Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities Questions in English

(D) Given the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$.
We know that $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting this into the equation: $1 - 2 \sin^2 \theta \cos^2 \theta + \lambda = 0$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4}$.
Thus,$1 - 2 \left(\frac{\sin^2 2\theta}{4}\right) + \lambda = 0$,which simplifies to $1 - \frac{\sin^2 2\theta}{2} + \lambda = 0$.
Therefore,$\lambda = \frac{\sin^2 2\theta}{2} - 1$.
Since $0 \leq \sin^2 2\theta \leq 1$,we have $0 \leq \frac{\sin^2 2\theta}{2} \leq \frac{1}{2}$.
Subtracting $1$ from all parts,we get $-1 \leq \frac{\sin^2 2\theta}{2} - 1 \leq -\frac{1}{2}$.
Hence,$\lambda \in \left[-1, -\frac{1}{2}\right]$.

(B) Given $A+B=\frac{\pi}{2}$,so $B=\frac{\pi}{2}-A$.
Substituting this into the expression,we get $\cos A \cdot \cos B = \cos A \cdot \cos(\frac{\pi}{2}-A)$.
Since $\cos(\frac{\pi}{2}-A) = \sin A$,the expression becomes $\cos A \cdot \sin A$.
Multiplying and dividing by $2$,we get $\frac{2 \sin A \cos A}{2} = \frac{\sin(2A)}{2}$.
The maximum value of $\sin(2A)$ is $1$.
Therefore,the maximum value of $\frac{\sin(2A)}{2}$ is $\frac{1}{2}$.

(B) In $\triangle ABC$,$A + B + C = \pi$,so $2A + 2B + 2C = 2\pi$.
Therefore,$2A + 2B = 2\pi - 2C$.
Taking tangent on both sides: $\tan(2A + 2B) = \tan(2\pi - 2C) = -\tan 2C$.
Using the formula $\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get:
$\frac{\tan 2A + \tan 2B}{1 - \tan 2A \tan 2B} = -\tan 2C$.
Multiplying both sides by $(1 - \tan 2A \tan 2B)$:
$\tan 2A + \tan 2B = -\tan 2C(1 - \tan 2A \tan 2B)$.
$\tan 2A + \tan 2B = -\tan 2C + \tan 2A \tan 2B \tan 2C$.
Rearranging the terms:
$\tan 2A + \tan 2B + \tan 2C = \tan 2A \tan 2B \tan 2C$.

(D) Let $X = \frac{\cot A}{1+\cot A} \cdot \frac{\cot B}{1+\cot B}$.
Converting to $\tan$ functions: $X = \frac{1/\tan A}{1+1/\tan A} \cdot \frac{1/\tan B}{1+1/\tan B} = \frac{1}{\tan A+1} \cdot \frac{1}{\tan B+1}$.
$X = \frac{1}{\tan A \tan B + \tan A + \tan B + 1}$.
Given $A+B = 225^{\circ}$,we have $\tan(A+B) = \tan(225^{\circ}) = 1$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$,we get $\tan A + \tan B = 1 - \tan A \tan B$.
Substituting this into the denominator: $\tan A \tan B + (1 - \tan A \tan B) + 1 = 2$.
Thus,$X = \frac{1}{2}$.

(B) Given that $A + B + C = \pi$.
Since $A + B = \pi - C$,we take the cotangent on both sides:
$\cot(A + B) = \cot(\pi - C)$.
Using the identity $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ and $\cot(\pi - C) = -\cot C$,we get:
$\frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C$.
Multiplying both sides by $(\cot A + \cot B)$,we obtain:
$\cot A \cot B - 1 = -\cot A \cot C - \cot B \cot C$.
Rearranging the terms,we get:
$\cot A \cot B + \cot B \cot C + \cot A \cot C = 1$.

(A) Given $A+B+C=180^{\circ}$,so $\frac{A+B+C}{2} = 90^{\circ}$.
Thus,$\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$.
Taking tangent on both sides: $\tan \left(\frac{A+B}{2}\right) = \tan \left(90^{\circ} - \frac{C}{2}\right) = \cot \left(\frac{C}{2}\right)$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get:
$\frac{\tan(A/2) + \tan(B/2)}{1 - \tan(A/2) \tan(B/2)} = \frac{1}{\tan(C/2)}$.
Cross-multiplying gives: $\tan(C/2) [\tan(A/2) + \tan(B/2)] = 1 - \tan(A/2) \tan(B/2)$.
Rearranging the terms: $\tan(A/2) \tan(B/2) + \tan(B/2) \tan(C/2) + \tan(C/2) \tan(A/2) = 1$.

(A) To find the maximum value of the function $f(x) = a \sin x + b \cos x$,we can rewrite it in the form $R \sin(x + \alpha)$.
We multiply and divide by $\sqrt{a^2 + b^2}$:
$f(x) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}} \cos x \right)$.
Let $\cos \alpha = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2 + b^2}}$.
Then $f(x) = \sqrt{a^2 + b^2} (\sin x \cos \alpha + \cos x \sin \alpha) = \sqrt{a^2 + b^2} \sin(x + \alpha)$.
Since the maximum value of $\sin(x + \alpha)$ is $1$,the maximum value of $f(x)$ is $\sqrt{a^2 + b^2}$.

(A) The expression $\sqrt{3} \sin x + \cos x$ can be written in the form $R \sin(x + \alpha)$,where $R = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2$.
Thus,the maximum value of $\sqrt{3} \sin x + \cos x$ is $2$.
Therefore,the maximum value of the function $y = e^{5 + \sqrt{3} \sin x + \cos x}$ is $e^{5 + 2} = e^{7}$.

(A) Given $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$.
This implies $(\cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n) = (\sin \alpha_1 \sin \alpha_2 \ldots \sin \alpha_n)$.
Let $P = \cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n$.
Then $P^2 = (\cos \alpha_1 \ldots \cos \alpha_n)(\sin \alpha_1 \ldots \sin \alpha_n)$.
$P^2 = \frac{1}{2^n} (2 \sin \alpha_1 \cos \alpha_1) (2 \sin \alpha_2 \cos \alpha_2) \ldots (2 \sin \alpha_n \cos \alpha_n)$.
$P^2 = \frac{1}{2^n} \sin(2\alpha_1) \sin(2\alpha_2) \ldots \sin(2\alpha_n)$.
Since $\sin(2\alpha_i) \leq 1$ for all $i$,we have $P^2 \leq \frac{1}{2^n}$.
Taking the square root,$P \leq \frac{1}{\sqrt{2^n}} = \frac{1}{2^{(n/2)}}$.
The maximum value is $\frac{1}{2^{(n/2)}}$.

(B) To find the maximum value of $f(x) = \sin x + \cos x$,we can rewrite the function by multiplying and dividing by $\sqrt{2}$:
$f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write:
$f(x) = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right)$
$f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right)$
Since the maximum value of the sine function $\sin \theta$ is $1$,the maximum value of $f(x)$ is $\sqrt{2} \times 1 = \sqrt{2}$.

(A) We know that the range of the sine function is $-1 \leq \sin x \leq 1$.
To find the minimum value of $f(x) = 1 - \sin x$,we need to subtract the maximum possible value of $\sin x$ from $1$.
Since the maximum value of $\sin x$ is $1$,we have:
$f_{\min} = 1 - (\sin x)_{\max} = 1 - 1 = 0$.

(C) Let $f(x) = 27^{\cos 2x} 81^{\sin 2x} = (3^3)^{\cos 2x} (3^4)^{\sin 2x} = 3^{3 \cos 2x + 4 \sin 2x}$.
To find the minimum value,we analyze the exponent $g(x) = 3 \cos 2x + 4 \sin 2x$.
The expression $a \cos \theta + b \sin \theta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$,so the range of $3 \cos 2x + 4 \sin 2x$ is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.
The minimum value of the exponent is $-5$.
Therefore,the minimum value of $f(x) = 3^{-5} = \frac{1}{3^5} = \frac{1}{243}$.

(C) Given,$f(x) = \sin^6 x + \cos^6 x$.
We can write this as $f(x) = (\sin^2 x)^3 + (\cos^2 x)^3$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,where $a = \sin^2 x$ and $b = \cos^2 x$:
$f(x) = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$.
Since $\sin^2 x + \cos^2 x = 1$,we have:
$f(x) = \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x$.
Adding and subtracting $2 \sin^2 x \cos^2 x$ inside the expression:
$f(x) = (\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x = 1 - 3 \sin^2 x \cos^2 x$.
Multiply and divide by $4$:
$f(x) = 1 - \frac{3}{4}(4 \sin^2 x \cos^2 x) = 1 - \frac{3}{4}(\sin 2x)^2$.
Since $0 \leq \sin^2 2x \leq 1$,we have:
$1 - \frac{3}{4}(1) \leq f(x) \leq 1 - \frac{3}{4}(0)$.
$\frac{1}{4} \leq f(x) \leq 1$.
Thus,$f(x) \in \left[\frac{1}{4}, 1\right]$.

(D) Let $f(\theta) = 5 \cos \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$.
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$f(\theta) = 5 \cos \theta + 3 \left(\cos \theta \cos \frac{\pi}{3} - \sin \theta \sin \frac{\pi}{3}\right) + 3$.
$f(\theta) = 5 \cos \theta + 3 \left(\frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta\right) + 3$.
$f(\theta) = \left(5 + \frac{3}{2}\right) \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta + 3 = \frac{13}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta + 3$.
The expression $a \cos \theta + b \sin \theta$ ranges between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.
Here,$a = \frac{13}{2}$ and $b = -\frac{3\sqrt{3}}{2}$.
$\sqrt{a^2 + b^2} = \sqrt{\frac{169}{4} + \frac{27}{4}} = \sqrt{\frac{196}{4}} = \sqrt{49} = 7$.
Thus,$-7 \leq \frac{13}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta \leq 7$.
Adding $3$ to all parts:
$-7 + 3 \leq f(\theta) \leq 7 + 3$.
$-4 \leq f(\theta) \leq 10$.

(A) Given $A=10^{\circ}, B=16^{\circ}, C=19^{\circ}$.
Then $2A=20^{\circ}, 2B=32^{\circ}, 2C=38^{\circ}$.
Sum $2A+2B+2C = 20^{\circ}+32^{\circ}+38^{\circ} = 90^{\circ}$.
For any three angles $X, Y, Z$ such that $X+Y+Z=90^{\circ}$,we have $\tan X \tan Y + \tan Y \tan Z + \tan Z \tan X = 1$.
Substituting $X=2A, Y=2B, Z=2C$,we get $\tan 2A \tan 2B + \tan 2B \tan 2C + \tan 2C \tan 2A = 1$.
Thus,Assertion $(A)$ is true.
The Reason $(R)$ states the general identity: if $X+Y+Z=90^{\circ}$,then $\tan X \tan Y + \tan Y \tan Z + \tan Z \tan X = 1$.
This is the correct mathematical principle used to prove the Assertion.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(B) Given that $A + B + C = \pi$ and $A, B, C \neq \frac{n\pi}{2}$.
We know the identity $\cot(B + C) = \frac{\cot B \cot C - 1}{\cot B + \cot C}$.
Since $B + C = \pi - A$,we have $\cot(B + C) = \cot(\pi - A) = -\cot A$.
Substituting this into the identity:
$-\cot A = \frac{\cot B \cot C - 1}{\cot B + \cot C}$.
Multiplying both sides by $(\cot B + \cot C)$:
$-\cot A(\cot B + \cot C) = \cot B \cot C - 1$.
$-\cot A \cot B - \cot A \cot C = \cot B \cot C - 1$.
Rearranging the terms gives:
$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.

(C) Given $\cos \alpha = \frac{a}{b+c}$.
Using the identity $\cos \alpha = \frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)}$,we have $\frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)} = \frac{a}{b+c}$.
Applying componendo and dividendo,$\frac{(1 + \tan^2(\alpha/2)) + (1 - \tan^2(\alpha/2))}{(1 + \tan^2(\alpha/2)) - (1 - \tan^2(\alpha/2))} = \frac{(b+c) + a}{(b+c) - a}$.
This simplifies to $\frac{2}{2 \tan^2(\alpha/2)} = \frac{a+b+c}{b+c-a}$,so $\tan^2(\alpha/2) = \frac{b+c-a}{a+b+c}$.
Similarly,$\tan^2(\beta/2) = \frac{c+a-b}{a+b+c}$ and $\tan^2(\gamma/2) = \frac{a+b-c}{a+b+c}$.
Summing these,$\tan^2(\alpha/2) + \tan^2(\beta/2) + \tan^2(\gamma/2) = \frac{(b+c-a) + (c+a-b) + (a+b-c)}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.

(D) Let $f(x) = \frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$.
Divide numerator and denominator by $\cos^2 x$ (assuming $\cos x \neq 0$):
$f(x) = \frac{\sec^2 x}{\tan^2 x + 3 \tan x + 5} = \frac{1 + \tan^2 x}{\tan^2 x + 3 \tan x + 5}$.
Let $t = \tan x$. Then $f(t) = \frac{1 + t^2}{t^2 + 3t + 5}$.
Let $y = \frac{1 + t^2}{t^2 + 3t + 5}$.
$y(t^2 + 3t + 5) = 1 + t^2 \implies (y-1)t^2 + 3yt + (5y-1) = 0$.
For $t$ to be real,the discriminant $D \geq 0$:
$(3y)^2 - 4(y-1)(5y-1) \geq 0$.
$9y^2 - 4(5y^2 - 6y + 1) \geq 0$.
$9y^2 - 20y^2 + 24y - 4 \geq 0$.
$-11y^2 + 24y - 4 \geq 0 \implies 11y^2 - 24y + 4 \leq 0$.
Solving $11y^2 - 24y + 4 = 0$ using the quadratic formula:
$y = \frac{24 \pm \sqrt{576 - 176}}{22} = \frac{24 \pm \sqrt{400}}{22} = \frac{24 \pm 20}{22}$.
$y_1 = \frac{4}{22} = \frac{2}{11}$ and $y_2 = \frac{44}{22} = 2$.
Thus,the range is $\left[\frac{2}{11}, 2\right]$.

(A) Given $\cos 2 A+\cos 2 B+\cos 2 C$.
Using the formula $\cos C+\cos D=2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get:
$=2 \cos (A+B) \cos (A-B)+\cos 2 C$.
Since $A+B+C=\frac{3 \pi}{2}$,we have $A+B=\frac{3 \pi}{2}-C$.
$=2 \cos \left(\frac{3 \pi}{2}-C\right) \cos (A-B)+\left(1-2 \sin ^2 C\right)$.
$=2(-\sin C) \cos (A-B)+1-2 \sin ^2 C$.
$=1-2 \sin C [\cos (A-B)+\sin C]$.
Since $\sin C = \sin \left(\frac{3 \pi}{2}-(A+B)\right) = -\cos (A+B)$,we have:
$=1-2 \sin C [\cos (A-B)-\cos (A+B)]$.
Using $\cos (A-B)-\cos (A+B)=2 \sin A \sin B$:
$=1-2 \sin C [2 \sin A \sin B] = 1-4 \sin A \sin B \sin C$.

(C) Let $f(x) = 12 \sin x - 5 \cos x + 3$.
We know that for any expression of the form $a \sin x + b \cos x$,the range is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 12$ and $b = -5$.
So,$\sqrt{a^2 + b^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
Thus,$-13 \leq 12 \sin x - 5 \cos x \leq 13$.
Adding $3$ to all parts,we get:
$-13 + 3 \leq 12 \sin x - 5 \cos x + 3 \leq 13 + 3$.
$-10 \leq f(x) \leq 16$.
Therefore,the maximum value of $f(x)$ is $16$.

(A) $(I) \ 3 \sin^2 x + 4 \cos^2 x - 2 = 3(\sin^2 x + \cos^2 x) + \cos^2 x - 2 = 3 + \cos^2 x - 2 = \cos^2 x + 1$. Since $0 \leq \cos^2 x \leq 1$,the range is $[1, 2]$. Thus,$(I) \rightarrow (c)$.
$(II) \ \cos^2 x + \sin^4 x = (1 - \sin^2 x) + \sin^4 x = \sin^4 x - \sin^2 x + 1$. Let $t = \sin^2 x$,where $t \in [0, 1]$. The function $f(t) = t^2 - t + 1$ has a vertex at $t = 1/2$. $f(0) = 1$,$f(1) = 1$,$f(1/2) = 1/4 - 1/2 + 1 = 3/4$. Thus,the range is $[3/4, 1]$. Thus,$(II) \rightarrow (d)$.
$(III) \ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 - 3 \sin^2 x \cos^2 x = 1 - \frac{3}{4} \sin^2(2x)$. Since $0 \leq \sin^2(2x) \leq 1$,the range is $[1 - 3/4, 1 - 0] = [1/4, 1]$. Thus,$(III) \rightarrow (a)$.
$(IV) \ \cos x \cos(\frac{2 \pi}{3} + x) \cos(\frac{2 \pi}{3} - x) = \cos x (\cos^2(\frac{2 \pi}{3}) \cos^2 x - \sin^2(\frac{2 \pi}{3}) \sin^2 x) = \cos x (\frac{1}{4} \cos^2 x - \frac{3}{4} \sin^2 x) = \frac{1}{4} \cos^3 x - \frac{3}{4} \cos x \sin^2 x = \frac{1}{4} \cos 3x$. Since $-1 \leq \cos 3x \leq 1$,the range is $[-1/4, 1/4]$. Thus,$(IV) \rightarrow (b)$.

(B) Given expression: $5 \tan^2 \alpha + \frac{9}{\tan^2 \alpha} + 4 \sec^2 \alpha$
Using the identity $\sec^2 \alpha = 1 + \tan^2 \alpha$,we get:
$= 5 \tan^2 \alpha + 9 \cot^2 \alpha + 4(1 + \tan^2 \alpha)$
$= 9 \tan^2 \alpha + 9 \cot^2 \alpha + 4$
Since $AM \geq GM$ for positive real numbers,we have:
$\frac{9 \tan^2 \alpha + 9 \cot^2 \alpha}{2} \geq \sqrt{9 \tan^2 \alpha \cdot 9 \cot^2 \alpha}$
$9 \tan^2 \alpha + 9 \cot^2 \alpha \geq 2 \cdot \sqrt{81 \cdot 1}$
$9 \tan^2 \alpha + 9 \cot^2 \alpha \geq 18$
Adding $4$ to both sides:
$9 \tan^2 \alpha + 9 \cot^2 \alpha + 4 \geq 18 + 4 = 22$
Therefore,the minimum value is $22$.

(C) The expression is of the form $A \cos \theta + B \sin \theta$,where $\theta = x + \frac{\pi}{3}$,$A = 1$,and $B = 2 \sqrt{2}$.
The range of $A \cos \theta + B \sin \theta$ is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.
Here,$A^2 + B^2 = (1)^2 + (2 \sqrt{2})^2 = 1 + 8 = 9$.
Thus,$\sqrt{A^2 + B^2} = \sqrt{9} = 3$.
The minimum value is $-3$ and the maximum value is $3$.
Therefore,the correct option is $C$.

(B) We use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,which states that for positive real numbers $a$ and $b$,$\frac{a+b}{2} \geq \sqrt{ab}$.
Let $a = 27 \tan^2 \theta$ and $b = 3 \cot^2 \theta$.
Then,$\frac{27 \tan^2 \theta + 3 \cot^2 \theta}{2} \geq \sqrt{27 \tan^2 \theta \cdot 3 \cot^2 \theta}$.
$\frac{27 \tan^2 \theta + 3 \cot^2 \theta}{2} \geq \sqrt{81 \tan^2 \theta \cdot \cot^2 \theta}$.
Since $\tan \theta \cdot \cot \theta = 1$,we have:
$\frac{27 \tan^2 \theta + 3 \cot^2 \theta}{2} \geq \sqrt{81 \cdot 1}$.
$\frac{27 \tan^2 \theta + 3 \cot^2 \theta}{2} \geq 9$.
$27 \tan^2 \theta + 3 \cot^2 \theta \geq 18$.
Thus,the minimum value is $18$.

(B) Given that $A, B, C$ are the angles of a triangle,so $A + B + C = \pi$.
Using the identity $\sin 2A + \sin 2C = 2 \sin(A+C) \cos(A-C)$.
Since $A+C = \pi - B$,we have $\sin(A+C) = \sin(\pi - B) = \sin B$.
Thus,$\sin 2A + \sin 2C = 2 \sin B \cos(A-C)$.
Now,the expression is:
$\sin 2A - \sin 2B + \sin 2C = (\sin 2A + \sin 2C) - \sin 2B$
$= 2 \sin B \cos(A-C) - 2 \sin B \cos B$
$= 2 \sin B [\cos(A-C) - \cos B]$
Since $B = \pi - (A+C)$,$\cos B = \cos(\pi - (A+C)) = -\cos(A+C)$.
$= 2 \sin B [\cos(A-C) + \cos(A+C)]$
Using $\cos(A-C) + \cos(A+C) = 2 \cos A \cos C$:
$= 2 \sin B [2 \cos A \cos C] = 4 \cos A \sin B \cos C$.

(B) Given,$A+B+C = \pi$.
We know that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
The given equation is $\sin \left(\frac{\pi}{4}-A\right) \sin \left(\frac{\pi}{4}-B\right) \sin \left(\frac{\pi}{4}-C\right) = -\frac{1}{2 \sqrt{2}}$.
Using $\sin \left(\frac{\pi}{4}-\theta\right) = \frac{1}{\sqrt{2}}(\cos \theta - \sin \theta) = \frac{\cos \theta}{\sqrt{2}}(1 - \tan \theta)$,we get:
$\frac{\cos A \cos B \cos C}{2 \sqrt{2}} (1 - \tan A)(1 - \tan B)(1 - \tan C) = -\frac{1}{2 \sqrt{2}}$.
$(1 - \tan A)(1 - \tan B)(1 - \tan C) = -\frac{1}{\cos A \cos B \cos C}$.
Expanding the left side: $1 - (\tan A + \tan B + \tan C) + (\tan A \tan B + \tan B \tan C + \tan C \tan A) - \tan A \tan B \tan C = -\frac{1}{\cos A \cos B \cos C}$.
Since $\tan A + \tan B + \tan C = \tan A \tan B \tan C$,the terms cancel out:
$1 + (\tan A \tan B + \tan B \tan C + \tan C \tan A) - 2(\tan A + \tan B + \tan C) = -\frac{1}{\cos A \cos B \cos C}$.
For $A+B+C = \pi$,$\cos(A+B+C) = \cos A \cos B \cos C (1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)) = -1$.
Thus,$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = -\frac{1}{\cos A \cos B \cos C}$.
Equating the two expressions for $-\frac{1}{\cos A \cos B \cos C}$:
$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 1 - (\tan A + \tan B + \tan C) + (\tan A \tan B + \tan B \tan C + \tan C \tan A) - \tan A \tan B \tan C$.
Since $\tan A + \tan B + \tan C = \tan A \tan B \tan C$,we get:
$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 1 - 2(\tan A + \tan B + \tan C) + (\tan A \tan B + \tan B \tan C + \tan C \tan A)$.
$2(\tan A \tan B + \tan B \tan C + \tan C \tan A) = 2(\tan A + \tan B + \tan C)$.
Therefore,$\tan A \tan B + \tan B \tan C + \tan C \tan A = \tan A + \tan B + \tan C$.

(A) Given the equation $a \cos x + a \sin x + a = 2K + 1$.
We can rewrite the expression as $a(\cos x + \sin x) + a = 2K + 1$.
Using the identity $\cos x + \sin x = \sqrt{2} \cos(x - \frac{\pi}{4})$,we get $a[\sqrt{2} \cos(x - \frac{\pi}{4}) + 1] = 2K + 1$.
Since $-1 \leq \cos(x - \frac{\pi}{4}) \leq 1$,the range of $\sqrt{2} \cos(x - \frac{\pi}{4}) + 1$ is $[1 - \sqrt{2}, 1 + \sqrt{2}]$.
Multiplying by $a$ (assuming $a > 0$),we have $a(1 - \sqrt{2}) \leq 2K + 1 \leq a(1 + \sqrt{2})$.
Solving for $K$: $a - a\sqrt{2} - 1 \leq 2K \leq a + a\sqrt{2} - 1$.
Thus,$K \in \left[\frac{a - a\sqrt{2} - 1}{2}, \frac{a + a\sqrt{2} - 1}{2}\right]$.
This matches option $A$.

(B) Given the equation $\sin^4 x + \cos^4 x = a$.
We can rewrite the expression as:
$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$
$= 1^2 - \frac{1}{2} (2 \sin x \cos x)^2$
$= 1 - \frac{1}{2} \sin^2(2x)$.
Since $0 \leq \sin^2(2x) \leq 1$,we multiply by $-\frac{1}{2}$:
$-\frac{1}{2} \leq -\frac{1}{2} \sin^2(2x) \leq 0$.
Adding $1$ to all parts:
$1 - \frac{1}{2} \leq 1 - \frac{1}{2} \sin^2(2x) \leq 1 + 0$
$\frac{1}{2} \leq a \leq 1$.
Thus,the equation has real solutions when $\frac{1}{2} \leq a \leq 1$.

(B) Given $A+B+C = \frac{\pi}{4}$,so $4(A+B+C) = \pi$.
Using the identity for $\sin x + \sin y + \sin z$ when $x+y+z = \pi$,we have $\sin x + \sin y + \sin z = 4 \sin \frac{x}{2} \sin \frac{y}{2} \sin \frac{z}{2}$.
Here,let $x=4A, y=4B, z=4C$.
Then $\sin 4A + \sin 4B + \sin 4C = 4 \sin 2A \sin 2B \sin 2C$.

(B) Let $t = \sin \frac{x}{4}$. Since $x \in R$,$t \in [-1, 1]$.
The expression becomes $f(t) = (1 - t^2) + t = -t^2 + t + 1$.
This is a downward-opening parabola with vertex at $t = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}$.
Since $t = \frac{1}{2}$ is within the interval $[-1, 1]$,the maximum value $\alpha = f(\frac{1}{2}) = -(\frac{1}{2})^2 + \frac{1}{2} + 1 = -\frac{1}{4} + \frac{1}{2} + 1 = \frac{5}{4}$.
The minimum value $\beta$ occurs at the endpoints of the interval $[-1, 1]$.
$f(-1) = -(-1)^2 + (-1) + 1 = -1 - 1 + 1 = -1$.
$f(1) = -(1)^2 + (1) + 1 = -1 + 1 + 1 = 1$.
Thus,$\beta = -1$.
Therefore,$\alpha - \beta = \frac{5}{4} - (-1) = \frac{5}{4} + 1 = \frac{9}{4}$.

(B) Given $\cos A \cos B + \sin A \sin B \sin C = 1$.
Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin C \le \cos A \cos B + \sin A \sin B = \cos(A-B)$.
Thus,$\cos(A-B) \ge 1$.
Since the maximum value of $\cos(A-B)$ is $1$,we must have $\cos(A-B) = 1$,which implies $A = B$.
Also,this requires $\sin C = 1$,so $C = 90^\circ$.
Since $A+B+C = 180^\circ$ and $A=B$,we have $2A + 90^\circ = 180^\circ$,so $A = B = 45^\circ$.
Now,$\sin A + \sin B + \sin C = \sin 45^\circ + \sin 45^\circ + \sin 90^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 = \frac{2}{\sqrt{2}} + 1 = \sqrt{2} + 1$.

(A) Given that $P+Q+R=\frac{\pi}{4}$.
Let $S = \cos \left(\frac{\pi}{8}-P\right)+\cos \left(\frac{\pi}{8}-Q\right)+\cos \left(\frac{\pi}{8}-R\right)$.
Using the formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$:
$S = 2 \cos \left(\frac{\pi}{8} - \frac{P+Q}{2}\right) \cos \left(\frac{Q-P}{2}\right) + \cos \left(\frac{\pi}{8}-R\right)$.
Since $P+Q = \frac{\pi}{4} - R$,then $\frac{P+Q}{2} = \frac{\pi}{8} - \frac{R}{2}$.
Substituting this:
$S = 2 \cos \left(\frac{\pi}{8} - (\frac{\pi}{8} - \frac{R}{2})\right) \cos \left(\frac{Q-P}{2}\right) + \cos \left(\frac{\pi}{8}-R\right)$
$S = 2 \cos \frac{R}{2} \cos \frac{Q-P}{2} + \cos \left(\frac{\pi}{8}-R\right)$.
Using $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$ or expansion:
$S = 2 \cos \frac{R}{2} \cos \frac{Q-P}{2} + 2 \cos^2 \left(\frac{\pi}{16} - \frac{R}{2}\right) - 1$ (This path is complex,let's use the identity for sum of cosines).
Actually,$S = 4 \cos \frac{P}{2} \cos \frac{Q}{2} \cos \frac{R}{2} - \cos \frac{\pi}{8}$ is the standard result derived from the expansion of the sum of cosines given the constraint.

(C) Given: $10 \sin^4 \alpha + 15 \cos^4 \alpha = 6$
Divide by $\cos^4 \alpha$:
$10 \tan^4 \alpha + 15 = 6 \sec^4 \alpha$
$10 \tan^4 \alpha + 15 = 6(1 + \tan^2 \alpha)^2$
$10 \tan^4 \alpha + 15 = 6(1 + 2 \tan^2 \alpha + \tan^4 \alpha)$
$10 \tan^4 \alpha + 15 = 6 + 12 \tan^2 \alpha + 6 \tan^4 \alpha$
$4 \tan^4 \alpha - 12 \tan^2 \alpha + 9 = 0$
$(2 \tan^2 \alpha - 3)^2 = 0$
$\tan^2 \alpha = \frac{3}{2}$
Then,$\cot^2 \alpha = \frac{2}{3}$
Now,$16 \tan^6 \alpha + 27 \cot^6 \alpha = 16(\frac{3}{2})^3 + 27(\frac{2}{3})^3$
$= 16(\frac{27}{8}) + 27(\frac{8}{27}) = 54 + 8 = 62$

(C) Given,$A+B+C=\pi$,which implies $B=\pi-(A+C)$.
Given the equation $\cos B=\cos A \cos C$.
Substituting $B$,we get $\cos(\pi-(A+C))=\cos A \cos C$.
Using the identity $\cos(\pi-\theta)=-\cos \theta$,we have $-\cos(A+C)=\cos A \cos C$.
Expanding the left side,$-(\cos A \cos C - \sin A \sin C) = \cos A \cos C$.
$-\cos A \cos C + \sin A \sin C = \cos A \cos C$.
Rearranging the terms,$\sin A \sin C = 2 \cos A \cos C$.
Dividing both sides by $\cos A \cos C$,we get $\frac{\sin A \sin C}{\cos A \cos C} = 2$.
Therefore,$\tan A \tan C = 2$.

(C) Given $A+B+C=270^{\circ}$,so $A+B=270^{\circ}-C$.
Using the formula $\cos 2A+\cos 2B=2 \cos(A+B) \cos(A-B)$,we have:
$\cos 2A+\cos 2B+\cos 2C = 2 \cos(A+B) \cos(A-B) + (2 \cos^2 C - 1)$
$= 2 \cos(270^{\circ}-C) \cos(A-B) + 2 \cos^2 C - 1$
$= 2(-\sin C) \cos(A-B) + 2 \cos^2 C - 1$
$= -2 \sin C \cos(A-B) + 2(1-\sin^2 C) - 1$
$= 1 - 2 \sin C [\cos(A-B) + \sin C]$
Since $C = 270^{\circ}-(A+B)$,$\sin C = \sin(270^{\circ}-(A+B)) = -\cos(A+B)$.
$= 1 - 2 \sin C [\cos(A-B) - \cos(A+B)]$
Using $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$:
$= 1 - 2 \sin C [2 \sin A \sin B] = 1 - 4 \sin A \sin B \sin C$.

(D) Given $x+y+z = \frac{\pi}{2}$ and $x \geq y \geq z \geq \frac{\pi}{12}$.
We want to minimize $f(x, y, z) = \cos x \sin y \cos z$.
Using the product-to-sum formula: $\cos x \cos z = \frac{1}{2}(\cos(x+z) + \cos(x-z))$.
Since $x+z = \frac{\pi}{2} - y$,we have $\cos(x+z) = \sin y$.
So,$f = \frac{1}{2}(\sin y + \cos(x-z)) \sin y = \frac{1}{2}(\sin^2 y + \sin y \cos(x-z))$.
To minimize this,we consider the boundary conditions. Since $x \geq y \geq z \geq \frac{\pi}{12}$ and $x+y+z = \frac{\pi}{2}$,the minimum occurs when $y$ and $z$ are as small as possible,i.e.,$y = z = \frac{\pi}{12}$.
Then $x = \frac{\pi}{2} - (y+z) = \frac{\pi}{2} - \frac{2\pi}{12} = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.
Substituting these values: $\cos(\frac{\pi}{3}) \sin(\frac{\pi}{12}) \cos(\frac{\pi}{12}) = \frac{1}{2} \cdot \frac{1}{2} \sin(\frac{\pi}{6}) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$.

(C) Given $A+B+C=0$,we have $A+B = -C$.
Taking the sine of both sides,$\sin(A+B) = \sin(-C) = -\sin(C)$.
Now,consider the expression $\sin(2A) + \sin(2B) + \sin(2C)$:
$\sin(2A) + \sin(2B) + \sin(2C) = 2 \sin(A+B) \cos(A-B) + 2 \sin(C) \cos(C)$
Since $A+B = -C$,then $\cos(A+B) = \cos(-C) = \cos(C)$.
Substituting $\sin(A+B) = -\sin(C)$ and $\cos(C) = \cos(A+B)$:
$= 2(-\sin(C)) \cos(A-B) + 2 \sin(C) \cos(A+B)$
$= -2 \sin(C) [\cos(A-B) - \cos(A+B)]$
Using the identity $\cos(x-y) - \cos(x+y) = 2 \sin(x) \sin(y)$:
$= -2 \sin(C) [2 \sin(A) \sin(B)]$
$= -4 \sin(A) \sin(B) \sin(C)$.

(B) For the first expression,we know that $-\sqrt{11^2 + 60^2} \leq 11 \cos 2x + 60 \sin 2x \leq \sqrt{11^2 + 60^2}$.
This simplifies to $-61 \leq 11 \cos 2x + 60 \sin 2x \leq 61$.
To maximize $\frac{1}{11 \cos 2x + 60 \sin 2x + 69}$,we need to minimize the denominator.
The minimum value of the denominator is $69 - 61 = 8$.
Thus,$M_1 = \frac{1}{8}$.
For the second expression,$3 \cos^2 5x + 4 \sin^2 5x = 3(\cos^2 5x + \sin^2 5x) + \sin^2 5x = 3 + \sin^2 5x$.
The maximum value occurs when $\sin^2 5x = 1$,so $M_2 = 3 + 1 = 4$.
Therefore,$\frac{M_1}{M_2} = \frac{1/8}{4} = \frac{1}{32}$.

(A) Given $f(x) = 5 \cos x + 3 \cos \left(x + \frac{\pi}{3}\right) + 8$.
Expanding the term $\cos(x + \frac{\pi}{3})$:
$f(x) = 5 \cos x + 3 \left[ \cos x \cdot \cos \frac{\pi}{3} - \sin x \cdot \sin \frac{\pi}{3} \right] + 8$
$f(x) = 5 \cos x + 3 \left[ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \right] + 8$
$f(x) = \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x + 8$.
For any expression of the form $A \cos x + B \sin x$,the range is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.
Here,$A = \frac{13}{2}$ and $B = -\frac{3\sqrt{3}}{2}$.
Calculating $\sqrt{A^2 + B^2} = \sqrt{\left(\frac{13}{2}\right)^2 + \left(-\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{169}{4} + \frac{27}{4}} = \sqrt{\frac{196}{4}} = \sqrt{49} = 7$.
Thus,the range of $\frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x$ is $[-7, 7]$.
Adding $8$ to the range,we get $f(x) \in [-7 + 8, 7 + 8]$,which is $[1, 15]$.
Therefore,the maximum value is $15$ and the minimum value is $1$.

(C) Given $f(x) = \sin x + \frac{1-\tan^2 x}{1+\tan^2 x}$.
Using the identity $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we have $f(x) = \sin x + \cos 2x$.
Since $\cos 2x = 1 - 2\sin^2 x$,we can write $f(x) = \sin x + 1 - 2\sin^2 x$.
Let $t = \sin x$,where $t \in [-1, 1]$. Then $g(t) = -2t^2 + t + 1$.
To find the maximum,we find the derivative $g'(t) = -4t + 1$.
Setting $g'(t) = 0$,we get $t = \frac{1}{4}$.
Since $g''(t) = -4 < 0$,$t = \frac{1}{4}$ is a point of local maxima.
We need to find the number of values of $x \in [0, 2\pi]$ such that $\sin x = \frac{1}{4}$.
Since $\frac{1}{4} > 0$,$\sin x = \frac{1}{4}$ has two solutions in the interval $[0, 2\pi]$ (one in the first quadrant and one in the second quadrant).
Thus,there are $2$ such values of $x$.

(D) Let $f(\alpha) = \left(1 + \frac{1}{\sin^n \alpha}\right) \left(1 + \frac{1}{\cos^n \alpha}\right)$.
By the $AM$-$GM$ inequality,for positive terms,the minimum value occurs when $\sin^n \alpha = \cos^n \alpha$,which implies $\sin \alpha = \cos \alpha$,so $\alpha = \frac{\pi}{4}$.
At $\alpha = \frac{\pi}{4}$,we have $\sin \alpha = \cos \alpha = \frac{1}{\sqrt{2}} = 2^{-1/2}$.
Then $\sin^n \alpha = \cos^n \alpha = (2^{-1/2})^n = 2^{-n/2}$.
Substituting this into the expression:
$f\left(\frac{\pi}{4}\right) = \left(1 + \frac{1}{2^{-n/2}}\right) \left(1 + \frac{1}{2^{-n/2}}\right) = (1 + 2^{n/2})(1 + 2^{n/2}) = (1 + 2^{n/2})^2$.
Thus,the minimum value is $(1 + 2^{n/2})^2$.

(B) Given $f(\theta) = \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} + \sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}$.
Squaring both sides,we get $[f(\theta)]^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta) + 2 \sqrt{(a^2 \cos^2 \theta + b^2 \sin^2 \theta)(a^2 \sin^2 \theta + b^2 \cos^2 \theta)}$.
Simplifying the terms,$[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^4 \sin^2 \theta \cos^2 \theta + a^2 b^2 \cos^4 \theta + a^2 b^2 \sin^4 \theta + b^4 \sin^2 \theta \cos^2 \theta}$.
$[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 (\sin^4 \theta + \cos^4 \theta) + (a^4 + b^4) \sin^2 \theta \cos^2 \theta}$.
Using $\sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$,we get $[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 (1 - 2 \sin^2 \theta \cos^2 \theta) + (a^4 + b^4) \sin^2 \theta \cos^2 \theta}$.
$[f(\theta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 + (a^4 + b^4 - 2 a^2 b^2) \sin^2 \theta \cos^2 \theta} = a^2 + b^2 + 2 \sqrt{a^2 b^2 + (a^2 - b^2)^2 \sin^2 \theta \cos^2 \theta}$.
Let $X = \sin^2 \theta \cos^2 \theta = \frac{\sin^2(2\theta)}{4}$. The range of $X$ is $[0, 1/4]$.
For maximum value $M$,$X = 1/4$: $M = a^2 + b^2 + 2 \sqrt{a^2 b^2 + \frac{(a^2 - b^2)^2}{4}} = a^2 + b^2 + 2 \sqrt{\frac{4a^2 b^2 + a^4 + b^4 - 2a^2 b^2}{4}} = a^2 + b^2 + 2 \sqrt{\frac{(a^2 + b^2)^2}{4}} = a^2 + b^2 + (a^2 + b^2) = 2(a^2 + b^2)$.
For minimum value $m$,$X = 0$: $m = a^2 + b^2 + 2 \sqrt{a^2 b^2} = a^2 + b^2 + 2ab = (a+b)^2$.
Thus,$M - m = 2a^2 + 2b^2 - (a^2 + b^2 + 2ab) = a^2 + b^2 - 2ab = (a-b)^2$.

(B) Given,$f(x) = \tan \left(x + \frac{2 \pi}{3} \right) - \tan \left(x + \frac{\pi}{6} \right) + \cos \left(x + \frac{\pi}{6} \right)$.
Using the identity $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$,we have:
$\tan \left(x + \frac{2 \pi}{3} \right) - \tan \left(x + \frac{\pi}{6} \right) = \frac{\sin \left( \frac{2 \pi}{3} - \frac{\pi}{6} \right)}{\cos \left(x + \frac{2 \pi}{3} \right) \cos \left(x + \frac{\pi}{6} \right)} = \frac{\sin \frac{\pi}{2}}{\cos \left(x + \frac{2 \pi}{3} \right) \cos \left(x + \frac{\pi}{6} \right)} = \frac{1}{\cos \left(x + \frac{2 \pi}{3} \right) \cos \left(x + \frac{\pi}{6} \right)}$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,the expression becomes:
$f(x) = \frac{2}{\cos \left( 2x + \frac{5 \pi}{6} \right) + \cos \frac{\pi}{2}} + \cos \left(x + \frac{\pi}{6} \right) = \frac{2}{\cos \left( 2x + \frac{5 \pi}{6} \right)} + \cos \left(x + \frac{\pi}{6} \right)$.
In the interval $\left[ -\frac{5 \pi}{12}, -\frac{\pi}{3} \right]$,the function is monotonically increasing.
Thus,the maximum value occurs at the right endpoint $x = -\frac{\pi}{3} = -60^{\circ}$.
$f(-60^{\circ}) = \tan \left( -60^{\circ} + 120^{\circ} \right) - \tan \left( -60^{\circ} + 30^{\circ} \right) + \cos \left( -60^{\circ} + 30^{\circ} \right) = \tan 60^{\circ} - \tan(-30^{\circ}) + \cos(-30^{\circ}) = \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \sqrt{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2} = \frac{6 \sqrt{3} + 2 \sqrt{3} + 3 \sqrt{3}}{6} = \frac{11 \sqrt{3}}{6}$.

(B) Given,$A+B=\frac{\pi}{3}$.
Let $y = \tan A \tan B$.
Since $B = \frac{\pi}{3} - A$,we have $y = \tan A \tan(\frac{\pi}{3} - A)$.
Using the $AM$-$GM$ inequality for positive values $\tan A$ and $\tan B$:
$\frac{\tan A + \tan B}{2} \geq \sqrt{\tan A \tan B}$.
We know that $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B} = \frac{\sin(\pi/3)}{\cos A \cos B} = \frac{\sqrt{3}/2}{\cos A \cos B}$.
Alternatively,using the identity $\tan A \tan B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)}$.
For a fixed sum $A+B = \frac{\pi}{3}$,the product $\tan A \tan B$ is maximized when $A=B$.
Thus,$A = B = \frac{\pi}{6}$.
Maximum value $= \tan(\frac{\pi}{6}) \tan(\frac{\pi}{6}) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.

(A) Given $y = 4 \sin^2 \theta - \cos 2 \theta$.
Using the identity $\cos 2 \theta = 1 - 2 \sin^2 \theta$,we get:
$y = 4 \sin^2 \theta - (1 - 2 \sin^2 \theta) = 6 \sin^2 \theta - 1$.
Since $0 \leq \sin^2 \theta \leq 1$,we have:
$0 \leq 6 \sin^2 \theta \leq 6$.
Subtracting $1$ from all parts:
$-1 \leq 6 \sin^2 \theta - 1 \leq 5$.
Thus,the minimum value $l = -1$ and the maximum value $m = 5$.
Now,$lm = (-1)(5) = -5$.
Also,$\frac{m}{l} = \frac{5}{-1} = -5$.
Therefore,$lm = \frac{m}{l}$.

(A) We know that $\sin^2 \theta + \cos^2 \theta = 1$.
Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$,we have:
$\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 = (\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1 - 3 \sin^2 \theta \cos^2 \theta$.
Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,we have:
$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta)^2 + (\cos^2 \theta)^2 = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting these into the expression:
$2(1 - 3 \sin^2 \theta \cos^2 \theta) - 3(1 - 2 \sin^2 \theta \cos^2 \theta)$
$= 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta$
$= -1$.