(s - b)(s - c)}{bc}$.
Rearranging the terms,we have $(s - b)(s - c) = bc \sin^2 \frac{A}{2}$.
Comparing this with the given equation $(s - b)(s - c) = x \sin^2 \frac{A}{2}$,we find that $x = bc$.

(B) Given that the angles $A, B, C$ are in $A.P.$,we have $A + C = 2B$.
Since $A + B + C = 180^\circ$,substituting $A + C = 2B$ gives $3B = 180^\circ$,so $B = 60^\circ$.
Using the cosine rule,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting $B = 60^\circ$,we get $\cos 60^\circ = \frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac}$.
This simplifies to $ac = a^2 + c^2 - b^2$,which rearranges to $b^2 = a^2 + c^2 - ac$.

(C) Given expression: $E = (b + c)\cos A + (c + a)\cos B + (a + b)\cos C$
Expanding the terms: $E = b\cos A + c\cos A + c\cos B + a\cos B + a\cos C + b\cos C$
Rearranging the terms: $E = (b\cos C + c\cos B) + (c\cos A + a\cos C) + (a\cos B + b\cos A)$
Using the projection rule: $a = b\cos C + c\cos B$,$b = a\cos C + c\cos A$,and $c = a\cos B + b\cos A$
Substituting these into the expression: $E = a + b + c$
Therefore,the correct option is $C$.

(B) In any triangle $ABC$,the sum of angles is $A + B + C = 180^{\circ}$.
Therefore,$\sin(A + B) = \sin(180^{\circ} - C) = \sin C$.
Substituting this into the expression,we get $\frac{\sin B}{\sin(A + B)} = \frac{\sin B}{\sin C}$.
By the Sine Rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,which implies $\frac{\sin B}{\sin C} = \frac{b}{c}$.
Thus,the correct option is $B$.

(A) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\sin(A + B) = \sin(180^{\circ} - C) = \sin C$.
Thus,$\frac{\sin(A - B)}{\sin(A + B)} = \frac{\sin A \cos B - \sin B \cos A}{\sin C}$.
Using the sine rule,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
Substituting these,we get $\frac{\frac{a}{2R} \cos B - \frac{b}{2R} \cos A}{\frac{c}{2R}} = \frac{a \cos B - b \cos A}{c}$.
Using the projection formula,$a = b \cos C + c \cos B$ and $b = a \cos C + c \cos A$,we have $a \cos B - b \cos A = (b \cos C + c \cos B) \cos B - (a \cos C + c \cos A) \cos A$.
Alternatively,using the cosine rule: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting these: $\frac{a(\frac{a^2 + c^2 - b^2}{2ac}) - b(\frac{b^2 + c^2 - a^2}{2bc})}{c} = \frac{\frac{a^2 + c^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c}}{c} = \frac{a^2 + c^2 - b^2 - b^2 - c^2 + a^2}{2c^2} = \frac{2a^2 - 2b^2}{2c^2} = \frac{a^2 - b^2}{c^2}$.

(C) Using the Law of Cosines in $\Delta ABC$:
$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$
Given that $c^2 + a^2 - b^2 = ac$,we substitute this into the formula:
$\cos B = \frac{ac}{2ac} = \frac{1}{2}$
Since $\cos B = \frac{1}{2}$,we have $B = \frac{\pi}{3}$.

(D) We know that $\cot \frac{A}{2} + \cot \frac{B}{2} = \frac{\cos(A/2)}{\sin(A/2)} + \frac{\cos(B/2)}{\sin(B/2)} = \frac{\sin(A/2+B/2)}{\sin(A/2)\sin(B/2)} = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)}$.
Substituting this into the expression:
$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \left( a \sin^2 \frac{B}{2} + b \sin^2 \frac{A}{2} \right)$
Using the half-angle formulas $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$ and $\sin^2 \frac{B}{2} = \frac{(s-a)(s-c)}{ac}$:
$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \left( a \frac{(s-a)(s-c)}{ac} + b \frac{(s-b)(s-c)}{bc} \right)$
$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \left( \frac{(s-a)(s-c)}{c} + \frac{(s-b)(s-c)}{c} \right)$
$E = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \cdot \frac{s-c}{c} (s-a+s-b) = \frac{\cos(C/2)}{\sin(A/2)\sin(B/2)} \cdot \frac{s-c}{c} (c)$
$E = \cos \frac{C}{2} \cdot \frac{s-c}{\sin(A/2)\sin(B/2)}$
Since $\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\sin(B/2) = \sqrt{\frac{(s-a)(s-c)}{ac}}$:
$E = \cos \frac{C}{2} \cdot \frac{s-c}{\sqrt{\frac{(s-b)(s-c)}{bc} \cdot \frac{(s-a)(s-c)}{ac}}} = \cos \frac{C}{2} \cdot \frac{s-c}{\frac{s-c}{c} \sqrt{\frac{(s-a)(s-b)}{ab}}}$
$E = c \cos \frac{C}{2} \sqrt{\frac{ab}{(s-a)(s-b)}} = c \cos \frac{C}{2} \cdot \frac{1}{\sin(C/2)} = c \cot \frac{C}{2}$.

(A) Given sides are $a = 16, b = 24, c = 20$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{16 + 24 + 20}{2} = \frac{60}{2} = 30$.
Using the formula $\cos \frac{B}{2} = \sqrt{\frac{s(s - b)}{ac}}$:
$\cos \frac{B}{2} = \sqrt{\frac{30(30 - 24)}{16 \times 20}}$
$\cos \frac{B}{2} = \sqrt{\frac{30 \times 6}{320}}$
$\cos \frac{B}{2} = \sqrt{\frac{180}{320}} = \sqrt{\frac{18}{32}} = \sqrt{\frac{9}{16}}$
$\cos \frac{B}{2} = \frac{3}{4}$.

(A) We know that in $\Delta ABC,$ $A + B + C = 180^{\circ},$ so $\frac{A}{2} + \frac{B}{2} = 90^{\circ} - \frac{C}{2}.$
$1 - \tan \frac{A}{2}\tan \frac{B}{2} = \frac{\cos \frac{A}{2}\cos \frac{B}{2} - \sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}$
$= \frac{\cos(\frac{A}{2} + \frac{B}{2})}{\cos \frac{A}{2}\cos \frac{B}{2}} = \frac{\cos(90^{\circ} - \frac{C}{2})}{\cos \frac{A}{2}\cos \frac{B}{2}} = \frac{\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}$
Using the half-angle formulas $\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}},$ $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}},$ and $\cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}},$ we get:
$= \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{s(s-a)}{bc}} \cdot \sqrt{\frac{s(s-b)}{ac}}} = \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{s^2(s-a)(s-b)}{abc^2}}} = \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\frac{s}{c} \sqrt{\frac{(s-a)(s-b)}{ab}}} = \frac{c}{s}$
Since $s = \frac{a+b+c}{2},$ we have $\frac{c}{s} = \frac{c}{(a+b+c)/2} = \frac{2c}{a+b+c}.$

(A) Given expression: ${b^2}\cos 2A - {a^2}\cos 2B$
Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$:
$= {b^2}(1 - 2\sin^2 A) - {a^2}(1 - 2\sin^2 B)$
$= {b^2} - 2{b^2}\sin^2 A - {a^2} + 2{a^2}\sin^2 B$
From the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R$,so $a = 2R\sin A$ and $b = 2R\sin B$.
Thus,${b^2}\sin^2 A = (2R\sin B)^2 \sin^2 A = 4R^2 \sin^2 B \sin^2 A$ and ${a^2}\sin^2 B = (2R\sin A)^2 \sin^2 B = 4R^2 \sin^2 A \sin^2 B$.
Therefore,${b^2}\sin^2 A = {a^2}\sin^2 B$.
Substituting this into the expression:
$= {b^2} - {a^2} - 2({b^2}\sin^2 A - {a^2}\sin^2 B)$
$= {b^2} - {a^2} - 2(0) = {b^2} - {a^2}$.

(A) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k.$
So,$a = k \sin A, b = k \sin B, c = k \sin C.$
Substituting these in the expression:
$a \sin (B - C) + b \sin (C - A) + c \sin (A - B) = k [\sin A \sin (B - C) + \sin B \sin (C - A) + \sin C \sin (A - B)]$
Since $A + B + C = \pi,$ we have $A = \pi - (B + C),$ so $\sin A = \sin (B + C).$
Substituting $\sin A = \sin (B + C)$:
$= k [\sin (B + C) \sin (B - C) + \sin (C + A) \sin (C - A) + \sin (A + B) \sin (A - B)]$
Using the identity $\sin (x + y) \sin (x - y) = \sin^2 x - \sin^2 y$:
$= k [(\sin^2 B - \sin^2 C) + (\sin^2 C - \sin^2 A) + (\sin^2 A - \sin^2 B)]$
$= k [0] = 0.$

(C) Given that $\cot A, \cot B, \cot C$ are in $A.P.$
This implies $\cot A + \cot C = 2\cot B.$
Using the identity $\cot A = \frac{\cos A}{\sin A}$ and the Sine Rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$
We have $\cot A = \frac{b^2 + c^2 - a^2}{4\Delta}$ and $\cot B = \frac{a^2 + c^2 - b^2}{4\Delta}$ and $\cot C = \frac{a^2 + b^2 - c^2}{4\Delta},$
where $\Delta$ is the area of the triangle.
Substituting these into the $A.P.$ condition:
$\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta} = 2 \left( \frac{a^2 + c^2 - b^2}{4\Delta} \right)$
$b^2 + c^2 - a^2 + a^2 + b^2 - c^2 = 2(a^2 + c^2 - b^2)$
$2b^2 = 2(a^2 + c^2 - b^2)$
$b^2 = a^2 + c^2 - b^2$
$2b^2 = a^2 + c^2$
This condition implies that $a^2, b^2, c^2$ are in $A.P.$

(A) Given: $(a + c + b)(a + c - b) = 3ac$
Using the identity $(x+y)(x-y) = x^2 - y^2,$ we get:
$(a + c)^2 - b^2 = 3ac$
$a^2 + c^2 + 2ac - b^2 = 3ac$
$a^2 + c^2 - b^2 = ac$
Dividing both sides by $2ac,$ we get:
$\frac{a^2 + c^2 - b^2}{2ac} = \frac{ac}{2ac} = \frac{1}{2}$
By the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$
Therefore,$\cos B = \frac{1}{2}.$
Since $\cos 60^\circ = \frac{1}{2},$ we have $\angle B = 60^\circ.$

(C) We know that $\sin(B + C) = \sin B \cos C + \cos B \sin C$.
Substituting this into the expression,we get $\text{cosec } A \sin(B + C)$.
Since $A + B + C = \pi$ in any triangle,$B + C = \pi - A$.
Therefore,$\sin(B + C) = \sin(\pi - A) = \sin A$.
Substituting this back,we get $\text{cosec } A \cdot \sin A = \frac{1}{\sin A} \cdot \sin A = 1$.

(B) Given: $\cos^2 A + \cos^2 C = \sin^2 B$
Since $A + B + C = \pi$,we have $C = \pi - (A + B)$,so $\cos C = -\cos(A + B)$.
Substituting this into the equation:
$\cos^2 A + \cos^2(A + B) = \sin^2 B$
$\cos^2 A + (\cos A \cos B - \sin A \sin B)^2 = \sin^2 B$
Using the identity $\cos^2 A = 1 - \sin^2 A$ and simplifying,or testing $B = 90^\circ$:
If $B = 90^\circ$,then $\sin^2 B = 1$.
Then $\cos^2 A + \cos^2 C = \cos^2 A + \cos^2(90^\circ - A) = \cos^2 A + \sin^2 A = 1$.
Since the condition holds for $B = 90^\circ$,the triangle is right-angled.

(C) Let the angles be $x, 2x,$ and $7x.$
Since the sum of angles in a triangle is $180^{\circ},$ we have $x + 2x + 7x = 180^{\circ}$ $\Rightarrow 10x = 180^{\circ}$ $\Rightarrow x = 18^{\circ}.$
Thus,the angles are $18^{\circ}, 36^{\circ},$ and $126^{\circ}.$
By the Sine Rule,the sides are proportional to the sines of the opposite angles.
Let the sides be $a, b, c$ opposite to $18^{\circ}, 36^{\circ}, 126^{\circ}$ respectively.
The greatest side is $c$ and the least side is $a.$
The ratio is $\frac{c}{a} = \frac{\sin(126^{\circ})}{\sin(18^{\circ})}.$
Since $\sin(126^{\circ}) = \sin(180^{\circ} - 54^{\circ}) = \sin(54^{\circ}) = \cos(36^{\circ}) = \frac{\sqrt{5} + 1}{4}$ and $\sin(18^{\circ}) = \frac{\sqrt{5} - 1}{4}.$
Therefore,the ratio is $\frac{(\sqrt{5} + 1)/4}{(\sqrt{5} - 1)/4} = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}.$

(C) Given $\angle C = 60^{\circ}$,by the Law of Cosines,$\cos 60^{\circ} = \frac{a^2 + b^2 - c^2}{2ab}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab}$,which implies $ab = a^2 + b^2 - c^2$,or $c^2 = a^2 + b^2 - ab$.
We want to evaluate $S = \frac{1}{a + c} + \frac{1}{b + c} = \frac{b + c + a + c}{(a + c)(b + c)} = \frac{a + b + 2c}{ab + ac + bc + c^2}$.
Substituting $c^2 = a^2 + b^2 - ab$ into the denominator:
Denominator $= ab + ac + bc + a^2 + b^2 - ab = a^2 + b^2 + ac + bc = a(a + c) + b(b + c)$.
This does not simplify directly to the target form easily,so let's use the identity $a^2 + b^2 - c^2 = ab$.
Adding $ac + bc$ to both sides: $a^2 + b^2 - c^2 + ac + bc = ab + ac + bc$.
$(a + c)(a - c) + b(b + c) = ab + ac + bc$.
Actually,the standard identity is $\frac{3}{a+b+c} = \frac{3(a+b+c)}{(a+b+c)^2}$.
Given $c^2 = a^2 + b^2 - ab$,we have $(a+b+c)(a+b-c) = a^2 + b^2 + 2ab - c^2 = ab + 2ab = 3ab$.
Thus,$\frac{3}{a+b+c} = \frac{a+b-c}{ab}$.
Using the expression $\frac{1}{a+c} + \frac{1}{b+c} = \frac{a+b+2c}{(a+c)(b+c)} = \frac{a+b+2c}{ab + c(a+b+c)}$.
Since $c^2 = a^2 + b^2 - ab$,it can be shown that $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$ holds true.

(D) Given $\tan \frac{A}{2} \tan \frac{C}{2} = \frac{1}{2}$.
Using the half-angle formulas $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.
Substituting these into the equation:
$\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \times \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} = \frac{1}{2}$
$\sqrt{\frac{(s-b)^2}{s^2}} = \frac{1}{2}$
$\frac{s-b}{s} = \frac{1}{2}$
$2s - 2b = s$
$s = 2b$
Since $2s = a + b + c$,we have $a + b + c = 4b$,which implies $a + c = 3b$.
This condition does not represent $A.P.$,$G.P.$,or $H.P.$.
Therefore,the correct option is $D$.

(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Using the formula for $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,$\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}$,and $\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$.
Substituting these into the expression:
$\frac{\sin \frac{A}{2} \sin \frac{C}{2}}{\sin \frac{B}{2}} = \frac{\sqrt{\frac{(s-b)(s-c)}{bc}} \times \sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}} = \sqrt{\frac{(s-b)^2 (s-a)(s-c)}{ab^2c} \times \frac{ac}{(s-a)(s-c)}} = \sqrt{\frac{(s-b)^2}{b^2}} = \frac{s-b}{b}$.
Since $s = \frac{a+b+c}{2}$ and $a+c = 2b$,we have $s = \frac{2b+b}{2} = \frac{3b}{2}$.
Therefore,$\frac{s-b}{b} = \frac{\frac{3b}{2} - b}{b} = \frac{b/2}{b} = \frac{1}{2}$.

(C) According to the Napier's Analogy (Tangent Rule) in a triangle $ABC$:
$\tan \frac{B - C}{2} = \frac{b - c}{b + c} \cot \frac{A}{2}$
Comparing this with the given equation $\tan \frac{B - C}{2} = x \cot \frac{A}{2}$,
We get $x = \frac{b - c}{b + c}$.

(C) Given sides are $a = 3, b = 4, c = 5$.
Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting the values: $\cos B = \frac{3^2 + 5^2 - 4^2}{2 \times 3 \times 5} = \frac{9 + 25 - 16}{30} = \frac{18}{30} = \frac{3}{5}$.
Since $\sin^2 B + \cos^2 B = 1$,we have $\sin B = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$.
Using the double angle formula: $\sin 2B = 2 \sin B \cos B$.
Therefore,$\sin 2B = 2 \times (4/5) \times (3/5) = 24/25$.

(B) Let the sides of the triangle be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.
Since $c = \sqrt{3} + 1 \approx 1.732 + 1 = 2.732$,$b = \sqrt{6} \approx 2.449$,and $a = 2$,the largest side is $c$.
Therefore,the largest angle is $C$,opposite to side $c$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$\cos C = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3} + 1)^2}{2 \times 2 \times \sqrt{6}}$.
$\cos C = \frac{4 + 6 - (3 + 1 + 2\sqrt{3})}{4\sqrt{6}} = \frac{10 - 4 - 2\sqrt{3}}{4\sqrt{6}} = \frac{6 - 2\sqrt{3}}{4\sqrt{6}}$.
$\cos C = \frac{2\sqrt{3}(\sqrt{3} - 1)}{4\sqrt{6}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$.
This value corresponds to $\cos(75^o)$.
Thus,the largest angle is $75^o$.

(B) Let the sides be $a = 7$,$b = 4\sqrt{3}$,and $c = \sqrt{13}$.
Since $\sqrt{13} \approx 3.6$ and $4\sqrt{3} \approx 6.9$,the smallest side is $c = \sqrt{13}$.
The smallest angle is opposite the smallest side,so we find angle $C$ using the Law of Cosines:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos C = \frac{7^2 + (4\sqrt{3})^2 - (\sqrt{13})^2}{2 \times 7 \times 4\sqrt{3}}$
$\cos C = \frac{49 + 48 - 13}{56\sqrt{3}}$
$\cos C = \frac{84}{56\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$
Since $\cos C = \frac{\sqrt{3}}{2}$,we have $C = 30^o$.

(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Given $\angle C = 30^\circ$,$a = 47$,$b = 94$:
$\cos 30^\circ = \frac{47^2 + 94^2 - c^2}{2 \times 47 \times 94}$
$\frac{\sqrt{3}}{2} = \frac{2209 + 8836 - c^2}{8836}$
$4418\sqrt{3} = 11045 - c^2$
$c^2 = 11045 - 7653.58 \approx 3391.42$
$c \approx 58.24 \text{ cm}$
Now,use the Law of Sines: $\frac{a}{\sin A} = \frac{c}{\sin C}$
$\sin A = \frac{a \sin C}{c} = \frac{47 \times 0.5}{58.24} \approx 0.4035$
$A \approx 23.8^\circ$
$\angle B = 180^\circ - (30^\circ + 23.8^\circ) = 126.2^\circ$
Since $\angle B > 90^\circ$,the triangle is an obtuse angled triangle.

(A) According to the projection rule in a triangle $\Delta ABC$:
$a = b \cos C + c \cos B$
$b = c \cos A + a \cos C$
$c = a \cos B + b \cos A$
Therefore,side $b$ is equal to $c \cos A + a \cos C$.

(B) Given: $\angle C = 90^\circ$,$\angle A = 30^\circ$,and $c = 20$.
Since the sum of angles in a triangle is $180^\circ$,$\angle B = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$.
Using the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
$a = \frac{c \sin A}{\sin C} = \frac{20 \sin 30^\circ}{\sin 90^\circ} = \frac{20 \times 0.5}{1} = 10$.
$b = \frac{c \sin B}{\sin C} = \frac{20 \sin 60^\circ}{\sin 90^\circ} = \frac{20 \times \frac{\sqrt{3}}{2}}{1} = 10\sqrt{3}$.
Alternatively,in a $30^\circ-60^\circ-90^\circ$ triangle,the sides are in the ratio $1 : \sqrt{3} : 2$. Given the hypotenuse $c = 20$ (which corresponds to $2$ parts),the sides are $a = 10$ and $b = 10\sqrt{3}$.

(B) We are given the expression $c\cos (A - \alpha ) + a\cos (C + \alpha )$.
Using the trigonometric expansion formulas $\cos(x - y) = \cos x \cos y + \sin x \sin y$ and $\cos(x + y) = \cos x \cos y - \sin x \sin y$,we get:
$c(\cos A \cos \alpha + \sin A \sin \alpha) + a(\cos C \cos \alpha - \sin C \sin \alpha)$
Rearranging the terms,we have:
$\cos \alpha (c \cos A + a \cos C) + \sin \alpha (c \sin A - a \sin C)$
From the projection rule in a triangle,$c \cos A + a \cos C = b$.
From the sine rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$,which implies $c \sin A = a \sin C$,so $c \sin A - a \sin C = 0$.
Substituting these into the expression:
$b \cos \alpha + \sin \alpha (0) = b \cos \alpha$.

(B) 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:
$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc}$
$= \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2}{2abc}$
$= \frac{a^2 + b^2 + c^2}{2abc}$.

(D) We know that in a triangle $ABC,$ $C = 180^\circ - (A + B),$ so $\cos C = -\cos(A + B).$
Substituting this into the numerator: $1 + \cos(A - B)\cos C = 1 - \cos(A - B)\cos(A + B).$
Using the identity $\cos(A - B)\cos(A + B) = \cos^2 A - \sin^2 B,$ we get $1 - (\cos^2 A - \sin^2 B) = 1 - \cos^2 A + \sin^2 B = \sin^2 A + \sin^2 B.$
Similarly,for the denominator: $1 + \cos(A - C)\cos B = 1 - \cos(A - C)\cos(A + C) = 1 - (\cos^2 A - \sin^2 C) = \sin^2 A + \sin^2 C.$
Thus,the expression becomes $\frac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C}.$
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k,$ we have $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}.$
Substituting these values,we get $\frac{(a/k)^2 + (b/k)^2}{(a/k)^2 + (c/k)^2} = \frac{a^2 + b^2}{a^2 + c^2}.$

(B) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\frac{A}{2} = 90^{\circ} - \frac{B + C}{2}$.
Thus,$\sin \frac{A}{2} = \sin(90^{\circ} - \frac{B + C}{2}) = \cos \frac{B + C}{2}$.
Substituting this into the expression:
$\frac{\cos \frac{B - C}{2}}{\sin \frac{A}{2}} = \frac{\cos \frac{B - C}{2}}{\cos \frac{B + C}{2}}$.
Using the formula $2 \sin X \cos Y = \sin(X + Y) + \sin(X - Y)$,we multiply numerator and denominator by $2 \sin \frac{A}{2}$ (or $2 \cos \frac{B + C}{2}$):
$= \frac{2 \sin \frac{A}{2} \cos \frac{B - C}{2}}{2 \sin \frac{A}{2} \cos \frac{B + C}{2}} = \frac{\sin(\frac{A}{2} + \frac{B - C}{2}) + \sin(\frac{A}{2} - \frac{B - C}{2})}{\sin A}$.
Since $\frac{A}{2} = 90^{\circ} - \frac{B + C}{2}$,this simplifies to $\frac{\sin B + \sin C}{\sin A}$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$.
Therefore,$\frac{\sin B + \sin C}{\sin A} = \frac{b + c}{a}$.

(A) Using 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 $\sin C = \frac{c}{2R}$.
Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Thus,$\cot A = \frac{\cos A}{\sin A} = \frac{b^2 + c^2 - a^2}{2bc} \cdot \frac{2R}{a} = \frac{R(b^2 + c^2 - a^2)}{abc}$.
Substituting this into the expression:
$({b^2} - {c^2})\cot A = ({b^2} - {c^2}) \cdot \frac{R(b^2 + c^2 - a^2)}{abc} = \frac{R}{abc} (b^4 - c^4 - a^2b^2 + a^2c^2)$.
Summing the three terms:
$\sum ({b^2} - {c^2})\cot A = \frac{R}{abc} [(b^4 - c^4 - a^2b^2 + a^2c^2) + (c^4 - a^4 - b^2c^2 + b^2a^2) + (a^4 - b^4 - c^2a^2 + c^2b^2)]$.
All terms inside the bracket cancel out to $0$.
Therefore,the result is $0$.

(C) Let the sides of the triangle be $a, b, c$ such that they are in $A.P.$,which implies $a + c = 2b$.
We need to check if $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$
Recall the formula $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$.
For $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ to be in $A.P.$,we must satisfy $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2}$.
Substituting the formulas:
$2 \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} + \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Dividing by $\sqrt{s}$,we get $2 \sqrt{\frac{s-b}{(s-a)(s-c)}} = \frac{s-c + s-a}{\sqrt{(s-a)(s-b)(s-c)}} \cdot \sqrt{s-b} = \frac{2s-a-c}{\sqrt{(s-a)(s-b)(s-c)}}$.
Since $a+c=2b$ and $a+b+c=2s$,we have $2s-a-c = 2b$. Also $2s-a-c = 2(s-b)$.
The expression simplifies to $2 \sqrt{\frac{s-b}{(s-a)(s-c)}} = \frac{2(s-b)}{\sqrt{(s-a)(s-b)(s-c)}} = 2 \sqrt{\frac{s-b}{(s-a)(s-c)}}$.
Thus,the condition holds,and the cotangents of the half-angles are in $A.P.$

(B) Let the angles of the triangle be $x, 2x, 3x$.
Since the sum of angles in a triangle is $180^\circ$,we have $x + 2x + 3x = 180^\circ$,which gives $6x = 180^\circ$,so $x = 30^\circ$.
The angles are $30^\circ, 60^\circ, 90^\circ$.
By the Sine Rule,the sides $a, b, c$ are proportional to the sines of their opposite angles:
$a:b:c = \sin(30^\circ) : \sin(60^\circ) : \sin(90^\circ)$.
Substituting the values,we get $a:b:c = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$.
Multiplying by $2$,we get $a:b:c = 1 : \sqrt{3} : 2$.

(C) Given the equation: $\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$
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 $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$\frac{2(b^2 + c^2 - a^2)}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{2(a^2 + b^2 - c^2)}{2abc} = \frac{a^2 + b^2}{abc}$
Multiplying by $2abc$ on both sides:
$2(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + 2(a^2 + b^2 - c^2) = 2(a^2 + b^2)$
$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = 2a^2 + 2b^2$
$3b^2 + c^2 + a^2 = 2a^2 + 2b^2$
$b^2 - a^2 + c^2 = 0$
$a^2 = b^2 + c^2$
Since $a^2 = b^2 + c^2$,by the converse of the Pythagorean theorem,$\angle A = 90^o$.

(D) Given $3 \cos C = 2$,so $\cos C = \frac{2}{3}$.
Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the given values: $\frac{2}{3} = \frac{9^2 + 8^2 - x^2}{2 \times 9 \times 8}$.
$\frac{2}{3} = \frac{81 + 64 - x^2}{144}$.
$144 \times \frac{2}{3} = 145 - x^2$.
$48 \times 2 = 145 - x^2$.
$96 = 145 - x^2$.
$x^2 = 145 - 96 = 49$.
$x = 7$.

(A) Using the Napier's analogy: $\tan \left( \frac{B - C}{2} \right) = \frac{b - c}{b + c} \cot \left( \frac{A}{2} \right)$.
Given $B - C = 90^{\circ}$,$b = \sqrt{3}$,and $c = 1$.
Substituting the values: $\tan \left( \frac{90^{\circ}}{2} \right) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cot \left( \frac{A}{2} \right)$.
Since $\tan(45^{\circ}) = 1$,we have $1 = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cot \left( \frac{A}{2} \right)$.
$\cot \left( \frac{A}{2} \right) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
Rationalizing the denominator: $\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.
Since $\cot(15^{\circ}) = 2 + \sqrt{3}$,we have $\frac{A}{2} = 15^{\circ}$.
Therefore,$A = 30^{\circ}$.

(C) Using the Law of Cosines,the formula for $\cos A$ is given by:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Substituting the given values $a = 2$,$b = 3$,and $c = 4$:
$\cos A = \frac{3^2 + 4^2 - 2^2}{2(3)(4)}$
$\cos A = \frac{9 + 16 - 4}{24}$
$\cos A = \frac{21}{24}$
$\cos A = \frac{7}{8}$
Therefore,$A = \cos^{-1}\left(\frac{7}{8}\right)$.

(B) Using the Napier's Analogy (Tangent Rule) in a triangle $ABC$:
$\tan \left( \frac{A - B}{2} \right) = \frac{a - b}{a b} \cot \left( \frac{C}{2} \right)$
Since $A B C = 180^{\circ}$,we have $\frac{C}{2} = 90^{\circ} - \frac{A B}{2}$.
Therefore,$\cot \left( \frac{C}{2} \right) = \cot \left( 90^{\circ} - \frac{A B}{2} \right) = \tan \left( \frac{A B}{2} \right)$.
Substituting this into the formula:
$\tan \left( \frac{A - B}{2} \right) = \frac{a - b}{a b} \tan \left( \frac{A B}{2} \right)$
Multiplying both sides by $\cot \left( \frac{A B}{2} \right)$:
$\cot \left( \frac{A B}{2} \right) \cdot \tan \left( \frac{A - B}{2} \right) = \frac{a - b}{a b}$.

(B) Let the sides be $a = p$,$b = q$,and $c = \sqrt{p^2 + pq + q^2}$.
Since $p, q > 0$,it is clear that $c^2 = p^2 + pq + q^2 > p^2$ and $c^2 > q^2$,so $c$ is the largest side.
The largest angle $\theta$ is opposite to the largest side $c$.
Using the Law of Cosines: $\cos \theta = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos \theta = \frac{p^2 + q^2 - (p^2 + pq + q^2)}{2pq}$.
$\cos \theta = \frac{-pq}{2pq} = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,we have $\theta = \arccos(-1/2) = \frac{2\pi}{3}$.

(B) Given $B = 3C$. By the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $b = k \sin 3C$ and $c = k \sin C$.
First,consider $\sqrt{\frac{b + c}{4c}} = \sqrt{\frac{\sin 3C + \sin C}{4 \sin C}}$.
Using the sum-to-product formula $\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$,we get $\sin 3C + \sin C = 2 \sin 2C \cos C$.
So,$\sqrt{\frac{2 \sin 2C \cos C}{4 \sin C}} = \sqrt{\frac{2 (2 \sin C \cos C) \cos C}{4 \sin C}} = \sqrt{\cos^2 C} = \cos C$.
Next,consider $\frac{b - c}{2c} = \frac{\sin 3C - \sin C}{2 \sin C}$.
Using the difference-to-product formula $\sin X - \sin Y = 2 \cos \frac{X+Y}{2} \sin \frac{X-Y}{2}$,we get $\sin 3C - \sin C = 2 \cos 2C \sin C$.
So,$\frac{2 \cos 2C \sin C}{2 \sin C} = \cos 2C$.
Since $A + B + C = 180^{\circ}$ and $B = 3C$,we have $A + 4C = 180^{\circ}$,so $2C = 90^{\circ} - \frac{A}{2}$.
Thus,$\cos 2C = \cos(90^{\circ} - \frac{A}{2}) = \sin \frac{A}{2}$.
Therefore,the values are $\cos C$ and $\sin \frac{A}{2}$.

(A) Let $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the expression:
$(b - c) \cot \frac{A}{2} = 2R(\sin B - \sin C) \cot \frac{A}{2}$
$= 2R \left( 2 \cos \frac{B+C}{2} \sin \frac{B-C}{2} \right) \cot \frac{A}{2}$
Since $\frac{B+C}{2} = 90^\circ - \frac{A}{2}$,$\cos \frac{B+C}{2} = \sin \frac{A}{2}$.
$= 2R \left( 2 \sin \frac{A}{2} \sin \frac{B-C}{2} \right) \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = 4R \sin \frac{B-C}{2} \cos \frac{A}{2}$
$= 4R \sin \frac{B-C}{2} \sin \frac{B+C}{2} = 2R(\cos(B-C) - \cos(B+C)) = 2R(\cos(B-C) + \cos A)$.
Summing this cyclically for all terms results in $0$.

(A) Given the equation: $2a^2b^2 + 2b^2c^2 = a^4 + b^4 + c^4$
Rearranging the terms: $a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 = 0$
We know the identity: $(a^2 - b^2 + c^2)^2 = a^4 + b^4 + c^4 - 2a^2b^2 + 2a^2c^2 - 2b^2c^2$
Substituting the given condition into the identity: $(a^2 - b^2 + c^2)^2 = 2a^2c^2$
Taking the square root on both sides: $a^2 - b^2 + c^2 = \pm \sqrt{2}ac$
Dividing by $2ac$: $\frac{a^2 + c^2 - b^2}{2ac} = \pm \frac{\sqrt{2}ac}{2ac} = \pm \frac{1}{\sqrt{2}}$
By the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,so $\cos B = \pm \frac{1}{\sqrt{2}}$
Thus,$B = 45^o$ or $B = 135^o$.

(B) Given: Area $\Delta = 10\sqrt{3}$,$C = 60^{\circ}$,and perimeter $a + b + c = 20$.
Using the area formula $\Delta = \frac{1}{2}ab \sin C$:
$10\sqrt{3} = \frac{1}{2}ab \sin 60^{\circ} = \frac{1}{2}ab \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}ab$.
Thus,$ab = 40$ ...$(i)$.
From the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \cos 60^{\circ} = \frac{1}{2}$.
So,$a^2 + b^2 - c^2 = ab$.
We know $(a+b)^2 = a^2 + b^2 + 2ab$,so $a^2 + b^2 = (a+b)^2 - 2ab$.
Substituting this into the cosine equation:
$(a+b)^2 - 2ab - c^2 = ab \Rightarrow (a+b)^2 - c^2 = 3ab$.
Since $a+b = 20-c$,we have:
$(20-c)^2 - c^2 = 3(40)$.
$400 - 40c + c^2 - c^2 = 120$.
$400 - 40c = 120$.
$40c = 280$.
$c = 7$.

(A) Given $A + C = 2B$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac}$,which implies $ac = a^2 + c^2 - b^2$,or $b^2 = a^2 - ac + c^2$.
Thus,$\sqrt{a^2 - ac + c^2} = b$.
The expression becomes $\frac{a + c}{b}$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = \frac{b}{\sin B} = 2R$,so $a = 2R \sin A$,$c = 2R \sin C$,and $b = 2R \sin B$.
Substituting these,we get $\frac{2R(\sin A + \sin C)}{2R \sin B} = \frac{\sin A + \sin C}{\sin B}$.
Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \frac{A + C}{2} \cos \frac{A - C}{2}$.
Since $A + C = 120^{\circ}$,$\frac{A + C}{2} = 60^{\circ}$,so $\sin \frac{A + C}{2} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
Also,$\sin B = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
Thus,the expression is $\frac{2 \sin 60^{\circ} \cos \frac{A - C}{2}}{\sin 60^{\circ}} = 2 \cos \frac{A - C}{2}$.