Trigonometry Questions in English

(A) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean ($AM$ $\ge$ $GM$).
Applying this to $\sqrt{\frac{x}{y}}$ and $\sqrt{\frac{y}{x}}$,we get:
$\frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right) \ge \sqrt{\sqrt{\frac{x}{y}} \cdot \sqrt{\frac{y}{x}}}$
$\Rightarrow \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right) \ge \sqrt{1} = 1$
Since $\sin \theta$ cannot exceed $1$,the expression must be equal to $1$.
$\sin \theta = 1 \Rightarrow \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right) = 1$
This equality holds if and only if $\sqrt{\frac{x}{y}} = \sqrt{\frac{y}{x}}$,which implies $\frac{x}{y} = 1$,or $x = y$.

(C) Given $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$.
From $\alpha + \beta = \pi$,we have $\alpha = \pi - \beta$,so $\tan \alpha = \tan(\pi - \beta) = -\tan \beta$.
Since $\alpha, \beta, \gamma$ are positive and $\alpha + \beta = \pi$,$\alpha$ must be in the second quadrant (as $\beta > 0$),so $\tan \alpha < 0$.
From $\beta + \gamma = \alpha$,we have $\tan \alpha = \tan(\beta + \gamma) = \frac{\tan \beta + \tan \gamma}{1 - \tan \beta \tan \gamma}$.
Substituting $\tan \beta = -\tan \alpha$ into the equation:
$\tan \alpha = \frac{-\tan \alpha + \tan \gamma}{1 - (-\tan \alpha) \tan \gamma} = \frac{-\tan \alpha + \tan \gamma}{1 + \tan \alpha \tan \gamma}$.
$\tan \alpha (1 + \tan \alpha \tan \gamma) = -\tan \alpha + \tan \gamma$.
$\tan \alpha + \tan^2 \alpha \tan \gamma = -\tan \alpha + \tan \gamma$.
$\tan^2 \alpha \tan \gamma = -2 \tan \alpha + \tan \gamma$.
This is a quadratic in $\tan \alpha$: $(\tan \gamma) \tan^2 \alpha + (2) \tan \alpha - \tan \gamma = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\tan \alpha = \frac{-2 \pm \sqrt{4 - 4(\tan \gamma)(-\tan \gamma)}}{2 \tan \gamma} = \frac{-2 \pm \sqrt{4 + 4 \tan^2 \gamma}}{2 \tan \gamma} = \frac{-2 \pm 2 \sqrt{1 + \tan^2 \gamma}}{2 \tan \gamma} = \frac{-1 \pm \sec \gamma}{\tan \gamma}$.
Alternatively,from $\tan^2 \alpha \tan \gamma = -2 \tan \alpha + \tan \gamma$,we get $\tan^2 \alpha = \frac{-2 \tan \alpha + \tan \gamma}{\tan \gamma}$. Since $\tan \alpha = -\tan \beta$,$\tan^2 \alpha = \frac{2 \tan \beta + \tan \gamma}{\tan \gamma}$.
Thus,$\tan \alpha = -\sqrt{\frac{2 \tan \beta + \tan \gamma}{\tan \gamma}}$ (taking the negative root because $\tan \alpha < 0$).

(D) We know that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Using these,$\tan \left( \frac{\pi}{4} + \theta \right) = \frac{1 + \tan \theta}{1 - \tan \theta}$ and $\tan \left( \frac{\pi}{4} - \theta \right) = \frac{1 - \tan \theta}{1 + \tan \theta}$.
Adding these two expressions:
$\frac{1 + \tan \theta}{1 - \tan \theta} + \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{(1 + \tan \theta)^2 + (1 - \tan \theta)^2}{(1 - \tan \theta)(1 + \tan \theta)}$
$= \frac{1 + \tan^2 \theta + 2 \tan \theta + 1 + \tan^2 \theta - 2 \tan \theta}{1 - \tan^2 \theta}$
$= \frac{2(1 + \tan^2 \theta)}{1 - \tan^2 \theta}$
Since $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$,we have $\sec 2\theta = \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta}$.
Thus,the expression becomes $2 \sec 2\theta$.
Comparing this with $\lambda \sec 2\theta$,we get $\lambda = 2$.

(A) We use the trigonometric identity $\cot \theta - \tan \theta = 2 \cot 2\theta$.
Step $1$: Simplify the first two terms: $\cot 5^o - \tan 5^o = 2 \cot 10^o$.
Step $2$: Now the expression becomes $2 \cot 10^o - 2 \tan 10^o - 4 \tan 20^o - 8 \cot 40^o$.
Step $3$: Factor out $2$: $2(\cot 10^o - \tan 10^o) - 4 \tan 20^o - 8 \cot 40^o = 2(2 \cot 20^o) - 4 \tan 20^o - 8 \cot 40^o = 4 \cot 20^o - 4 \tan 20^o - 8 \cot 40^o$.
Step $4$: Factor out $4$: $4(\cot 20^o - \tan 20^o) - 8 \cot 40^o = 4(2 \cot 40^o) - 8 \cot 40^o = 8 \cot 40^o - 8 \cot 40^o = 0$.

(B) Given expression is $E = \frac{25 \sec^4 x - 50 \sec^2 x + 74}{\tan^2 x}$.
We know that $\sec^2 x = 1 + \tan^2 x$.
Substituting this,we get $E = \frac{25(1 + \tan^2 x)^2 - 50(1 + \tan^2 x) + 74}{\tan^2 x}$.
Let $t = \tan^2 x$. Then $E = \frac{25(1 + t)^2 - 50(1 + t) + 74}{t}$.
$E = \frac{25(1 + 2t + t^2) - 50 - 50t + 74}{t} = \frac{25 + 50t + 25t^2 - 50 - 50t + 74}{t}$.
$E = \frac{25t^2 + 49}{t} = 25t + \frac{49}{t}$.
Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,we have $\frac{25t + \frac{49}{t}}{2} \ge \sqrt{25t \cdot \frac{49}{t}}$.
$25t + \frac{49}{t} \ge 2 \sqrt{25 \cdot 49} = 2 \cdot 5 \cdot 7 = 70$.
Thus,the minimum value is $70$.

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

(B) We know that for any real numbers $x$ and $y$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{x+y}{2} \geq \sqrt{xy}$.
Squaring both sides,we get $\frac{(x+y)^2}{4} \geq xy$,which implies $\frac{4xy}{(x+y)^2} \leq 1$.
Since $cosec^2 \theta \geq 1$ for all real $\theta$,the equation $cosec^2 \theta = \frac{4xy}{(x+y)^2}$ can only hold if both sides are equal to $1$.
This requires $\frac{4xy}{(x+y)^2} = 1$,which simplifies to $4xy = (x+y)^2$,or $(x-y)^2 = 0$,meaning $x = y$.
Additionally,since $cosec^2 \theta$ is undefined if the denominator is zero,we must have $x+y \neq 0$. Given $x=y$,this implies $2x \neq 0$,so $x \neq 0$.
Thus,the condition is $x = y$ and $x \neq 0$.

(C) Given: $\tan 80^{\circ} = a$ and $\tan 47^{\circ} = b$.
We know that $\tan 80^{\circ} = \cot 10^{\circ} = \frac{1}{\tan 10^{\circ}}$,therefore $\tan 10^{\circ} = \frac{1}{a}$.
We need to find $\tan 37^{\circ}$.
We can write $\tan 37^{\circ} = \tan(47^{\circ} - 10^{\circ})$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$\tan 37^{\circ} = \frac{\tan 47^{\circ} - \tan 10^{\circ}}{1 + \tan 47^{\circ} \tan 10^{\circ}}$.
Substituting the values,we have:
$\tan 37^{\circ} = \frac{b - \frac{1}{a}}{1 + b \cdot \frac{1}{a}}$.
Multiplying the numerator and denominator by $a$,we get:
$\tan 37^{\circ} = \frac{ab - 1}{a + b}$.

(D) Given expression: $E = \sin \theta + \cos \theta + \sin 2\theta$.
We know that $\sin 2\theta = 2 \sin \theta \cos \theta$.
Let $t = \sin \theta + \cos \theta$. Squaring both sides,we get $t^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + \sin 2\theta$.
Thus,$\sin 2\theta = t^2 - 1$.
The range of $t = \sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$ is $[-\sqrt{2}, \sqrt{2}]$.
Substituting $t$ into the expression: $E = t + t^2 - 1 = t^2 + t - 1$.
To find the maximum value of $E$ on the interval $[-\sqrt{2}, \sqrt{2}]$,we evaluate the quadratic $f(t) = t^2 + t - 1$ at the boundaries.
At $t = \sqrt{2}$,$E = (\sqrt{2})^2 + \sqrt{2} - 1 = 2 + \sqrt{2} - 1 = 1 + \sqrt{2}$.
At $t = -\sqrt{2}$,$E = (-\sqrt{2})^2 - \sqrt{2} - 1 = 2 - \sqrt{2} - 1 = 1 - \sqrt{2}$.
Since $1 + \sqrt{2} \approx 2.414$ and $\tan(\frac{3\pi}{8}) = \tan(67.5^\circ) = \sqrt{2} + 1$,the maximum value is $1 + \sqrt{2}$,which corresponds to $\tan \frac{3\pi}{8}$.

(B) We use the identity $\cot \alpha - \tan \alpha = 2\cot 2\alpha$,which implies $\tan \alpha = \cot \alpha - 2\cot 2\alpha$.
Given expression: $E = \tan 3^{\circ} + 2\tan 6^{\circ} + 4\tan 12^{\circ} + 8\cot 24^{\circ}$.
Using $\tan 3^{\circ} = \cot 3^{\circ} - 2\cot 6^{\circ}$:
$E = (\cot 3^{\circ} - 2\cot 6^{\circ}) + 2\tan 6^{\circ} + 4\tan 12^{\circ} + 8\cot 24^{\circ}$.
Using $2\tan 6^{\circ} = 2(\cot 6^{\circ} - 2\cot 12^{\circ}) = 2\cot 6^{\circ} - 4\cot 12^{\circ}$:
$E = \cot 3^{\circ} - 2\cot 6^{\circ} + 2\cot 6^{\circ} - 4\cot 12^{\circ} + 4\tan 12^{\circ} + 8\cot 24^{\circ}$.
Simplifying:
$E = \cot 3^{\circ} - 4\cot 12^{\circ} + 4\tan 12^{\circ} + 8\cot 24^{\circ}$.
Using $4\tan 12^{\circ} = 4(\cot 12^{\circ} - 2\cot 24^{\circ}) = 4\cot 12^{\circ} - 8\cot 24^{\circ}$:
$E = \cot 3^{\circ} - 4\cot 12^{\circ} + 4\cot 12^{\circ} - 8\cot 24^{\circ} + 8\cot 24^{\circ} = \cot 3^{\circ}$.
Thus,$\cot \theta^{\circ} = \cot 3^{\circ}$,so $\theta = 3$.
Checking options: $\cot (15 \times 3)^{\circ} = \cot 45^{\circ} = 1$.

(D) Let $f(x) = \tan(e^x)$ and $g(x) = e^x + e^{-x}$.
We are looking for the number of intersection points of $y = \tan(e^x)$ and $y = e^x + e^{-x}$ for $x > 0$.
Note that $e^x + e^{-x} = 2 \cosh(x)$,which is always $\ge 2$ for all real $x$.
As $x$ increases from $0$ to $\infty$,$e^x$ increases from $1$ to $\infty$.
The function $\tan(e^x)$ is periodic with respect to $e^x$ and has vertical asymptotes at $e^x = (2n+1)\frac{\pi}{2}$ for $n = 0, 1, 2, \dots$.
Since $e^x$ takes all values in $(1, \infty)$ as $x$ ranges from $(0, \infty)$,the function $\tan(e^x)$ will oscillate between $-\infty$ and $+\infty$ infinitely many times.
Specifically,in every interval where $e^x$ increases by $\pi$,$\tan(e^x)$ covers the range $(-\infty, \infty)$.
Since $e^x + e^{-x} \ge 2$,the graph of $y = e^x + e^{-x}$ will intersect each branch of the tangent function $\tan(e^x)$ at least once as long as the branch reaches values $\ge 2$.
Because there are infinitely many such branches as $x \to \infty$,there are infinitely many solutions.
Therefore,the correct answer is 'none of these'.

(B) Given $y = \frac{7 + 6 \tan x - \tan^2 x}{1 + \tan^2 x}$.
Since $1 + \tan^2 x = \sec^2 x$,we have $y = (7 + 6 \tan x - \tan^2 x) \cos^2 x$.
$y = 7 \cos^2 x + 6 \sin x \cos x - \sin^2 x$.
Using trigonometric identities $\cos^2 x = \frac{1 + \cos 2x}{2}$,$\sin^2 x = \frac{1 - \cos 2x}{2}$,and $\sin x \cos x = \frac{\sin 2x}{2}$:
$y = 7 \left( \frac{1 + \cos 2x}{2} \right) + 6 \left( \frac{\sin 2x}{2} \right) - \left( \frac{1 - \cos 2x}{2} \right)$.
$y = \frac{7 + 7 \cos 2x + 6 \sin 2x - 1 + \cos 2x}{2} = \frac{6 + 8 \cos 2x + 6 \sin 2x}{2} = 3 + 4 \cos 2x + 3 \sin 2x$.
The expression $a \sin \theta + b \cos \theta$ has a maximum value of $\sqrt{a^2 + b^2}$.
Here,$3 \sin 2x + 4 \cos 2x$ has a maximum value of $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
Thus,the maximum value of $y$ is $\lambda = 3 + 5 = 8$.
We need to find $\log_{\sqrt{2}}(\lambda) = \log_{\sqrt{2}}(8)$.
Since $8 = 2^3 = (\sqrt{2})^6$,we have $\log_{\sqrt{2}}((\sqrt{2})^6) = 6$.

(C) Let $3\cos \theta + 4\sin \theta = 5$ be equation $(1)$ and $3\sin \theta - 4\cos \theta = x$ be equation $(2)$.
Squaring both equations and adding them:
$(3\cos \theta + 4\sin \theta)^2 + (3\sin \theta - 4\cos \theta)^2 = 5^2 + x^2$
$(9\cos^2 \theta + 16\sin^2 \theta + 24\sin \theta \cos \theta) + (9\sin^2 \theta + 16\cos^2 \theta - 24\sin \theta \cos \theta) = 25 + x^2$
$9(\cos^2 \theta + \sin^2 \theta) + 16(\sin^2 \theta + \cos^2 \theta) = 25 + x^2$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$9(1) + 16(1) = 25 + x^2$
$25 = 25 + x^2$
$x^2 = 0$
Therefore,$x = 0$.

(B) Given equations are:
$1) \cos A + \cos B = \cos C$
$2) \sin A + \sin B = \sin C$
Let $z_1 = e^{iA}$,$z_2 = e^{iB}$,and $z_3 = e^{iC}$.
Adding the two equations: $(\cos A + \cos B) + i(\sin A + \sin B) = \cos C + i\sin C$
This implies $e^{iA} + e^{iB} = e^{iC}$.
Taking the conjugate of the equations:
$\cos A + \cos B = \cos C$
$-\sin A - \sin B = -\sin C$
Adding these gives $e^{-iA} + e^{-iB} = e^{-iC}$.
From $e^{iA} + e^{iB} = e^{iC}$,we have $e^{iA} + e^{iB} = e^{iC}$.
Dividing the two conjugate forms:
$\frac{e^{iA} + e^{iB}}{e^{-iA} + e^{-iB}} = \frac{e^{iC}}{e^{-iC}}$
$\frac{e^{iA} + e^{iB}}{e^{-i(A+B)}(e^{iB} + e^{iA})} = e^{2iC}$
$e^{i(A+B)} = e^{2iC}$
Equating the imaginary parts:
$\sin(A + B) = \sin 2C$
Therefore,$\frac{\sin(A + B)}{\sin 2C} = 1$.

(A) Given $3 \tan A - 4 = 0$,so $\tan A = \frac{4}{3}$.
Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.
Using $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}$,we find the hypotenuse $r = \sqrt{3^2 + 4^2} = 5$.
Thus,$\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.
Now,substitute these values into the expression $5 \sin 2A + 3 \sin A + 4 \cos A$:
$5(2 \sin A \cos A) + 3 \sin A + 4 \cos A$
$= 10 \left(-\frac{4}{5}\right) \left(-\frac{3}{5}\right) + 3 \left(-\frac{4}{5}\right) + 4 \left(-\frac{3}{5}\right)$
$= 10 \left(\frac{12}{25}\right) - \frac{12}{5} - \frac{12}{5}$
$= \frac{120}{25} - \frac{24}{5} = \frac{24}{5} - \frac{24}{5} = 0$.

(C) For $f(x)$ to be an even function, $f(x) = f(-x)$ must hold for all $x$ in the domain.
Given $f(x) = Ax^3 - Bx - \tan x \cdot \text{sgn}(x)$.
Since $\tan(-x) = -\tan x$ and $\text{sgn}(-x) = -\text{sgn}(x)$, we have $\tan(-x) \cdot \text{sgn}(-x) = (-\tan x) \cdot (-\text{sgn}(x)) = \tan x \cdot \text{sgn}(x)$.
Thus, $f(-x) = A(-x)^3 - B(-x) - \tan(-x) \cdot \text{sgn}(-x) = -Ax^3 + Bx - \tan x \cdot \text{sgn}(x)$.
For $f(x) = f(-x)$, we require $Ax^3 - Bx - \tan x \cdot \text{sgn}(x) = -Ax^3 + Bx - \tan x \cdot \text{sgn}(x)$.
This simplifies to $2Ax^3 - 2Bx = 0$ for all $x$, which implies $A = 0$ and $B = 0$ simultaneously.
$A = \sin^2 \alpha - \sin \alpha + \frac{1}{4} = (\sin \alpha - \frac{1}{2})^2 = 0 \implies \sin \alpha = \frac{1}{2}$.
$B = \tan^2 \alpha + \frac{2}{\sqrt{3}} \tan \alpha + \frac{1}{3} = (\tan \alpha + \frac{1}{\sqrt{3}})^2 = 0 \implies \tan \alpha = -\frac{1}{\sqrt{3}}$.
We need $\alpha \in [-\frac{3\pi}{2}, 2\pi]$ such that $\sin \alpha = \frac{1}{2}$ and $\tan \alpha = -\frac{1}{\sqrt{3}}$.
From $\sin \alpha = \frac{1}{2}$, $\alpha = \frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}, -\frac{11\pi}{6}$.
From $\tan \alpha = -\frac{1}{\sqrt{3}}$, $\alpha = \frac{5\pi}{6}, -\frac{\pi}{6}, \frac{11\pi}{6}, -\frac{7\pi}{6}$.
The common values are $\alpha = \frac{5\pi}{6}$ and $\alpha = -\frac{7\pi}{6}$.
Thus, there are $2$ such values.

(C) Let $E = \log_{\frac{1}{8}\csc^2 \frac{\pi}{8}} \sin^2 \frac{3\pi}{8}$.
First,simplify the base: $\csc^2 \frac{\pi}{8} = \frac{1}{\sin^2 \frac{\pi}{8}} = \frac{1}{\frac{1-\cos(\pi/4)}{2}} = \frac{2}{1-\frac{1}{\sqrt{2}}} = \frac{2\sqrt{2}}{\sqrt{2}-1} = 2\sqrt{2}(\sqrt{2}+1) = 4+2\sqrt{2}$.
So,the base is $\frac{1}{8}(4+2\sqrt{2}) = \frac{2+\sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{2}+1)}{4} = \frac{\sqrt{2}+1}{2\sqrt{2}}$.
Next,simplify the argument: $\sin^2 \frac{3\pi}{8} = \frac{1-\cos(3\pi/4)}{2} = \frac{1-(-\frac{1}{\sqrt{2}})}{2} = \frac{1+\frac{1}{\sqrt{2}}}{2} = \frac{\sqrt{2}+1}{2\sqrt{2}}$.
Therefore,$E = \log_{\frac{\sqrt{2}+1}{2\sqrt{2}}} \left( \frac{\sqrt{2}+1}{2\sqrt{2}} \right) = 1$.

(C) Let the expression be $E = \frac{4 \sin 9^{\circ} \sin 21^{\circ} \sin 39^{\circ} \sin 51^{\circ} \sin 69^{\circ} \sin 81^{\circ}}{\sin 54^{\circ}}$.
Using the identity $\sin \theta \sin(60^{\circ}-\theta) \sin(60^{\circ}+\theta) = \frac{1}{4} \sin 3\theta$,we group the terms:
$E = \frac{4}{\sin 54^{\circ}} [(\sin 9^{\circ} \sin 51^{\circ} \sin 69^{\circ}) \cdot (\sin 21^{\circ} \sin 39^{\circ} \sin 81^{\circ})]$.
Note that $\sin 51^{\circ} = \sin(60^{\circ}-9^{\circ})$ and $\sin 69^{\circ} = \sin(60^{\circ}+9^{\circ})$,so $\sin 9^{\circ} \sin 51^{\circ} \sin 69^{\circ} = \frac{1}{4} \sin(3 \times 9^{\circ}) = \frac{1}{4} \sin 27^{\circ}$.
Similarly,$\sin 21^{\circ} \sin 39^{\circ} \sin 81^{\circ} = \sin 21^{\circ} \sin(60^{\circ}-21^{\circ}) \sin(60^{\circ}+21^{\circ}) = \frac{1}{4} \sin(3 \times 21^{\circ}) = \frac{1}{4} \sin 63^{\circ}$.
Substituting these back into $E$:
$E = \frac{4}{\sin 54^{\circ}} \cdot \frac{1}{4} \sin 27^{\circ} \cdot \frac{1}{4} \sin 63^{\circ} = \frac{\sin 27^{\circ} \sin 63^{\circ}}{4 \sin 54^{\circ}}$.
Since $\sin 63^{\circ} = \cos 27^{\circ}$,we have $E = \frac{\sin 27^{\circ} \cos 27^{\circ}}{4 \sin 54^{\circ}} = \frac{2 \sin 27^{\circ} \cos 27^{\circ}}{8 \sin 54^{\circ}} = \frac{\sin 54^{\circ}}{8 \sin 54^{\circ}} = \frac{1}{8}$.

(D) Given $2\cos^2 \theta + \sin \theta \le 2$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get $2(1 - \sin^2 \theta) + \sin \theta \le 2$.
$2 - 2\sin^2 \theta + \sin \theta \le 2$.
$-2\sin^2 \theta + \sin \theta \le 0$.
Multiplying by $-1$,we get $2\sin^2 \theta - \sin \theta \ge 0$.
$\sin \theta(2\sin \theta - 1) \ge 0$.
This inequality holds when $\sin \theta \le 0$ or $\sin \theta \ge \frac{1}{2}$.
For $\sin \theta \le 0$,$\theta \in [\pi, 2\pi]$.
For $\sin \theta \ge \frac{1}{2}$,$\theta \in [\frac{\pi}{6}, \frac{5\pi}{6}]$.
Thus,$A = [\frac{\pi}{6}, \frac{5\pi}{6}] \cup [\pi, 2\pi]$.
Given $B = [\frac{\pi}{2}, \frac{3\pi}{2}]$.
Taking the intersection $A \cap B$:
$A \cap B = ([\frac{\pi}{6}, \frac{5\pi}{6}] \cup [\pi, 2\pi]) \cap [\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A \cap B = [\frac{\pi}{2}, \frac{5\pi}{6}] \cup [\pi, \frac{3\pi}{2}]$.

(D) Given expression: $E = \frac{\sin 81^\circ + \cos 81^\circ}{\sin 81^\circ - \cos 81^\circ}$
Divide the numerator and denominator by $\cos 81^\circ$:
$E = \frac{\tan 81^\circ + 1}{\tan 81^\circ - 1}$
Since $\tan 45^\circ = 1$,we can write:
$E = \frac{\tan 81^\circ + \tan 45^\circ}{\tan 81^\circ - \tan 45^\circ}$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,this does not directly match. Let us rewrite the expression as:
$E = -\frac{1 + \tan 81^\circ}{1 - \tan 81^\circ} = -\tan(45^\circ + 81^\circ) = -\tan(126^\circ)$
Alternatively,using $\sin 81^\circ = \cos 9^\circ$ and $\cos 81^\circ = \sin 9^\circ$:
$E = \frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ} = \frac{1 + \tan 9^\circ}{1 - \tan 9^\circ} = \tan(45^\circ + 9^\circ) = \tan 54^\circ$

(A) Let $S = \cos 2\theta + \cos \theta$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$S = 2\cos^2 \theta - 1 + \cos \theta$.
To find the minimum value,we complete the square for the quadratic expression in $\cos \theta$:
$S = 2(\cos^2 \theta + \frac{1}{2}\cos \theta) - 1$.
$S = 2(\cos^2 \theta + \frac{1}{2}\cos \theta + \frac{1}{16} - \frac{1}{16}) - 1$.
$S = 2(\cos \theta + \frac{1}{4})^2 - \frac{1}{8} - 1$.
$S = 2(\cos \theta + \frac{1}{4})^2 - \frac{9}{8}$.
Since the square term $2(\cos \theta + \frac{1}{4})^2 \geq 0$,the minimum value of $S$ is attained when $\cos \theta = -1/4$.
Thus,the minimum value is $-9/8$.

(A) Given that $\sin \alpha$ and $\cos \alpha$ are roots of $ax^2 + bx + c = 0$.
From the properties of quadratic equations,the sum of roots is $\sin \alpha + \cos \alpha = -\frac{b}{a}$ and the product of roots is $\sin \alpha \cos \alpha = \frac{c}{a}$.
We know the identity $\sin^2 \alpha + \cos^2 \alpha = 1$.
This can be written as $(\sin \alpha + \cos \alpha)^2 - 2 \sin \alpha \cos \alpha = 1$.
Substituting the values,we get $(-\frac{b}{a})^2 - 2(\frac{c}{a}) = 1$.
$\frac{b^2}{a^2} - \frac{2c}{a} = 1$.
Multiplying by $a^2$,we get $b^2 - 2ac = a^2$.
Rearranging the terms,we get $a^2 - b^2 + 2ac = 0$.

(A) We know that $A = \sin 45^{\circ} + \cos 45^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Alternatively,$A = \sqrt{2} (\frac{1}{\sqrt{2}} \sin 45^{\circ} + \frac{1}{\sqrt{2}} \cos 45^{\circ}) = \sqrt{2} \sin(45^{\circ} + 45^{\circ}) = \sqrt{2} \sin 90^{\circ} = \sqrt{2}(1) = \sqrt{2}$.
Now,$B = \sin 44^{\circ} + \cos 44^{\circ} = \sqrt{2} (\frac{1}{\sqrt{2}} \sin 44^{\circ} + \frac{1}{\sqrt{2}} \cos 44^{\circ})$.
Using the formula $\sin(x + y) = \sin x \cos y + \cos x \sin y$,we get $B = \sqrt{2} \sin(44^{\circ} + 45^{\circ}) = \sqrt{2} \sin 89^{\circ}$.
Since $\sin 89^{\circ} < \sin 90^{\circ}$,it follows that $\sqrt{2} \sin 89^{\circ} < \sqrt{2} \sin 90^{\circ}$.
Therefore,$B < A$ or $A > B$.

(A) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for four positive terms: $\sin \theta, \sin \theta, \sin \theta$,and $\csc^3 \theta$.
$\frac{\sin \theta + \sin \theta + \sin \theta + \csc^3 \theta}{4} \ge \sqrt[4]{(\sin \theta \cdot \sin \theta \cdot \sin \theta \cdot \csc^3 \theta)}$
$\frac{3\sin \theta + \csc^3 \theta}{4} \ge \sqrt[4]{(\sin^3 \theta \cdot \frac{1}{\sin^3 \theta})}$
$\frac{3\sin \theta + \csc^3 \theta}{4} \ge \sqrt[4]{1}$
$3\sin \theta + \csc^3 \theta \ge 4 \times 1$
Thus,the minimum value is $4$.

(C) Let $P = \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$.
Multiply and divide by $2 \sin \frac{\pi}{7}$:
$P = \frac{1}{2 \sin \frac{\pi}{7}} \left( 2 \sin \frac{\pi}{7} \cos \frac{\pi}{7} \right) \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$
Using $2 \sin \theta \cos \theta = \sin 2\theta$:
$P = \frac{1}{2 \sin \frac{\pi}{7}} \left( \sin \frac{2\pi}{7} \cos \frac{2\pi}{7} \right) \cos \frac{3\pi}{7}$
Multiply and divide by $2$:
$P = \frac{1}{4 \sin \frac{\pi}{7}} \left( 2 \sin \frac{2\pi}{7} \cos \frac{2\pi}{7} \right) \cos \frac{3\pi}{7} = \frac{1}{4 \sin \frac{\pi}{7}} \sin \frac{4\pi}{7} \cos \frac{3\pi}{7}$
Since $\sin \frac{4\pi}{7} = \sin \left( \pi - \frac{3\pi}{7} \right) = \sin \frac{3\pi}{7}$:
$P = \frac{1}{4 \sin \frac{\pi}{7}} \sin \frac{3\pi}{7} \cos \frac{3\pi}{7}$
Multiply and divide by $2$:
$P = \frac{1}{8 \sin \frac{\pi}{7}} \left( 2 \sin \frac{3\pi}{7} \cos \frac{3\pi}{7} \right) = \frac{1}{8 \sin \frac{\pi}{7}} \sin \frac{6\pi}{7}$
Since $\sin \frac{6\pi}{7} = \sin \left( \pi - \frac{\pi}{7} \right) = \sin \frac{\pi}{7}$:
$P = \frac{\sin \frac{\pi}{7}}{8 \sin \frac{\pi}{7}} = \frac{1}{8}$.

(D) Given that $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$.
Since the maximum value of the $\sin$ function is $1$,the only way for the sum of three $\sin$ values to be $3$ is if each individual term is equal to $1$.
Therefore,$\sin \theta_1 = 1, \sin \theta_2 = 1,$ and $\sin \theta_3 = 1$.
This implies that $\theta_1 = \theta_2 = \theta_3 = \frac{\pi}{2}$.
Now,we calculate the sum of the cosines: $\cos \theta_1 + \cos \theta_2 + \cos \theta_3$.
Since $\cos(\frac{\pi}{2}) = 0$,we have $\cos \theta_1 = 0, \cos \theta_2 = 0,$ and $\cos \theta_3 = 0$.
Thus,$0 + 0 + 0 = 0$.

(B) We know that $\sin(\pi - \theta) = \sin \theta$.
Therefore,$\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$ and $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$.
The given expression becomes: $2 \sin^2 \frac{\pi}{8} + 2 \sin^2 \frac{3\pi}{8} = 2 [\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8}]$.
Since $\sin \frac{3\pi}{8} = \sin(\frac{\pi}{2} - \frac{\pi}{8}) = \cos \frac{\pi}{8}$,we have $\sin^2 \frac{3\pi}{8} = \cos^2 \frac{\pi}{8}$.
Substituting this into the expression: $2 [\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8}] = 2(1) = 2$.

(D) Given expression: $E = \frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$
Convert to $\sin$ and $\cos$: $E = \frac{3 + \frac{\cos 76^{\circ} \cos 16^{\circ}}{\sin 76^{\circ} \sin 16^{\circ}}}{\frac{\cos 76^{\circ}}{\sin 76^{\circ}} + \frac{\cos 16^{\circ}}{\sin 16^{\circ}}} = \frac{3 \sin 76^{\circ} \sin 16^{\circ} + \cos 76^{\circ} \cos 16^{\circ}}{\cos 76^{\circ} \sin 16^{\circ} + \sin 76^{\circ} \cos 16^{\circ}}$
Using $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
Numerator: $2 \sin 76^{\circ} \sin 16^{\circ} + (\sin 76^{\circ} \sin 16^{\circ} + \cos 76^{\circ} \cos 16^{\circ}) = 2 \sin 76^{\circ} \sin 16^{\circ} + \cos(76^{\circ} - 16^{\circ}) = 2 \sin 76^{\circ} \sin 16^{\circ} + \cos 60^{\circ} = 2 \sin 76^{\circ} \sin 16^{\circ} + \frac{1}{2}$
Denominator: $\sin(76^{\circ} + 16^{\circ}) = \sin 92^{\circ}$
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$E = \frac{\cos(76^{\circ}-16^{\circ}) - \cos(76^{\circ}+16^{\circ}) + \frac{1}{2}}{\sin 92^{\circ}} = \frac{\cos 60^{\circ} - \cos 92^{\circ} + \frac{1}{2}}{\sin 92^{\circ}} = \frac{\frac{1}{2} - \cos 92^{\circ} + \frac{1}{2}}{\sin 92^{\circ}} = \frac{1 - \cos 92^{\circ}}{\sin 92^{\circ}}$
Using half-angle formulas $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$ and $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$:
$E = \frac{2 \sin^2 46^{\circ}}{2 \sin 46^{\circ} \cos 46^{\circ}} = \tan 46^{\circ} = \cot(90^{\circ} - 46^{\circ}) = \cot 44^{\circ}$

(B) Given expression: $5 \cos \theta + 3 \cos \left( \theta + \frac{\pi}{3} \right) - 1$
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$= 5 \cos \theta + 3 \left( \cos \theta \cdot \cos \frac{\pi}{3} - \sin \theta \cdot \sin \frac{\pi}{3} \right) - 1$
Substituting $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:
$= 5 \cos \theta + 3 \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) - 1$
$= 5 \cos \theta + \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta - 1$
$= \frac{13}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta - 1$
The expression is of the form $a \cos \theta + b \sin \theta + c$,where the maximum value is $\sqrt{a^2 + b^2} + c$.
Here,$a = \frac{13}{2}$,$b = -\frac{3\sqrt{3}}{2}$,and $c = -1$.
Maximum value $= \sqrt{\left( \frac{13}{2} \right)^2 + \left( -\frac{3\sqrt{3}}{2} \right)^2} - 1$
$= \sqrt{\frac{169}{4} + \frac{27}{4}} - 1$
$= \sqrt{\frac{196}{4}} - 1$
$= \sqrt{49} - 1 = 7 - 1 = 6$.

(C) Given that $A + B + C = \frac{\pi}{2} = 90^\circ$.
Therefore,$A + B = 90^\circ - C$.
Taking the tangent on both sides:
$\tan(A + B) = \tan(90^\circ - C)$
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(90^\circ - C) = \cot C = \frac{1}{\tan C}$,we get:
$\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1}{\tan C}$
Cross-multiplying:
$\tan C(\tan A + \tan B) = 1 - \tan A \tan B$
Expanding the left side:
$\tan C \tan A + \tan C \tan B = 1 - \tan A \tan B$
Rearranging the terms:
$\tan A \tan B + \tan B \tan C + \tan C \tan A = 1$.

(D) We know that $1 \text{ radian} \approx 57.296^\circ$.
$\sin(1 \text{ rad}) = \sin(57.296^\circ)$
$\sin(2 \text{ rad}) = \sin(114.592^\circ) = \sin(180^\circ - 114.592^\circ) = \sin(65.408^\circ)$
$\sin(3 \text{ rad}) = \sin(171.887^\circ) = \sin(180^\circ - 171.887^\circ) = \sin(8.113^\circ)$
Since the function $y = \sin x$ is increasing in the interval $[0, \pi/2]$,we compare the angles in the first quadrant:
$8.113^\circ < 57.296^\circ < 65.408^\circ$
Therefore,$\sin(8.113^\circ) < \sin(57.296^\circ) < \sin(65.408^\circ)$.
Thus,$\sin 3 < \sin 1 < \sin 2$.

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

(B) Given that $\tan A$ and $\tan B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$.
From the properties of quadratic equations,we have $\tan A + \tan B = \frac{10}{3}$ and $\tan A \tan B = -\frac{25}{3}$.
We know that $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{10/3}{1 - (-25/3)} = \frac{10/3}{28/3} = \frac{10}{28} = \frac{5}{14}$.
Now,the expression is $E = 3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$.
Dividing the entire expression by $\cos^2 (A + B)$,we get $E = \cos^2 (A + B) [3 \tan^2 (A + B) - 10 \tan (A + B) - 25]$.
Since $\tan (A + B) = \frac{5}{14}$,the term inside the bracket is $3(\frac{5}{14})^2 - 10(\frac{5}{14}) - 25 = 3(\frac{25}{196}) - \frac{50}{14} - 25 = \frac{75}{196} - \frac{700}{196} - \frac{4900}{196} = \frac{-5525}{196}$.
Also,$\cos^2 (A + B) = \frac{1}{1 + \tan^2 (A + B)} = \frac{1}{1 + (5/14)^2} = \frac{1}{1 + 25/196} = \frac{196}{221}$.
Therefore,$E = \frac{196}{221} \times \frac{-5525}{196} = -\frac{5525}{221} = -25$.

(A) For statement $p$: We have $\sin 120^\circ = \frac{\sqrt{3}}{2}$,so $2 \sin 120^\circ = \sqrt{3}$.
Substituting $\theta = 240^\circ$ in the $RHS$: $\sqrt{1 + \sin 240^\circ} - \sqrt{1 - \sin 240^\circ} = \sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}} - \sqrt{\frac{2 + \sqrt{3}}{2}}$.
Since $\sqrt{2 \pm \sqrt{3}} = \sqrt{\frac{4 \pm 2\sqrt{3}}{2}} = \frac{\sqrt{3} \pm 1}{\sqrt{2}}$,the expression becomes $\frac{\sqrt{3}-1}{2} - \frac{\sqrt{3}+1}{2} = -1 \neq \sqrt{3}$. Thus,statement $p$ is False.
For statement $q$: In any quadrilateral $ABCD$,$A + B + C + D = 360^\circ = 2\pi$. Therefore,$\frac{A+C}{2} + \frac{B+D}{2} = \pi$. Let $\alpha = \frac{A+C}{2}$,then $\frac{B+D}{2} = \pi - \alpha$. The equation becomes $\cos \alpha + \cos(\pi - \alpha) = \cos \alpha - \cos \alpha = 0$. Thus,statement $q$ is True.
Therefore,the truth values are $F, T$.

(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.
Using $\sin^2 2x = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we get:
$f(x) = 4 + \frac{1}{2} [4(1 - \cos^2 x) \cos^2 x] - 2 \cos^4 x$
$f(x) = 4 + 2 \cos^2 x - 2 \cos^4 x - 2 \cos^4 x$
$f(x) = 4 + 2 \cos^2 x - 4 \cos^4 x$.
Let $t = \cos^2 x$,where $0 \le t \le 1$.
$f(t) = -4t^2 + 2t + 4$.
This is a downward-opening parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-4)} = \frac{1}{4}$.
Since $\frac{1}{4} \in [0, 1]$,the maximum value $M$ occurs at $t = \frac{1}{4}$:
$M = f(\frac{1}{4}) = -4(\frac{1}{16}) + 2(\frac{1}{4}) + 4 = -\frac{1}{4} + \frac{1}{2} + 4 = \frac{1}{4} + 4 = \frac{17}{4}$.
The minimum value $m$ occurs at the boundaries $t=0$ or $t=1$:
$f(0) = 4$.
$f(1) = -4(1)^2 + 2(1) + 4 = 2$.
Thus,$m = 2$.
$M - m = \frac{17}{4} - 2 = \frac{17 - 8}{4} = \frac{9}{4}$.

(B) Given $A + B = \frac{\pi}{6}$. Let $y = \tan A + \tan B$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\tan(\frac{\pi}{6}) = \frac{y}{1 - \tan A \tan B} = \frac{1}{\sqrt{3}}$.
Thus,$1 - \tan A \tan B = \sqrt{3}y$,which implies $\tan A \tan B = 1 - \sqrt{3}y$.
Since $A, B > 0$ and $A+B = \frac{\pi}{6}$,both $\tan A$ and $\tan B$ are positive. By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality:
$\frac{\tan A + \tan B}{2} \ge \sqrt{\tan A \tan B}$
$\frac{y}{2} \ge \sqrt{1 - \sqrt{3}y}$
Squaring both sides: $\frac{y^2}{4} \ge 1 - \sqrt{3}y \Rightarrow y^2 + 4\sqrt{3}y - 4 \ge 0$.
Solving the quadratic equation $y^2 + 4\sqrt{3}y - 4 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{-4\sqrt{3} \pm \sqrt{48 - 4(1)(-4)}}{2} = \frac{-4\sqrt{3} \pm \sqrt{64}}{2} = -2\sqrt{3} \pm 4$.
Since $y > 0$,we take $y \ge 4 - 2\sqrt{3}$.
Thus,the minimum value is $4 - 2\sqrt{3}$.

(C) Given that $f$ is an odd function,by definition,$f(-x) = -f(x)$ for all $x$ in the domain.
We are given $f(x) = 3 \sin x + 4 \cos x$ for $x \geq 0$.
To find $f\left(-\frac{11\pi}{6}\right)$,we use the property of odd functions: $f(-x) = -f(x)$.
Therefore,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right)$.
First,calculate $f\left(\frac{11\pi}{6}\right)$:
$f\left(\frac{11\pi}{6}\right) = 3 \sin\left(\frac{11\pi}{6}\right) + 4 \cos\left(\frac{11\pi}{6}\right)$.
Since $\frac{11\pi}{6} = 2\pi - \frac{\pi}{6}$,we have:
$\sin\left(\frac{11\pi}{6}\right) = \sin\left(2\pi - \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$.
$\cos\left(\frac{11\pi}{6}\right) = \cos\left(2\pi - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.
Thus,$f\left(\frac{11\pi}{6}\right) = 3\left(-\frac{1}{2}\right) + 4\left(\frac{\sqrt{3}}{2}\right) = -\frac{3}{2} + 2\sqrt{3}$.
Finally,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right) = -\left(-\frac{3}{2} + 2\sqrt{3}\right) = \frac{3}{2} - 2\sqrt{3}$.

(A) Given that $\sec(\theta - \phi)$,$\sec\theta$,and $\sec(\theta + \phi)$ are in $A.P.$
Therefore,$2 \sec\theta = \sec(\theta - \phi) + \sec(\theta + \phi)$.
$\frac{2}{\cos\theta} = \frac{1}{\cos(\theta - \phi)} + \frac{1}{\cos(\theta + \phi)}$
$\frac{2}{\cos\theta} = \frac{\cos(\theta + \phi) + \cos(\theta - \phi)}{\cos(\theta - \phi) \cos(\theta + \phi)}$
Using the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$:
$\frac{2}{\cos\theta} = \frac{2 \cos\theta \cos\phi}{\cos^2\theta - \sin^2\phi}$
$\cos^2\theta - \sin^2\phi = \cos^2\theta \cos\phi$
$\cos^2\theta(1 - \cos\phi) = \sin^2\phi$
$\cos^2\theta(1 - \cos\phi) = 1 - \cos^2\phi = (1 - \cos\phi)(1 + \cos\phi)$
Since $\phi \neq 0$,$1 - \cos\phi \neq 0$,so we can divide by $(1 - \cos\phi)$:
$\cos^2\theta = 1 + \cos\phi = 2 \cos^2(\frac{\phi}{2})$
Taking the square root on both sides:
$\cos\theta = \pm \sqrt{2} \cos(\frac{\phi}{2})$
Comparing this with $\cos\theta = k \cos(\frac{\phi}{2})$,we get $k = \pm \sqrt{2}$.

(C) Given expression: $\cos 255^o + \sin 195^o$
We can write the angles as:
$\cos 255^o = \cos(270^o - 15^o) = -\sin 15^o$
$\sin 195^o = \sin(180^o + 15^o) = -\sin 15^o$
Substituting these values:
$-\sin 15^o - \sin 15^o = -2 \sin 15^o$
We know that $\sin 15^o = \sin(45^o - 30^o) = \sin 45^o \cos 30^o - \cos 45^o \sin 30^o = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$
Therefore,$-2 \sin 15^o = -2 \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) = -\frac{\sqrt{3} - 1}{\sqrt{2}}$

(B) Let the expression be $E = 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4\sin^6 \theta$.
We know that $(\sin \theta - \cos \theta)^2 = 1 - \sin 2\theta$ and $(\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta$.
Thus,$E = 3(1 - \sin 2\theta)^2 + 6(1 + \sin 2\theta) + 4\sin^6 \theta$.
Expanding the terms: $E = 3(1 - 2\sin 2\theta + \sin^2 2\theta) + 6 + 6\sin 2\theta + 4\sin^6 \theta$.
$E = 3 - 6\sin 2\theta + 3\sin^2 2\theta + 6 + 6\sin 2\theta + 4\sin^6 \theta$.
$E = 9 + 3\sin^2 2\theta + 4\sin^6 \theta$.
Using $\sin 2\theta = 2\sin \theta \cos \theta$,we get $E = 9 + 3(4\sin^2 \theta \cos^2 \theta) + 4\sin^6 \theta = 9 + 12\sin^2 \theta \cos^2 \theta + 4\sin^6 \theta$.
Substitute $\sin^2 \theta = 1 - \cos^2 \theta$:
$E = 9 + 12(1 - \cos^2 \theta)\cos^2 \theta + 4(1 - \cos^2 \theta)^3$.
$E = 9 + 12\cos^2 \theta - 12\cos^4 \theta + 4(1 - 3\cos^2 \theta + 3\cos^4 \theta - \cos^6 \theta)$.
$E = 9 + 12\cos^2 \theta - 12\cos^4 \theta + 4 - 12\cos^2 \theta + 12\cos^4 \theta - 4\cos^6 \theta$.
$E = 13 - 4\cos^6 \theta$.

(C) Given equation: $\sin^2 2\theta + \cos^4 2\theta = \frac{3}{4}$.
Using the identity $\sin^2 2\theta = 1 - \cos^2 2\theta$,we get:
$1 - \cos^2 2\theta + \cos^4 2\theta = \frac{3}{4}$.
Let $t = \cos^2 2\theta$. Then the equation becomes $t^2 - t + 1 = \frac{3}{4}$,which simplifies to $t^2 - t + \frac{1}{4} = 0$.
This is $(t - \frac{1}{2})^2 = 0$,so $t = \frac{1}{2}$.
Thus,$\cos^2 2\theta = \frac{1}{2}$,which implies $\cos 4\theta = 2\cos^2 2\theta - 1 = 2(\frac{1}{2}) - 1 = 0$.
For $\theta \in (0, \frac{\pi}{2})$,$4\theta \in (0, 2\pi)$.
$\cos 4\theta = 0$ implies $4\theta = \frac{\pi}{2}, \frac{3\pi}{2}$.
Therefore,$\theta = \frac{\pi}{8}, \frac{3\pi}{8}$.
The sum of these values is $\frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2}$.

(A) We use the identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \dots \cdot \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$.
Let $A = \frac{\pi}{2^{10}}$. The given expression is $P = \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \dots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$.
Note that $\frac{\pi}{2^2} = 2^8 \cdot \frac{\pi}{2^{10}}$,$\frac{\pi}{2^3} = 2^7 \cdot \frac{\pi}{2^{10}}$,and so on.
This is equivalent to $\cos(2^8 A) \cdot \cos(2^7 A) \cdot \dots \cdot \cos(A) \cdot \sin A$.
Rearranging the product: $(\cos A \cdot \cos 2A \cdot \dots \cdot \cos 2^8 A) \cdot \sin A$.
Using the formula with $n=9$: $\frac{\sin(2^9 A)}{2^9 \sin A} \cdot \sin A = \frac{\sin(2^9 \cdot \frac{\pi}{2^{10}})}{2^9} = \frac{\sin(\frac{\pi}{2})}{512} = \frac{1}{512}$.

(A) Given ${f_k}(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.
For $k=4$,${f_4}(x) = \frac{\sin^4 x + \cos^4 x}{4}$.
Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$,we get:
${f_4}(x) = \frac{1 - 2\sin^2 x \cos^2 x}{4} = \frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x$.
For $k=6$,${f_6}(x) = \frac{\sin^6 x + \cos^6 x}{6}$.
Using the identity $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 - 3\sin^2 x \cos^2 x$,we get:
${f_6}(x) = \frac{1 - 3\sin^2 x \cos^2 x}{6} = \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x$.
Now,${f_4}(x) - {f_6}(x) = (\frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x) - (\frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x) = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$.

(A) Let $f(\theta) = 3 \cos \theta + 5 \sin \left( \theta - \frac{\pi}{6} \right)$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$f(\theta) = 3 \cos \theta + 5 \left( \sin \theta \cos \frac{\pi}{6} - \cos \theta \sin \frac{\pi}{6} \right)$
$f(\theta) = 3 \cos \theta + 5 \left( \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta \right)$
$f(\theta) = \frac{5\sqrt{3}}{2} \sin \theta + \left( 3 - \frac{5}{2} \right) \cos \theta$
$f(\theta) = \frac{5\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta$.
The expression is of the form $a \sin \theta + b \cos \theta$,which has a maximum value of $\sqrt{a^2 + b^2}$.
Maximum value $= \sqrt{\left( \frac{5\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2}$
$= \sqrt{\frac{25 \times 3}{4} + \frac{1}{4}} = \sqrt{\frac{75 + 1}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19}$.

(C) Given $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha + \beta < \frac{\pi}{2}$ and $-\frac{\pi}{4} < \alpha - \beta < \frac{\pi}{4}$.
Since $\cos(\alpha + \beta) = \frac{3}{5}$,we find $\tan(\alpha + \beta) = \frac{\sqrt{1 - (3/5)^2}}{3/5} = \frac{4/5}{3/5} = \frac{4}{3}$.
Since $\sin(\alpha - \beta) = \frac{5}{13}$,we find $\tan(\alpha - \beta) = \frac{5/13}{\sqrt{1 - (5/13)^2}} = \frac{5/13}{12/13} = \frac{5}{12}$.
Now,$\tan(2\alpha) = \tan((\alpha + \beta) + (\alpha - \beta))$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan(2\alpha) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3} \cdot \frac{5}{12})} = \frac{\frac{16+5}{12}}{1 - \frac{20}{36}} = \frac{21/12}{16/36} = \frac{21}{12} \cdot \frac{36}{16} = \frac{21 \cdot 3}{16} = \frac{63}{16}$.

(B) Let $E = \cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$.
Multiply and divide by $2$:
$E = \frac{1}{2} [2 \cos^2 10^\circ - 2 \cos 10^\circ \cos 50^\circ + 2 \cos^2 50^\circ]$
Using the identities $2 \cos^2 \theta = 1 + \cos 2\theta$ and $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \frac{1}{2} [(1 + \cos 20^\circ) - (\cos 60^\circ + \cos(-40^\circ)) + (1 + \cos 100^\circ)]$
$E = \frac{1}{2} [1 + \cos 20^\circ - \frac{1}{2} - \cos 40^\circ + 1 + \cos 100^\circ]$
$E = \frac{1}{2} [\frac{3}{2} + \cos 20^\circ - (\cos 40^\circ - \cos 100^\circ)]$
Using $\cos C - \cos D = 2 \sin(\frac{C+D}{2}) \sin(\frac{D-C}{2})$:
$\cos 40^\circ - \cos 100^\circ = 2 \sin(\frac{140^\circ}{2}) \sin(\frac{60^\circ}{2}) = 2 \sin 70^\circ \sin 30^\circ = 2 \sin 70^\circ (\frac{1}{2}) = \sin 70^\circ = \cos 20^\circ$.
Substituting back:
$E = \frac{1}{2} [\frac{3}{2} + \cos 20^\circ - \cos 20^\circ] = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.

(D) We use the identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta$.
Given expression: $E = \sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ$.
Rearranging the terms: $E = \sin 30^\circ [\sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)]$.
Using the identity with $\theta = 10^\circ$: $E = \sin 30^\circ [\frac{1}{4} \sin(3 \times 10^\circ)]$.
Since $\sin 30^\circ = \frac{1}{2}$,we have $E = \frac{1}{2} [\frac{1}{4} \sin 30^\circ]$.
Substituting $\sin 30^\circ = \frac{1}{2}$ again: $E = \frac{1}{2} [\frac{1}{4} \times \frac{1}{2}] = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}$.

(B) The expression is of the form $y = a\,sin\,x + b\,cos\,x$.
The maximum value of this expression 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) The given equation is $(k+1) \tan^2 x - (\sqrt{2} \lambda) \tan x + (k-1) = 0$.
Let $t = \tan x$. The roots of the quadratic equation $(k+1)t^2 - (\sqrt{2} \lambda)t + (k-1) = 0$ are $\tan \alpha$ and $\tan \beta$.
From the properties of quadratic equations,the sum of roots is $\tan \alpha + \tan \beta = \frac{\sqrt{2} \lambda}{k+1}$ and the product of roots is $\tan \alpha \tan \beta = \frac{k-1}{k+1}$.
Using the tangent addition formula,$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.
Substituting the values,$\tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k+1}}{1 - \frac{k-1}{k+1}} = \frac{\sqrt{2} \lambda}{k+1 - k + 1} = \frac{\sqrt{2} \lambda}{2} = \frac{\lambda}{\sqrt{2}}$.
Given $\tan^2(\alpha + \beta) = 50$,we have $\left(\frac{\lambda}{\sqrt{2}}\right)^2 = 50$.
$\frac{\lambda^2}{2} = 50 \Rightarrow \lambda^2 = 100 \Rightarrow \lambda = \pm 10$.
Thus,a possible value of $\lambda$ is $10$.

(A) Given $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}} = \frac{1}{7}$.
Using the identity $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we get $\frac{\sqrt{2} \sin \alpha}{\sqrt{2 \cos^2 \alpha}} = \frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha} = \tan \alpha = \frac{1}{7}$.
Given $\sqrt{\frac{1-\cos 2 \beta}{2}} = \frac{1}{\sqrt{10}}$.
Using the identity $1-\cos 2 \beta = 2 \sin^2 \beta$,we get $\sqrt{\frac{2 \sin^2 \beta}{2}} = \sin \beta = \frac{1}{\sqrt{10}}$.
Since $\sin \beta = \frac{1}{\sqrt{10}}$,we have $\cos \beta = \sqrt{1 - \frac{1}{10}} = \frac{3}{\sqrt{10}}$,so $\tan \beta = \frac{1}{3}$.
Now,$\tan 2 \beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2(1/3)}{1 - (1/9)} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$.
Finally,$\tan (\alpha + 2 \beta) = \frac{\tan \alpha + \tan 2 \beta}{1 - \tan \alpha \tan 2 \beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{(4+21)/28}{(28-3)/28} = \frac{25}{25} = 1$.