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

(D) By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:
$\frac{2^{\sin \theta} + 2^{\cos \theta}}{2} \ge \sqrt{2^{\sin \theta} \cdot 2^{\cos \theta}}$
$2^{\sin \theta} + 2^{\cos \theta} \ge 2 \cdot 2^{(\sin \theta + \cos \theta)/2} = 2^{1 + \frac{\sin \theta + \cos \theta}{2}}$
Since $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4)$,the minimum value of $\sin \theta + \cos \theta$ is $-\sqrt{2}$.
Therefore,$2^{\sin \theta} + 2^{\cos \theta} \ge 2^{1 + \frac{-\sqrt{2}}{2}} = 2^{1 - \frac{1}{\sqrt{2}}}$.

(B) We are given the equation $(a + b)^2 = 4ab \sin^2 \theta$.
Since the range of $\sin^2 \theta$ is $[0, 1]$,we must have $\sin^2 \theta = \frac{(a + b)^2}{4ab} \le 1$.
This implies $(a + b)^2 \le 4ab$.
Rearranging the terms,we get $(a + b)^2 - 4ab \le 0$,which simplifies to $(a - b)^2 \le 0$.
Since the square of any real number cannot be negative,the only possibility is $(a - b)^2 = 0$,which means $a = b$.

(A) We know that for any real $\theta$,$\sec^2 \theta \ge 1$.
Given the equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$,it must satisfy $\frac{4xy}{(x + y)^2} \ge 1$.
Since $(x + y)^2 > 0$ (assuming $x, y \neq 0$),we have $4xy \ge (x + y)^2$.
$4xy \ge x^2 + 2xy + y^2$.
$0 \ge x^2 - 2xy + y^2$.
$0 \ge (x - y)^2$.
Since the square of a real number cannot be negative,$(x - y)^2$ must be $0$.
Therefore,$x - y = 0$,which implies $x = y$.

(D) We know that for any real $\theta$,$0 \le \sin^2 \theta \le 1$.
Given $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,we must have $\frac{x^2 + y^2 + 1}{2x} \le 1$.
Assuming $x > 0$,we get $x^2 + y^2 + 1 \le 2x$,which simplifies to $(x - 1)^2 + y^2 \le 0$.
Since squares of real numbers are non-negative,this inequality holds only if $(x - 1)^2 = 0$ and $y^2 = 0$,which implies $x = 1$ and $y = 0$.
If $x < 0$,the expression $\frac{x^2 + y^2 + 1}{2x}$ would be negative,which contradicts $\sin^2 \theta \ge 0$.
Since the value of $x$ depends on $y$ (specifically $y$ must be $0$ for $x=1$),$x$ is not uniquely determined as a constant independent of $y$ in all cases,or rather,the condition requires $y=0$ and $x=1$. Given the options,$x=1$ is a specific case,but since $y$ is not defined,the most accurate choice is $D$.

(B) Given $\cos 2B = \frac{\cos(A + C)}{\cos(A - C)}$.
Using the expansion of $\cos(A+C)$ and $\cos(A-C)$,we get $\cos 2B = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$.
Dividing the numerator and denominator by $\cos A \cos C$,we get $\cos 2B = \frac{1 - \tan A \tan C}{1 + \tan A \tan C}$.
We know that $\cos 2B = \frac{1 - \tan^2 B}{1 + \tan^2 B}$.
Equating the two expressions: $\frac{1 - \tan^2 B}{1 + \tan^2 B} = \frac{1 - \tan A \tan C}{1 + \tan A \tan C}$.
By componendo and dividendo or cross-multiplication,we get $1 - \tan^2 B + \tan A \tan C - \tan^2 B \tan A \tan C = 1 + \tan^2 B - \tan A \tan C - \tan^2 B \tan A \tan C$.
This simplifies to $2 \tan^2 B = 2 \tan A \tan C$,which implies $\tan^2 B = \tan A \tan C$.
Thus,$\tan A, \tan B, \tan C$ are in $G.P.$

(B) Given: $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$
Using the expansion formulas $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \left( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \right)$
$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - 2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$
Rearranging the terms:
$3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$
Dividing both sides by $3 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$:
$\frac{\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \frac{1}{3}$
$\tan \frac{\alpha}{2} \tan \frac{\beta}{2} = \frac{1}{3}$

(D) Let $f(\theta) = a \cos \theta + b \sin \theta$.
We can write this as $f(\theta) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos \theta + \frac{b}{\sqrt{a^2 + b^2}} \sin \theta \right)$.
Let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$.
Then $f(\theta) = \sqrt{a^2 + b^2} (\cos \theta \cos \phi + \sin \theta \sin \phi) = \sqrt{a^2 + b^2} \cos(\theta - \phi)$.
Since the range of $\cos(\theta - \phi)$ is $[-1, 1]$,the range of $f(\theta)$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Therefore,the value of $a \cos \theta + b \sin \theta$ lies between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.

(C) The expression is of the form $a\cos \theta + b\sin \theta$,where $a = 3$ and $b = -4$.
The maximum value of the expression $a\cos \theta + b\sin \theta$ is given by $\sqrt{a^2 + b^2}$.
Substituting the values,we get:
Maximum value $= \sqrt{3^2 + (-4)^2}$
$= \sqrt{9 + 16}$
$= \sqrt{25}$
$= 5$.
Therefore,the maximum value is $5$.

(D) Let $f(\theta) = 5 \sin^2 \theta + 4 \cos^2 \theta$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we can write $\cos^2 \theta = 1 - \sin^2 \theta$.
Substituting this into the expression:
$f(\theta) = 5 \sin^2 \theta + 4(1 - \sin^2 \theta)$
$f(\theta) = 5 \sin^2 \theta + 4 - 4 \sin^2 \theta$
$f(\theta) = 4 + \sin^2 \theta$.
Since the minimum value of $\sin^2 \theta$ is $0$,the minimum value of $f(\theta)$ is $4 + 0 = 4$.

(C) Using the identity $\cos^2 A - \cos^2 B = \sin(B - A) \sin(B + A)$:
Let $A = \frac{\pi}{3} - x$ and $B = \frac{\pi}{3} + x$.
Then $B - A = (\frac{\pi}{3} + x) - (\frac{\pi}{3} - x) = 2x$.
And $B + A = (\frac{\pi}{3} + x) + (\frac{\pi}{3} - x) = \frac{2\pi}{3}$.
Therefore,$\cos^2 \left( \frac{\pi}{3} - x \right) - \cos^2 \left( \frac{\pi}{3} + x \right) = \sin(2x) \sin\left( \frac{2\pi}{3} \right)$.
Since $\sin\left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}$,the expression becomes $\frac{\sqrt{3}}{2} \sin(2x)$.
The maximum value of $\sin(2x)$ is $1$.
Thus,the maximum value of the expression is $\frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2}$.

(A) We know that for any real number $x \neq 0$,$(x - \frac{1}{x})^2 \ge 0$.
Expanding this,we get $x^2 + \frac{1}{x^2} - 2 \ge 0$,which implies $x^2 + \frac{1}{x^2} \ge 2$.
Let $x = \tan \theta$. Since $\cot \theta = \frac{1}{\tan \theta}$,we substitute $x = \tan \theta$ into the inequality:
$\tan^2 \theta + \frac{1}{\tan^2 \theta} \ge 2$
$\tan^2 \theta + \cot^2 \theta \ge 2$.
Thus,the correct option is $A$.

(D) Let $f(x) = a \cos x + b \sin x$.
We can write this expression as $f(x) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos x + \frac{b}{\sqrt{a^2 + b^2}} \sin x \right)$.
Let $\cos \alpha = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2 + b^2}}$.
Then $f(x) = \sqrt{a^2 + b^2} (\cos \alpha \cos x + \sin \alpha \sin x) = \sqrt{a^2 + b^2} \cos(x - \alpha)$.
Since the maximum value of $\cos(x - \alpha)$ is $1$,the maximum value of $f(x)$ is $\sqrt{a^2 + b^2}$.

(D) The expression is of the form $a \cos x + b \sin x + c$,where $a = 3$,$b = 4$,and $c = 5$.
The range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Calculating the value: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Thus,the minimum value of $3 \cos x + 4 \sin x$ is $-5$.
Therefore,the minimum value of $3 \cos x + 4 \sin x + 5$ is $-5 + 5 = 0$.

(B) Let $f(x) = \sin x \cos x$.
Multiplying and dividing by $2$,we get $f(x) = \frac{2 \sin x \cos x}{2} = \frac{\sin 2x}{2}$.
We know that the range of $\sin 2x$ is $[-1, 1]$,so $-1 \le \sin 2x \le 1$.
Dividing the inequality by $2$,we get $-\frac{1}{2} \le \frac{\sin 2x}{2} \le \frac{1}{2}$.
Thus,the greatest value is $\frac{1}{2}$ and the least value is $-\frac{1}{2}$.

(B) Let $f(\theta) = \cos \theta + \sin \theta$.
We can rewrite this as $f(\theta) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta \right) = \sqrt{2} \cos \left( \theta - \frac{\pi}{4} \right)$.
Since the range of the cosine function is $[-1, 1]$,we have $-1 \le \cos \left( \theta - \frac{\pi}{4} \right) \le 1$.
Multiplying by $\sqrt{2}$,we get $-\sqrt{2} \le \sqrt{2} \cos \left( \theta - \frac{\pi}{4} \right) \le \sqrt{2}$.
Thus,the minimum value of $f(\theta)$ is $-\sqrt{2}$.

(B) Let $f(x) = 4\sin^2 x + 3\cos^2 x$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we can write $\cos^2 x = 1 - \sin^2 x$.
Substituting this into the expression: $f(x) = 4\sin^2 x + 3(1 - \sin^2 x) = 4\sin^2 x + 3 - 3\sin^2 x = \sin^2 x + 3$.
Since the range of $\sin^2 x$ is $[0, 1]$,the maximum value of $f(x)$ occurs when $\sin^2 x = 1$.
Therefore,the maximum value is $1 + 3 = 4$.

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$.
We can rewrite this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \left( x + \frac{\pi}{6} \right) + \frac{1}{\sqrt{2}} \cos \left( x + \frac{\pi}{6} \right) \right)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{6} + \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x + \frac{5\pi}{12} \right)$.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = \frac{\pi}{2}$.
Setting $x + \frac{5\pi}{12} = \frac{\pi}{2}$,we get $x = \frac{\pi}{2} - \frac{5\pi}{12} = \frac{6\pi - 5\pi}{12} = \frac{\pi}{12}$.
Since $\frac{\pi}{12}$ lies in the interval $\left( 0, \frac{\pi}{2} \right)$,the maximum value is attained at $x = \frac{\pi}{12}$.

(D) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for two positive numbers $a$ and $b$,which states $\frac{a+b}{2} \ge \sqrt{ab}$.
Let $a = 9 \tan^2 \theta$ and $b = 4 \cot^2 \theta$.
Then,$\frac{9 \tan^2 \theta + 4 \cot^2 \theta}{2} \ge \sqrt{9 \tan^2 \theta \cdot 4 \cot^2 \theta}$.
Since $\tan^2 \theta \cdot \cot^2 \theta = 1$,we have $\frac{9 \tan^2 \theta + 4 \cot^2 \theta}{2} \ge \sqrt{36}$.
$\frac{9 \tan^2 \theta + 4 \cot^2 \theta}{2} \ge 6$.
$9 \tan^2 \theta + 4 \cot^2 \theta \ge 12$.
Therefore,the minimum value is $12$.

(C) Given $\alpha + \beta + \gamma = \pi$.
For a triangle with angles $\alpha, \beta, \gamma$,the sum of sines is given by $\sin \alpha + \sin \beta + \sin \gamma = 4 \cos(\frac{\alpha}{2}) \cos(\frac{\beta}{2}) \cos(\frac{\gamma}{2})$.
Since $\alpha, \beta, \gamma$ are angles of a triangle,each angle must be in the interval $(0, \pi)$,which implies $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2} \in (0, \frac{\pi}{2})$.
In the interval $(0, \frac{\pi}{2})$,the cosine function is always positive.
Therefore,$4 \cos(\frac{\alpha}{2}) \cos(\frac{\beta}{2}) \cos(\frac{\gamma}{2}) > 0$.
Thus,the expression $\sin \alpha + \sin \beta + \sin \gamma$ is always positive.

(D) The expression is of the form $a\sin \theta + b\cos \theta$,where $a = 3$ and $b = 4$.
The range of the function $f(\theta) = a\sin \theta + b\cos \theta$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Substituting the values,we get the range as $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$.
Therefore,the minimum value is $-5$.

(A) The expression is of the form $a \sin x + b \cos x$,where $a = 1$ and $b = -1$.
The maximum value of the expression $a \sin x + b \cos x$ is given by $\sqrt{a^2 + b^2}$.
Substituting the values,we get $\sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,the maximum value is $\sqrt{2}$.

(D) Given $A = \cos^2 \theta + \sin^4 \theta$.
Since $\sin^2 \theta \le 1$,we have $A = \cos^2 \theta + \sin^2 \theta \cdot \sin^2 \theta \le \cos^2 \theta + \sin^2 \theta = 1$.
Thus,$A \le 1$.
Now,express $A$ in terms of $\sin^2 \theta$:
$A = (1 - \sin^2 \theta) + \sin^4 \theta = \sin^4 \theta - \sin^2 \theta + 1$.
Let $x = \sin^2 \theta$,where $0 \le x \le 1$. Then $A = x^2 - x + 1$.
Completing the square: $A = (x - 1/2)^2 + 3/4$.
Since $(x - 1/2)^2 \ge 0$,the minimum value of $A$ is $3/4$ when $x = 1/2$ (i.e.,$\sin^2 \theta = 1/2$).
Therefore,$3/4 \le A \le 1$.

(B) Given $A = \sin^2 \theta + \cos^4 \theta$.
Since $\cos^2 \theta \le 1$,we have $\cos^4 \theta \le \cos^2 \theta$.
Therefore,$A = \sin^2 \theta + \cos^4 \theta \le \sin^2 \theta + \cos^2 \theta = 1$.
Thus,$A \le 1$.
Now,express $A$ in terms of $\cos^2 \theta$:
$A = (1 - \cos^2 \theta) + \cos^4 \theta = \cos^4 \theta - \cos^2 \theta + 1$.
Let $x = \cos^2 \theta$,where $0 \le x \le 1$.
Then $A = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.
Since $(x - \frac{1}{2})^2 \ge 0$,the minimum value is $\frac{3}{4}$ when $x = \frac{1}{2}$.
Thus,$\frac{3}{4} \le A \le 1$.

(C) Let $f(x) = \sqrt{3} \sin x + \cos x$.
We can write this as $f(x) = 2 \left( \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \right)$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we have $f(x) = 2 \sin(x + 30^\circ)$.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = 90^\circ$.
Therefore,$x + 30^\circ = 90^\circ$.
$x = 90^\circ - 30^\circ = 60^\circ$.

(A) Given $\alpha + \beta - \gamma = \pi ,$ so $\gamma = \alpha + \beta - \pi .$
Consider the expression $E = \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma .$
Using the identity $\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B),$
$E = \sin^2 \alpha + \sin(\beta - \gamma)\sin(\beta + \gamma).$
Since $\beta - \gamma = \pi - \alpha,$ we have $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha.$
Also,$\beta + \gamma = \beta + (\alpha + \beta - \pi) = \alpha + 2\beta - \pi.$
Alternatively,using $\gamma = \alpha + \beta - \pi,$
$E = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta - \pi) = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta).$
Using $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B),$
$E = \sin^2 \alpha - \sin(\alpha + \beta - \beta)\sin(\alpha + \beta + \beta) = \sin^2 \alpha - \sin \alpha \sin(\alpha + 2\beta).$
$E = \sin \alpha [\sin \alpha - \sin(\alpha + 2\beta)].$
Using $\sin C - \sin D = 2\cos(\frac{C+D}{2})\sin(\frac{C-D}{2}),$
$E = \sin \alpha [2\cos(\alpha + \beta)\sin(-\beta)] = -2\sin \alpha \sin \beta \cos(\alpha + \beta).$
Since $\alpha + \beta = \pi + \gamma,$ $\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma.$
Therefore,$E = -2\sin \alpha \sin \beta (-\cos \gamma) = 2\sin \alpha \sin \beta \cos \gamma.$

(C) Given $A + B + C = \pi$.
Let the expression be $E = \frac{\cos A}{\sin B \sin C} + \frac{\cos B}{\sin C \sin A} + \frac{\cos C}{\sin A \sin B}$.
Taking the common denominator $\sin A \sin B \sin C$,we get:
$E = \frac{\cos A \sin A + \cos B \sin B + \cos C \sin C}{\sin A \sin B \sin C}$.
Multiplying numerator and denominator by $2$:
$E = \frac{2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C}{2 \sin A \sin B \sin C} = \frac{\sin 2A + \sin 2B + \sin 2C}{2 \sin A \sin B \sin C}$.
Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$ for $A + B + C = \pi$:
$E = \frac{4 \sin A \sin B \sin C}{2 \sin A \sin B \sin C} = 2$.

(B) In $\Delta ABC, A + B + C = 180^\circ.$
$\sin A + \sin B + \sin C = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$
Since $\frac{A+B}{2} = 90^\circ - \frac{C}{2}$,we have $\sin \frac{A+B}{2} = \cos \frac{C}{2}$.
Also,$\sin \frac{C}{2} = \cos \frac{A+B}{2}$.
Substituting these,we get:
$= 2\cos \frac{C}{2} \cos \frac{A-B}{2} + 2\cos \frac{C}{2} \cos \frac{A+B}{2}$
$= 2\cos \frac{C}{2} \left[ \cos \frac{A-B}{2} + \cos \frac{A+B}{2} \right]$
Using the identity $\cos(x-y) + \cos(x+y) = 2\cos x \cos y$:
$= 2\cos \frac{C}{2} \left( 2\cos \frac{A}{2} \cos \frac{B}{2} \right)$
$= 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

(B) Given $x + y + z = 180^\circ$,so $x + y = 180^\circ - z$.
Consider the expression $\cos 2x + \cos 2y - \cos 2z$.
Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$\cos 2x + \cos 2y = 2 \cos(x + y) \cos(x - y)$.
Since $x + y = 180^\circ - z$,$\cos(x + y) = \cos(180^\circ - z) = -\cos z$.
Using the identity $\cos 2z = 2 \cos^2 z - 1$:
$\cos 2x + \cos 2y - \cos 2z = 2(-\cos z) \cos(x - y) - (2 \cos^2 z - 1)$
$= 1 - 2 \cos z \cos(x - y) - 2 \cos^2 z$
$= 1 - 2 \cos z (\cos(x - y) + \cos z)$.
Since $z = 180^\circ - (x + y)$,$\cos z = \cos(180^\circ - (x + y)) = -\cos(x + y)$.
$= 1 - 2 \cos z (\cos(x - y) - \cos(x + y))$.
Using the identity $\cos(A - B) - \cos(A + B) = 2 \sin A \sin B$:
$= 1 - 2 \cos z (2 \sin x \sin y)$
$= 1 - 4 \sin x \sin y \cos z$.

(A) Given $\alpha + \beta + \gamma = 2\pi$.
Dividing by $2$,we get $\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$.
Taking tangent on both sides,$\tan(\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2}) = \tan(\pi) = 0$.
Using the identity $\tan(A+B+C) = \frac{\sum \tan A - \prod \tan A}{1 - \sum \tan A \tan B}$,we have:
$\frac{\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} - \tan \frac{\alpha}{2}\tan \frac{\beta}{2}\tan \frac{\gamma}{2}}{1 - (\tan \frac{\alpha}{2}\tan \frac{\beta}{2} + \tan \frac{\beta}{2}\tan \frac{\gamma}{2} + \tan \frac{\gamma}{2}\tan \frac{\alpha}{2})} = 0$.
Since the denominator is not zero,the numerator must be zero:
$\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} - \tan \frac{\alpha}{2}\tan \frac{\beta}{2}\tan \frac{\gamma}{2} = 0$.
Therefore,$\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} = \tan \frac{\alpha}{2}\tan \frac{\beta}{2}\tan \frac{\gamma}{2}$.

(C) Given $A + B + C = \pi$,so $A + B = \pi - C$.
$L.H.S. = \cos 2A + \cos 2B + \cos 2C$
$= 2 \cos (A + B) \cos (A - B) + (2 \cos^2 C - 1)$
$= 2 \cos (\pi - C) \cos (A - B) + 2 \cos^2 C - 1$
$= - 2 \cos C \cos (A - B) + 2 \cos^2 C - 1$
$= - 1 - 2 \cos C [\cos (A - B) - \cos C]$
Since $C = \pi - (A + B)$,$\cos C = - \cos (A + B)$.
$= - 1 - 2 \cos C [\cos (A - B) + \cos (A + B)]$
$= - 1 - 2 \cos C [2 \cos A \cos B]$
$= - 1 - 4 \cos A \cos B \cos C$.

(B) Given $A + B + C = 180^\circ$.
Numerator: $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C = 4(2 \sin \frac{A}{2} \cos \frac{A}{2})(2 \sin \frac{B}{2} \cos \frac{B}{2})(2 \sin \frac{C}{2} \cos \frac{C}{2}) = 32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}$.
Denominator: $\cos A + \cos B + \cos C - 1 = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Dividing the numerator by the denominator:
$\frac{32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

(C) In any triangle $ABC$,the identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ holds true.
Given $\tan A + \tan B + \tan C = 6$ and $\tan A \tan B = 2$.
Substituting these into the identity: $6 = 2 \tan C \Rightarrow \tan C = 3$.
Now,$\tan A + \tan B = 6 - \tan C = 6 - 3 = 3$.
We have the system $\tan A + \tan B = 3$ and $\tan A \tan B = 2$.
The quadratic equation with roots $\tan A$ and $\tan B$ is $x^2 - 3x + 2 = 0$.
Solving $(x - 1)(x - 2) = 0$,we get $x = 1$ or $x = 2$.
Thus,the values are $(\tan A, \tan B, \tan C) = (1, 2, 3)$ or $(2, 1, 3)$.

(B) Let $x = \tan \frac{A}{2}, y = \tan \frac{B}{2}, z = \tan \frac{C}{2}$.
Since $A + B + C = \pi$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$.
Using the identity $\tan(\frac{A}{2} + \frac{B}{2} + \frac{C}{2}) = \frac{x+y+z-xyz}{1-(xy+yz+zx)} = \tan \frac{\pi}{2} = \infty$.
This implies $1 - (xy + yz + zx) = 0$,so $xy + yz + zx = 1$.
We know the inequality $(x-y)^2 + (y-z)^2 + (z-x)^2 \ge 0$.
Expanding this,we get $2(x^2 + y^2 + z^2) - 2(xy + yz + zx) \ge 0$.
$x^2 + y^2 + z^2 \ge xy + yz + zx$.
Since $xy + yz + zx = 1$,we have $x^2 + y^2 + z^2 \ge 1$.
Thus,$\tan ^2 \frac{A}{2} + \tan ^2 \frac{B}{2} + \tan ^2 \frac{C}{2} \ge 1$.

(C) Given that $A + B + C = 180^\circ$.
We know that for a triangle or any three angles summing to $180^\circ$,the identity $\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$ holds true.
Therefore,the expression $\frac{\tan A + \tan B + \tan C}{\tan A \cdot \tan B \cdot \tan C} = \frac{\tan A \cdot \tan B \cdot \tan C}{\tan A \cdot \tan B \cdot \tan C} = 1$.

(D) Given that $\cos A = \cos B \cos C$ and $A + B + C = \pi.$
Since $A + B + C = \pi,$ we have $B + C = \pi - A.$
Taking cosine on both sides,$\cos(B + C) = \cos(\pi - A).$
Using the identity $\cos(B + C) = \cos B \cos C - \sin B \sin C$ and $\cos(\pi - A) = -\cos A,$ we get:
$\cos B \cos C - \sin B \sin C = -\cos A.$
Substitute $\cos A = \cos B \cos C$ into the equation:
$\cos B \cos C - \sin B \sin C = -\cos B \cos C.$
Rearranging the terms,we get:
$2 \cos B \cos C = \sin B \sin C.$
Dividing both sides by $\sin B \sin C,$ we obtain:
$\frac{\cos B \cos C}{\sin B \sin C} = \frac{1}{2}.$
Therefore,$\cot B \cot C = \frac{1}{2}.$

(B) Given $A + B + C = 180^o.$
We know that $\cot B + \cot C = \frac{\sin C \cos B + \sin B \cos C}{\sin B \sin C} = \frac{\sin(B + C)}{\sin B \sin C}.$
Since $B + C = 180^o - A,$ we have $\sin(B + C) = \sin(180^o - A) = \sin A.$
Thus,$\cot B + \cot C = \frac{\sin A}{\sin B \sin C}.$
Similarly,$\cot C + \cot A = \frac{\sin B}{\sin C \sin A}$ and $\cot A + \cot B = \frac{\sin C}{\sin A \sin B}.$
Multiplying these three expressions:
$(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B) = \left(\frac{\sin A}{\sin B \sin C}\right) \left(\frac{\sin B}{\sin C \sin A}\right) \left(\frac{\sin C}{\sin A \sin B}\right)$
$= \frac{\sin A \sin B \sin C}{(\sin A \sin B \sin C)^2} = \frac{1}{\sin A \sin B \sin C} = \csc A \csc B \csc C.$

(C) Given $A + B + C = 180^o$,we have $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.
Taking $\cot$ on both sides: $\cot(\frac{A}{2} + \frac{B}{2}) = \cot(90^o - \frac{C}{2})$.
Using the formula $\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$,we get $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = \tan \frac{C}{2}$.
Since $\tan \frac{C}{2} = \frac{1}{\cot \frac{C}{2}}$,we have $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = \frac{1}{\cot \frac{C}{2}}$.
Cross-multiplying gives $(\cot \frac{A}{2} \cot \frac{B}{2} - 1) \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.
Expanding this,we get $\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} - \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.
Rearranging the terms,we find $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$.

(B) Given $A + B + C = 270^o.$ Let $A = B = C = 90^o.$
Then,the expression becomes:
$\cos 2A + \cos 2B + \cos 2C + 4\sin A \sin B \sin C$
$= \cos 180^o + \cos 180^o + \cos 180^o + 4\sin 90^o \sin 90^o \sin 90^o$
$= (-1) + (-1) + (-1) + 4(1)(1)(1)$
$= -3 + 4 = 1.$

(B) Given $A + B + C = 180^o$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^o$.
Therefore,$\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.
Taking tangent on both sides: $\tan(\frac{A}{2} + \frac{B}{2}) = \tan(90^o - \frac{C}{2}) = \cot \frac{C}{2}$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get:
$\frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \frac{1}{\tan \frac{C}{2}}$.
Cross-multiplying: $\tan \frac{C}{2} (\tan \frac{A}{2} + \tan \frac{B}{2}) = 1 - \tan \frac{A}{2} \tan \frac{B}{2}$.
Rearranging the terms: $\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1$.
Thus,$\sum \tan \frac{A}{2} \tan \frac{B}{2} = 1$.

(B) Given $A + B + C = \pi$,so $A + B = \pi - C$.
Taking tangent on both sides: $\tan(A + B) = \tan(\pi - C)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$.
Since $C$ is an obtuse angle $(90^{\circ} < C < 180^{\circ})$,$\tan C < 0$,which implies $-\tan C > 0$.
Therefore,$\frac{\tan A + \tan B}{1 - \tan A \tan B} > 0$.
Since $A, B, C$ are angles of a triangle and $C$ is obtuse,$A$ and $B$ must be acute,so $\tan A > 0$ and $\tan B > 0$,implying $\tan A + \tan B > 0$.
For the fraction to be positive,the denominator must be positive: $1 - \tan A \tan B > 0$.
Thus,$\tan A \tan B < 1$.

(A) Given $A + B + C = \pi$,we know that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Since $A, B, C$ are acute,$\tan A, \tan B, \tan C > 0$. By the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality:
$\frac{\tan A + \tan B + \tan C}{3} \ge (\tan A \tan B \tan C)^{1/3}$
Substituting $\tan A + \tan B + \tan C = \tan A \tan B \tan C$:
$\frac{\tan A \tan B \tan C}{3} \ge (\tan A \tan B \tan C)^{1/3}$
Let $P = \tan A \tan B \tan C$. Then $\frac{P}{3} \ge P^{1/3}$ $\Rightarrow P^{2/3} \ge 3$ $\Rightarrow P \ge 3^{3/2} = 3\sqrt{3}$.
Since $K = \cot A \cot B \cot C = \frac{1}{\tan A \tan B \tan C} = \frac{1}{P}$,we have:
$K = \frac{1}{P} \le \frac{1}{3\sqrt{3}}$.

(D) Given: $A + B + C = \frac{3\pi}{2} \implies A + B = \frac{3\pi}{2} - C$.
Consider the expression: $\cos 2A + \cos 2B + \cos 2C$.
Using the formula $\cos X + \cos Y = 2\cos\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)$:
$\cos 2A + \cos 2B = 2\cos(A+B)\cos(A-B)$.
Substitute $A+B = \frac{3\pi}{2} - C$:
$2\cos\left(\frac{3\pi}{2} - C\right)\cos(A-B) = 2(-\sin C)\cos(A-B) = -2\sin C \cos(A-B)$.
Now,$\cos 2C = 1 - 2\sin^2 C$.
So,$\cos 2A + \cos 2B + \cos 2C = 1 - 2\sin C \cos(A-B) - 2\sin^2 C$.
$= 1 - 2\sin C [\cos(A-B) + \sin C]$.
Since $C = \frac{3\pi}{2} - (A+B)$,$\sin C = \sin\left(\frac{3\pi}{2} - (A+B)\right) = -\cos(A+B)$.
$= 1 - 2\sin C [\cos(A-B) - \cos(A+B)]$.
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) The given function is $f(x) = \sin x + \cos x$.
We can rewrite this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$.
Using the identity $\sin(x + \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}$,we get $f(x) = \sqrt{2} \sin(x + \frac{\pi}{4})$.
Since the maximum value of $\sin \theta$ is $1$,the maximum value of $f(x)$ is $\sqrt{2} \times 1 = \sqrt{2}$.

(B) The function is given by $f(x) = \sqrt{3} \sin x + \cos x$.
To find the maximum distance from the $x$-axis,we need to find the maximum absolute value of the function,which is $|f(x)|_{max}$.
For a function of the form $f(x) = a \sin x + b \cos x$,the maximum value is $\sqrt{a^2 + b^2}$ and the minimum value is $-\sqrt{a^2 + b^2}$.
Here,$a = \sqrt{3}$ and $b = 1$.
Maximum value $= \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Minimum value $= -\sqrt{(\sqrt{3})^2 + (1)^2} = -2$.
The maximum distance from the $x$-axis is the maximum value of $|f(x)|$,which is $|2| = 2$ or $|-2| = 2$.
Therefore,the maximum distance is $2$.

(C) The function is of the form $f(x) = a \sin x + b \cos x$.
The maximum value of this function is given by $\sqrt{a^2 + b^2}$.
Here,$a = 3$ and $b = 4$.
Therefore,the maximum value is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(B) Let $f(\theta) = \sin \theta + \cos \theta = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta \right) = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right)$.
Since the maximum value of $\sin(\alpha)$ is $1$,the function $f(\theta)$ attains its maximum value when $\sin \left( \theta + \frac{\pi}{4} \right) = 1$.
This occurs when $\theta + \frac{\pi}{4} = \frac{\pi}{2}$.
Solving for $\theta$,we get $\theta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
Converting to degrees,$\theta = 45^o$.

(D) We know that for any real number $a$,$(a - \frac{1}{a})^2 \ge 0$.
This implies $a^2 + \frac{1}{a^2} - 2 \ge 0$,or $a^2 + \frac{1}{a^2} \ge 2$.
Let $a = \cos x$. Since $\sec x = \frac{1}{\cos x}$,we have $f(x) = \cos^2 x + \frac{1}{\cos^2 x}$.
By the Arithmetic Mean-Geometric Mean inequality ($AM$ $\ge$ $GM$),for positive real numbers $u$ and $v$,$\frac{u+v}{2} \ge \sqrt{uv}$.
Here,$f(x) = \cos^2 x + \sec^2 x \ge 2 \sqrt{\cos^2 x \cdot \sec^2 x} = 2 \sqrt{1} = 2$.
Thus,$f(x) \ge 2$.

(C) Let $f(\theta) = \cos 2\theta + 2\cos \theta$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$f(\theta) = 2\cos^2 \theta - 1 + 2\cos \theta$
$f(\theta) = 2(\cos^2 \theta + \cos \theta) - 1$
Completing the square:
$f(\theta) = 2(\cos^2 \theta + \cos \theta + \frac{1}{4}) - 1 - \frac{1}{2}$
$f(\theta) = 2(\cos \theta + \frac{1}{2})^2 - \frac{3}{2}$
Since $-1 \le \cos \theta \le 1$,the minimum value occurs when $\cos \theta = -\frac{1}{2}$,giving $f(\theta) = -\frac{3}{2}$.
The maximum value occurs at the boundaries of $\cos \theta$,i.e.,$\cos \theta = 1$,giving $f(\theta) = 2(1 + \frac{1}{2})^2 - \frac{3}{2} = 2(\frac{9}{4}) - \frac{3}{2} = \frac{9}{2} - \frac{3}{2} = 3$.
Thus,$-\frac{3}{2} \le \cos 2\theta + 2\cos \theta \le 3$.

(B) Given $A + B + C = \pi,$ so $A = \pi - (B + C).$
Taking cosine on both sides,$\cos A = \cos(\pi - (B + C)).$
Since $\cos(\pi - \theta) = -\cos \theta,$ we have $\cos A = -\cos(B + C).$
Given $\cos A = \cos B \cos C,$ we substitute this into the equation:
$-\cos(B + C) = \cos B \cos C.$
Using the expansion formula $\cos(B + C) = \cos B \cos C - \sin B \sin C,$
$-(\cos B \cos C - \sin B \sin C) = \cos B \cos C.$
$-\cos B \cos C + \sin B \sin C = \cos B \cos C.$
$\sin B \sin C = 2 \cos B \cos C.$
Dividing both sides by $\cos B \cos C$ (assuming $\cos B \neq 0$ and $\cos C \neq 0$),
$\frac{\sin B \sin C}{\cos B \cos C} = 2.$
$\tan B \tan C = 2.$

(B) Let $u = \cos \theta \left\{ \sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha} \right\}$.
Rearranging the terms,we get $u - \sin \theta \cos \theta = \cos \theta \sqrt{\sin^2 \theta + \sin^2 \alpha}$.
Squaring both sides:
$(u - \sin \theta \cos \theta)^2 = \cos^2 \theta (\sin^2 \theta + \sin^2 \alpha)$.
$u^2 - 2u \sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta = \sin^2 \theta \cos^2 \theta + \cos^2 \theta \sin^2 \alpha$.
$u^2 - 2u \sin \theta \cos \theta = \cos^2 \theta \sin^2 \alpha$.
Dividing by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$):
$u^2 \sec^2 \theta - 2u \tan \theta = \sin^2 \alpha$.
Using $\sec^2 \theta = 1 + \tan^2 \theta$:
$u^2(1 + \tan^2 \theta) - 2u \tan \theta - \sin^2 \alpha = 0$.
$u^2 \tan^2 \theta - 2u \tan \theta + (u^2 - \sin^2 \alpha) = 0$.
Since $\tan \theta$ is real,the discriminant $D \ge 0$:
$(-2u)^2 - 4(u^2)(u^2 - \sin^2 \alpha) \ge 0$.
$4u^2 - 4u^4 + 4u^2 \sin^2 \alpha \ge 0$.
Dividing by $4u^2$ (assuming $u \neq 0$):
$1 - u^2 + \sin^2 \alpha \ge 0$.
$u^2 \le 1 + \sin^2 \alpha$.
Therefore,$|u| \le \sqrt{1 + \sin^2 \alpha}$.
Thus,$k = \sqrt{1 + \sin^2 \alpha}$.