Mix Examples-Trigonometrical Ratios, Functions and Identities Questions in English

(A) For statement $p$: $\sin 120^\circ = \frac{\sqrt{3}}{2}$,so $2\sin 120^\circ = \sqrt{3}$.
Substituting $\theta = 240^\circ$ in the $RHS$: $\sqrt{1 + \sin 240^\circ} - \sqrt{1 - \sin 240^\circ} = \sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}} - \sqrt{\frac{2 + \sqrt{3}}{2}} = \frac{\sqrt{3}-1}{2} - \frac{\sqrt{3}+1}{2} = -1 \neq \sqrt{3}$. Thus,statement $p$ is False.
For statement $q$: In any quadrilateral $ABCD$,$A + B + C + D = 360^\circ$,so $\frac{A+C}{2} + \frac{B+D}{2} = 180^\circ = \pi$.
Let $\alpha = \frac{A+C}{2}$,then $\frac{B+D}{2} = \pi - \alpha$.
Thus,$\cos(\alpha) + \cos(\pi - \alpha) = \cos(\alpha) - \cos(\alpha) = 0$. Thus,statement $q$ is True.
Therefore,the truth values are $F, T$.

(D) Let $f(x) = \sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}$.
We are given $|f(x)| = 1$,which implies $f(x) = 1$ or $f(x) = -1$.
Using $\cos^2 x = 1 - \sin^2 x$,we have $2 \sin^4 x + 18(1 - \sin^2 x) = 2 \sin^4 x - 18 \sin^2 x + 18 = 2(\sin^2 x - \frac{9}{2})^2 + 18 - \frac{81}{2} = 2(\sin^2 x - 4.5)^2 - 22.5$ (not helpful).
Let $u = \sin^2 x$,then $v = \cos^2 x = 1 - u$. The expression is $\sqrt{2u^2 + 18(1-u)} - \sqrt{2(1-u)^2 + 18u} = \pm 1$.
$\sqrt{2u^2 - 18u + 18} - \sqrt{2u^2 - 14u + 20} = \pm 1$.
Squaring and simplifying leads to $8$ distinct values of $x$ in the interval $[0, 2\pi]$.

(C) For set $P$,we have $\sin \theta - \cos \theta = \sqrt{2} \cos \theta$.
Rearranging gives $\sin \theta = (\sqrt{2} + 1) \cos \theta$.
Multiplying by $(\sqrt{2} - 1)$,we get $(\sqrt{2} - 1) \sin \theta = (\sqrt{2} - 1)(\sqrt{2} + 1) \cos \theta$,which simplifies to $(\sqrt{2} - 1) \sin \theta = \cos \theta$.
Thus,$\sin \theta = \cos \theta + \sqrt{2} \sin \theta - \sin \theta$,which leads to $\sin \theta + \cos \theta = \sqrt{2} \sin \theta$.
This is the defining condition for set $Q$.
Similarly,starting from $Q$,we can derive the condition for $P$.
Therefore,$P = Q$.

(B) Given $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$.
Using sum-to-product formulas:
$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{3}{2}$ $(i)$
$2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{2}$ $(ii)$
Dividing $(ii)$ by $(i)$ gives $\tan \frac{\alpha+\beta}{2} = \frac{1}{3}$.
Since $\theta = \frac{\alpha+\beta}{2}$,we have $\tan \theta = \frac{1}{3}$.
We need to find $\sin 2\theta + \cos 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} + \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$.
Substituting $\tan \theta = \frac{1}{3}$:
$= \frac{2(1/3)}{1 + 1/9} + \frac{1 - 1/9}{1 + 1/9} = \frac{2/3}{10/9} + \frac{8/9}{10/9} = \frac{6}{10} + \frac{8}{10} = \frac{14}{10} = \frac{7}{5}$.

(D) Given $2 \cos \theta + \sin \theta = 1$.
Squaring both sides,we get $(2 \cos \theta + \sin \theta)^2 = 1^2$.
$4 \cos^2 \theta + \sin^2 \theta + 4 \sin \theta \cos \theta = 1$.
Since $\sin^2 \theta = 1 - \cos^2 \theta$,we have $4 \cos^2 \theta + (1 - \cos^2 \theta) + 4 \sin \theta \cos \theta = 1$.
$3 \cos^2 \theta + 4 \sin \theta \cos \theta = 0$.
$\cos \theta (3 \cos \theta + 4 \sin \theta) = 0$.
Since $\theta \neq \frac{\pi}{2}$,$\cos \theta \neq 0$,so $3 \cos \theta + 4 \sin \theta = 0$,which implies $\tan \theta = -\frac{3}{4}$.
Using $\tan \theta = -\frac{3}{4}$,we can form a right triangle with opposite side $3$ and adjacent side $4$.
Since $\tan \theta$ is negative,$\theta$ is in the second or fourth quadrant.
Given $2 \cos \theta + \sin \theta = 1$,if $\cos \theta = \frac{4}{5}$ and $\sin \theta = -\frac{3}{5}$,then $2(\frac{4}{5}) + (-\frac{3}{5}) = \frac{8-3}{5} = 1$ (Satisfied).
If $\cos \theta = -\frac{4}{5}$ and $\sin \theta = \frac{3}{5}$,then $2(-\frac{4}{5}) + \frac{3}{5} = -1 \neq 1$.
Thus,$\cos \theta = \frac{4}{5}$ and $\sin \theta = -\frac{3}{5}$.
Therefore,$7 \cos \theta + 6 \sin \theta = 7(\frac{4}{5}) + 6(-\frac{3}{5}) = \frac{28}{5} - \frac{18}{5} = \frac{10}{5} = 2$.

(A) Since $\sec(\theta - \phi)$,$\sec \theta$,and $\sec(\theta + \phi)$ are in $A.P.$,we have:
$2 \sec \theta = \sec(\theta - \phi) + \sec(\theta + \phi)$
$\frac{2}{\cos \theta} = \frac{\cos(\theta + \phi) + \cos(\theta - \phi)}{\cos(\theta - \phi) \cos(\theta + \phi)}$
Using the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$ and $\cos(A-B) \cos(A+B) = \cos^2 A - \sin^2 B$:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \phi}{\cos^2 \theta - \sin^2 \phi}$
$\cos^2 \theta - \sin^2 \phi = \cos^2 \theta \cos \phi$
$\cos^2 \theta (1 - \cos \phi) = \sin^2 \phi$
$\cos^2 \theta (1 - \cos \phi) = 1 - \cos^2 \phi = (1 - \cos \phi)(1 + \cos \phi)$
Since $\phi \neq 0$,$1 - \cos \phi \neq 0$,so:
$\cos^2 \theta = 1 + \cos \phi = 2 \cos^2(\frac{\phi}{2})$
$\cos \theta = \pm \sqrt{2} \cos(\frac{\phi}{2})$
Given $\cos \theta = k \cos(\frac{\phi}{2})$,we find $k = \pm \sqrt{2}$.

(B) Given expression: $E = 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4\sin^6 \theta$
Using $(\sin \theta - \cos \theta)^2 = 1 - \sin 2\theta$ and $(\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta$:
$E = 3(1 - \sin 2\theta)^2 + 6(1 + \sin 2\theta) + 4\sin^6 \theta$
$E = 3(1 - 2\sin 2\theta + \sin^2 2\theta) + 6 + 6\sin 2\theta + 4\sin^6 \theta$
$E = 3 - 6\sin 2\theta + 3\sin^2 2\theta + 6 + 6\sin 2\theta + 4\sin^6 \theta$
$E = 9 + 3\sin^2 2\theta + 4\sin^6 \theta$
Substitute $\sin 2\theta = 2\sin \theta \cos \theta$:
$E = 9 + 3(4\sin^2 \theta \cos^2 \theta) + 4\sin^6 \theta$
$E = 9 + 12\sin^2 \theta \cos^2 \theta + 4\sin^6 \theta$
Since $\sin^2 \theta = 1 - \cos^2 \theta$:
$E = 9 + 12(1 - \cos^2 \theta)\cos^2 \theta + 4(1 - \cos^2 \theta)^3$
$E = 9 + 12\cos^2 \theta - 12\cos^4 \theta + 4(1 - 3\cos^2 \theta + 3\cos^4 \theta - \cos^6 \theta)$
$E = 9 + 12\cos^2 \theta - 12\cos^4 \theta + 4 - 12\cos^2 \theta + 12\cos^4 \theta - 4\cos^6 \theta$
$E = 13 - 4\cos^6 \theta$

(A) $f_4(x) = \frac{\sin^4 x + \cos^4 x}{4} = \frac{(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x}{4} = \frac{1 - 2\sin^2 x \cos^2 x}{4} = \frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x$
$f_6(x) = \frac{\sin^6 x + \cos^6 x}{6} = \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)}{6} = \frac{1 - 3\sin^2 x \cos^2 x}{6} = \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x$
$f_4(x) - f_6(x) = (\frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x) - (\frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x)$
$f_4(x) - f_6(x) = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$

(D) Using the $AM \geq GM$ inequality for the four terms $\sin^4 \alpha, 4 \cos^4 \beta, 1, 1$:
$\frac{\sin^4 \alpha + 4 \cos^4 \beta + 1 + 1}{4} \geq (\sin^4 \alpha \cdot 4 \cos^4 \beta \cdot 1 \cdot 1)^{1/4}$
$\frac{\sin^4 \alpha + 4 \cos^4 \beta + 2}{4} \geq (4 \sin^4 \alpha \cos^4 \beta)^{1/4} = \sqrt{2} \sin \alpha \cos \beta$
$\sin^4 \alpha + 4 \cos^4 \beta + 2 \geq 4\sqrt{2} \sin \alpha \cos \beta$
Given the equality holds,$AM = GM$,so $\sin^4 \alpha = 4 \cos^4 \beta = 1$.
Thus,$\sin^4 \alpha = 1 \Rightarrow \sin \alpha = 1$ (since $\alpha \in [0, \pi]$).
And $4 \cos^4 \beta = 1$ $\Rightarrow \cos^2 \beta = 1/2$ $\Rightarrow \cos \beta = \pm 1/\sqrt{2}$.
Since $\beta \in [0, \pi]$,$\sin \beta = 1/\sqrt{2}$.
If $\cos \beta = 1/\sqrt{2}$,then $\alpha = \pi/2, \beta = \pi/4 \Rightarrow \cos(\alpha + \beta) = \cos(3\pi/4) = -1/\sqrt{2}$.
If $\cos \beta = -1/\sqrt{2}$,then $\alpha = \pi/2, \beta = 3\pi/4 \Rightarrow \cos(\alpha + \beta) = \cos(5\pi/4) = -1/\sqrt{2}$.

(B) Let $E = \cos^2 10^o - \cos 10^o \cos 50^o + \cos^2 50^o$.
Multiplying by $\frac{2}{2}$,we get $E = \frac{1}{2} (2 \cos^2 10^o - 2 \cos 10^o \cos 50^o + 2 \cos^2 50^o)$.
Using the identities $2 \cos^2 \theta = 1 + \cos 2\theta$ and $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \frac{1}{2} [(1 + \cos 20^o) - (\cos 60^o + \cos(-40^o)) + (1 + \cos 100^o)]$.
$E = \frac{1}{2} [2 + \cos 20^o - \frac{1}{2} - \cos 40^o + \cos 100^o]$.
$E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - 2 \sin(\frac{100^o+40^o}{2}) \sin(\frac{100^o-40^o}{2})]$.
$E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - 2 \sin 70^o \sin 30^o]$.
Since $\sin 30^o = \frac{1}{2}$ and $\sin 70^o = \cos 20^o$:
$E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - 2 \cos 20^o \cdot \frac{1}{2}]$.
$E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - \cos 20^o] = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$.

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

(C) Given equation: $y = \sin \,x \sin \,(x + 2) - \sin^2 \,(x + 1)$
Multiply by $2$: $2y = 2 \sin \,x \sin \,(x + 2) - 2 \sin^2 \,(x + 1)$
Using the identity $2 \sin \,A \sin \,B = \cos(A - B) - \cos(A + B)$:
$2 \sin \,x \sin \,(x + 2) = \cos(x - (x + 2)) - \cos(x + x + 2) = \cos(-2) - \cos(2x + 2) = \cos \,2 - \cos(2x + 2)$
Using the identity $2 \sin^2 \,A = 1 - \cos(2A)$:
$2 \sin^2 \,(x + 1) = 1 - \cos(2(x + 1)) = 1 - \cos(2x + 2)$
Substituting these into the equation:
$2y = (\cos \,2 - \cos(2x + 2)) - (1 - \cos(2x + 2))$
$2y = \cos \,2 - 1$
Since $\cos \,2 \approx -0.416$,we have $2y \approx -1.416$,so $y \approx -0.708$.
This represents a horizontal line $y = k$ where $k < 0$.
$A$ horizontal line with a negative $y$-intercept passes through the third and fourth quadrants.

(C) Given equation: $\cos 2x + \alpha \sin x = 2\alpha - 7$
Using $\cos 2x = 1 - 2 \sin^2 x$,we get:
$1 - 2 \sin^2 x + \alpha \sin x = 2\alpha - 7$
$2 \sin^2 x - \alpha \sin x + 2\alpha - 8 = 0$
This is a quadratic equation in $\sin x$. Using the quadratic formula $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\sin x = \frac{\alpha \pm \sqrt{\alpha^2 - 4(2)(2\alpha - 8)}}{4}$
$\sin x = \frac{\alpha \pm \sqrt{\alpha^2 - 16\alpha + 64}}{4}$
$\sin x = \frac{\alpha \pm \sqrt{(\alpha - 8)^2}}{4}$
$\sin x = \frac{\alpha \pm (\alpha - 8)}{4}$
Case $1$: $\sin x = \frac{\alpha + \alpha - 8}{4} = \frac{2\alpha - 8}{4} = \frac{\alpha - 4}{2}$
Case $2$: $\sin x = \frac{\alpha - (\alpha - 8)}{4} = \frac{8}{4} = 2$ (Not possible as $\sin x \in [-1, 1]$)
For a solution to exist,we must have $-1 \leq \frac{\alpha - 4}{2} \leq 1$
$-2 \leq \alpha - 4 \leq 2$
$2 \leq \alpha \leq 6$
Thus,$S = [2, 6]$.

(B) The given equation is $(k+1) \tan^{2} x - (\sqrt{2} \lambda) \tan x + (k-1) = 0$.
Let $t = \tan x$. Then $(k+1) t^{2} - (\sqrt{2} \lambda) t + (k-1) = 0$.
Since $\alpha$ and $\beta$ are roots,$\tan \alpha$ and $\tan \beta$ are roots of this quadratic equation.
Sum of roots: $\tan \alpha + \tan \beta = \frac{\sqrt{2} \lambda}{k+1}$.
Product of roots: $\tan \alpha \tan \beta = \frac{k-1}{k+1}$.
Using the formula $\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$,we get:
$\tan(\alpha+\beta) = \frac{\frac{\sqrt{2} \lambda}{k+1}}{1 - \frac{k-1}{k+1}} = \frac{\sqrt{2} \lambda}{k+1 - (k-1)} = \frac{\sqrt{2} \lambda}{2} = \frac{\lambda}{\sqrt{2}}$.
Given $\tan^{2}(\alpha+\beta) = 50$,so $\left(\frac{\lambda}{\sqrt{2}}\right)^{2} = 50$.
$\frac{\lambda^{2}}{2} = 50 \implies \lambda^{2} = 100 \implies \lambda = \pm 10$.
Thus,a possible value of $\lambda$ is $10$.

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

(C) Let $\theta = \frac{\pi}{8}$. Then $\frac{3\pi}{8} = 3\theta$.
Given expression is $\cos^{3}(\theta) \cos(3\theta) + \sin^{3}(\theta) \sin(3\theta)$.
Using the identities $\cos(3\theta) = 4\cos^{3}(\theta) - 3\cos(\theta)$ and $\sin(3\theta) = 3\sin(\theta) - 4\sin^{3}(\theta)$,we can rewrite the expression.
Alternatively,note that $\cos(\frac{3\pi}{8}) = \sin(\frac{\pi}{8})$ and $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$.
Substituting these: $\cos^{3}(\frac{\pi}{8}) \sin(\frac{\pi}{8}) + \sin^{3}(\frac{\pi}{8}) \cos(\frac{\pi}{8})$.
Factor out $\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8})$:
$= \sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}) [\cos^{2}(\frac{\pi}{8}) + \sin^{2}(\frac{\pi}{8})]$.
Since $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$,this simplifies to $\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8})$.
Using $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$,we get $\frac{1}{2} \sin(2 \cdot \frac{\pi}{8}) = \frac{1}{2} \sin(\frac{\pi}{4})$.
$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.

(A) $L.H.S. = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x$
$= \cot x \cot 2x - \cot 3x(\cot 2x + \cot x)$
$= \cot x \cot 2x - \cot(2x + x)(\cot 2x + \cot x)$
$= \cot x \cot 2x - \left[ \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x} \right](\cot 2x + \cot x)$
$\left[ \because \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \right]$
$= \cot x \cot 2x - (\cot 2x \cot x - 1) = 1 = R.H.S.$

(N/A) $L.H.S.$ $= (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x$
$= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$
$= (\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x)$
$= \cos(3x - x) - \cos 2x$
$= \cos 2x - \cos 2x$
$= 0$
$= R.H.S.$

(A) Let the coordinates of $A, B, C$ be $(R\cos \alpha, R\sin \alpha), (R\cos \beta, R\sin \beta)$ and $(R\cos \gamma, R\sin \gamma)$ respectively,where $R$ is the circumradius.
Since the circumcentre is at the origin $(0,0)$,the distance of $A, B, C$ from the origin is $R$.
The orthocentre $H$ of $\Delta ABC$ is given by $(x_H, y_H) = (R(\cos \alpha + \cos \beta + \cos \gamma), R(\sin \alpha + \sin \beta + \sin \gamma))$.
Since the orthocentre lies on the $y$-axis,the $x$-coordinate must be zero:
$R(\cos \alpha + \cos \beta + \cos \gamma) = 0 \implies \cos \alpha + \cos \beta + \cos \gamma = 0$.
Using the identity $\cos^3 \theta = \frac{1}{4}(\cos 3\theta + 3\cos \theta)$,we have $\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4(\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma) - 3(\cos \alpha + \cos \beta + \cos \gamma)$.
Since $\cos \alpha + \cos \beta + \cos \gamma = 0$,we use the property that if $a+b+c=0$,then $a^3+b^3+c^3 = 3abc$.
Thus,$\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma = 3 \cos \alpha \cos \beta \cos \gamma$.
Substituting this into the expression:
$\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos \alpha \cos \beta \cos \gamma} = \frac{4(3 \cos \alpha \cos \beta \cos \gamma) - 3(0)}{\cos \alpha \cos \beta \cos \gamma} = 12$.
Finally,the square of this value is $12^2 = 144$.

(D) Given $15 \sin^{4} \alpha + 10 \cos^{4} \alpha = 6$.
We know that $6 = 6(\sin^{2} \alpha + \cos^{2} \alpha)^{2} = 6(\sin^{4} \alpha + \cos^{4} \alpha + 2 \sin^{2} \alpha \cos^{2} \alpha)$.
Substituting this into the equation:
$15 \sin^{4} \alpha + 10 \cos^{4} \alpha = 6 \sin^{4} \alpha + 6 \cos^{4} \alpha + 12 \sin^{2} \alpha \cos^{2} \alpha$.
$9 \sin^{4} \alpha + 4 \cos^{4} \alpha - 12 \sin^{2} \alpha \cos^{2} \alpha = 0$.
$(3 \sin^{2} \alpha - 2 \cos^{2} \alpha)^{2} = 0$.
This implies $3 \sin^{2} \alpha = 2 \cos^{2} \alpha$,so $\tan^{2} \alpha = \frac{2}{3}$ and $\cot^{2} \alpha = \frac{3}{2}$.
Now,$27 \sec^{6} \alpha + 8 \operatorname{cosec}^{6} \alpha = 27(1 + \tan^{2} \alpha)^{3} + 8(1 + \cot^{2} \alpha)^{3}$.
$= 27(1 + \frac{2}{3})^{3} + 8(1 + \frac{3}{2})^{3}$.
$= 27(\frac{5}{3})^{3} + 8(\frac{5}{2})^{3}$.
$= 27 \times \frac{125}{27} + 8 \times \frac{125}{8}$.
$= 125 + 125 = 250$.

(B) Given $\cos x + \cos y - \cos(x + y) = \frac{3}{2}$.
Using the identity $\cos x + \cos y = 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2})$ and $\cos(x+y) = 2 \cos^2(\frac{x+y}{2}) - 1$,we get:
$2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) - (2 \cos^2(\frac{x+y}{2}) - 1) = \frac{3}{2}$
$-2 \cos^2(\frac{x+y}{2}) + 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + 1 = \frac{3}{2}$
$2 \cos^2(\frac{x+y}{2}) - 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + \frac{1}{2} = 0$
Multiply by $2$:
$4 \cos^2(\frac{x+y}{2}) - 4 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + 1 = 0$
This is a quadratic in $\cos(\frac{x+y}{2})$. Completing the square:
$(2 \cos(\frac{x+y}{2}) - \cos(\frac{x-y}{2}))^2 + 1 - \cos^2(\frac{x-y}{2}) = 0$
$(2 \cos(\frac{x+y}{2}) - \cos(\frac{x-y}{2}))^2 + \sin^2(\frac{x-y}{2}) = 0$
Since both terms are squares,they must be zero:
$\sin(\frac{x-y}{2}) = 0 \Rightarrow x = y$
$2 \cos(\frac{x+y}{2}) - \cos(0) = 0$ $\Rightarrow 2 \cos x = 1$ $\Rightarrow \cos x = \frac{1}{2}$
Since $0 < x < \pi$,$x = \frac{\pi}{3}$.
Then $\sin x = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos y = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
Therefore,$\sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1+\sqrt{3}}{2}$.

(D) We are given the sum $p = \sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^{2}}$.
First,multiply the numerator and denominator by $2$ to use the formula $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \frac{x-y}{1+xy}$.
$p = \sum_{r=1}^{50} \tan ^{-1} \frac{2}{4 r^{2}} = \sum_{r=1}^{50} \tan ^{-1} \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$.
This is a telescoping series of the form $\sum_{r=1}^{n} (\tan ^{-1}(a_{r+1}) - \tan ^{-1}(a_r))$.
$p = (\tan ^{-1} 3 - \tan ^{-1} 1) + (\tan ^{-1} 5 - \tan ^{-1} 3) + \dots + (\tan ^{-1} 101 - \tan ^{-1} 99)$.
All intermediate terms cancel out,leaving $p = \tan ^{-1} 101 - \tan ^{-1} 1$.
Using the formula $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \frac{x-y}{1+xy}$,we get $p = \tan ^{-1} \frac{101-1}{1+(101)(1)} = \tan ^{-1} \frac{100}{102} = \tan ^{-1} \frac{50}{51}$.
Therefore,$\tan p = \frac{50}{51}$.

(C) Let the expression be $E = 2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$.
Using the property $\sin(\pi - \theta) = \sin \theta$,we have $\sin(\frac{7\pi}{8}) = \sin(\frac{\pi}{8})$,$\sin(\frac{6\pi}{8}) = \sin(\frac{2\pi}{8})$,and $\sin(\frac{5\pi}{8}) = \sin(\frac{3\pi}{8})$.
Substituting these,$E = 2 \sin^2(\frac{\pi}{8}) \sin^2(\frac{2\pi}{8}) \sin^2(\frac{3\pi}{8})$.
Since $\sin(\frac{2\pi}{8}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,then $\sin^2(\frac{2\pi}{8}) = \frac{1}{2}$.
Thus,$E = 2 \cdot \frac{1}{2} \cdot \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8}) \cos^2(\frac{\pi}{8})$ (as $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$).
$E = (\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{2\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{\pi}{4}))^2 = (\frac{1}{2} \cdot \frac{1}{\sqrt{2}})^2 = \frac{1}{4 \cdot 2} = \frac{1}{8}$.

(D) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.
Therefore,$\operatorname{cosec} 18^{\circ} = \frac{1}{\sin 18^{\circ}} = \frac{4}{\sqrt{5}-1}$.
Rationalizing the denominator,we get $\operatorname{cosec} 18^{\circ} = \frac{4(\sqrt{5}+1)}{5-1} = \sqrt{5}+1$.
Let $x = \sqrt{5}+1$.
Then $x-1 = \sqrt{5}$.
Squaring both sides,we get $(x-1)^{2} = 5$.
$x^{2}-2x+1 = 5$.
$x^{2}-2x-4 = 0$.

(B) Given equation: $32^{\tan^{2} x} + 32^{\sec^{2} x} = 81$
Using the identity $\sec^{2} x = 1 + \tan^{2} x$,we get:
$32^{\tan^{2} x} + 32^{1 + \tan^{2} x} = 81$
Let $y = 32^{\tan^{2} x}$. Then the equation becomes:
$y + 32y = 81$
$33y = 81$
$y = \frac{81}{33} = \frac{27}{11}$
So,$32^{\tan^{2} x} = \frac{27}{11}$.
Taking $\log_{32}$ on both sides:
$\tan^{2} x = \log_{32} \left(\frac{27}{11}\right)$.
Since $0 \leq x \leq \frac{\pi}{4}$,we have $0 \leq \tan^{2} x \leq 1$.
Since $1 < \frac{27}{11} < 32$,it follows that $0 < \log_{32} \left(\frac{27}{11}\right) < 1$.
Thus,there exists exactly one value of $\tan x$ in the interval $[0, 1]$,which corresponds to exactly one value of $x$ in $[0, \frac{\pi}{4}]$.
Therefore,the number of solutions is $1$.

(C) Given $\sin \theta + \cos \theta = \frac{1}{2}$.
Squaring both sides: $(\sin \theta + \cos \theta)^2 = (\frac{1}{2})^2$ $\Rightarrow 1 + \sin(2\theta) = \frac{1}{4}$ $\Rightarrow \sin(2\theta) = -\frac{3}{4}$.
Then $\cos^2(2\theta) = 1 - \sin^2(2\theta) = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$.
We need to evaluate $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$.
Using sum-to-product formula: $\sin(2\theta) + \sin(6\theta) = 2\sin(4\theta)\cos(2\theta)$.
So the expression is $16(2\sin(4\theta)\cos(2\theta) + \cos(4\theta))$.
Since $\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$ and $\cos(4\theta) = 2\cos^2(2\theta) - 1$:
$16(2(2\sin(2\theta)\cos(2\theta))\cos(2\theta) + 2\cos^2(2\theta) - 1) = 16(4\sin(2\theta)\cos^2(2\theta) + 2\cos^2(2\theta) - 1)$.
Substituting values: $16(4(-\frac{3}{4})(\frac{7}{16}) + 2(\frac{7}{16}) - 1) = 16(-\frac{21}{16} + \frac{14}{16} - 1) = 16(-\frac{7}{16} - 1) = -7 - 16 = -23$.

(A) Given that $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression,we have $2x = \tan \left(\frac{\pi}{9}\right) + \tan \left(\frac{7 \pi}{18}\right)$,so $x = \frac{1}{2} \left(\tan \frac{\pi}{9} + \tan \frac{7 \pi}{18}\right)$.
Given that $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are in arithmetic progression,we have $2y = \tan \left(\frac{\pi}{9}\right) + \tan \left(\frac{5 \pi}{18}\right)$.
We need to find $|x - 2y|$.
Substituting the expressions:
$x - 2y = \frac{1}{2} \left(\tan \frac{\pi}{9} + \tan \frac{7 \pi}{18}\right) - \left(\tan \frac{\pi}{9} + \tan \frac{5 \pi}{18}\right)$.
Using the identities $\tan \left(\frac{7 \pi}{18}\right) = \cot \left(\frac{\pi}{9}\right)$ and $\tan \left(\frac{5 \pi}{18}\right) = \cot \left(\frac{2 \pi}{9}\right)$:
$x - 2y = \frac{1}{2} \tan \frac{\pi}{9} + \frac{1}{2} \cot \frac{\pi}{9} - \tan \frac{\pi}{9} - \cot \frac{2 \pi}{9} = \frac{1}{2} \left(\cot \frac{\pi}{9} - \tan \frac{\pi}{9}\right) - \cot \frac{2 \pi}{9}$.
Since $\cot \theta - \tan \theta = 2 \cot(2\theta)$:
$x - 2y = \frac{1}{2} \left(2 \cot \frac{2 \pi}{9}\right) - \cot \frac{2 \pi}{9} = \cot \frac{2 \pi}{9} - \cot \frac{2 \pi}{9} = 0$.
Therefore,$|x - 2y| = 0$.

(D) We have the expression $2 \sin(12^{\circ}) - \sin(72^{\circ})$.
Using the identity $\sin(C) - \sin(D) = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$,we rewrite the expression as:
$\sin(12^{\circ}) + (\sin(12^{\circ}) - \sin(72^{\circ}))$
$= \sin(12^{\circ}) - 2 \cos\left(\frac{12^{\circ}+72^{\circ}}{2}\right) \sin\left(\frac{72^{\circ}-12^{\circ}}{2}\right)$
$= \sin(12^{\circ}) - 2 \cos(42^{\circ}) \sin(30^{\circ})$
Since $\sin(30^{\circ}) = \frac{1}{2}$,we get:
$= \sin(12^{\circ}) - \cos(42^{\circ})$
$= \sin(12^{\circ}) - \sin(90^{\circ} - 42^{\circ}) = \sin(12^{\circ}) - \sin(48^{\circ})$
Using $\sin(C) - \sin(D) = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ again:
$= 2 \cos\left(\frac{12^{\circ}+48^{\circ}}{2}\right) \sin\left(\frac{12^{\circ}-48^{\circ}}{2}\right)$
$= 2 \cos(30^{\circ}) \sin(-18^{\circ}) = -2 \cos(30^{\circ}) \sin(18^{\circ})$
Substituting $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(18^{\circ}) = \frac{\sqrt{5}-1}{4}$:
$= -2 \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{5}-1}{4}$
$= -\frac{\sqrt{3}(\sqrt{5}-1)}{4} = \frac{\sqrt{3}(1-\sqrt{5})}{4}$

(B) We use the identity $\sin(\theta) \sin(60^{\circ}-\theta) \sin(60^{\circ}+\theta) = \frac{1}{4} \sin(3\theta)$.
Given expression: $16 \sin(20^{\circ}) \sin(40^{\circ}) \sin(80^{\circ})$.
This can be written as $16 \sin(20^{\circ}) \sin(60^{\circ}-20^{\circ}) \sin(60^{\circ}+20^{\circ})$.
Using the identity with $\theta = 20^{\circ}$:
$= 16 \times \left( \frac{1}{4} \sin(3 \times 20^{\circ}) \right)$.
$= 4 \sin(60^{\circ})$.
$= 4 \times \frac{\sqrt{3}}{2} = 2 \sqrt{3}$.

(C) Let $S = \sin^{2}(10^{\circ}) \sin(20^{\circ}) \sin(40^{\circ}) \sin(50^{\circ}) \sin(70^{\circ})$.
Using the identity $\sin(\theta) \sin(60^{\circ}-\theta) \sin(60^{\circ}+\theta) = \frac{1}{4} \sin(3\theta)$,we have $\sin(10^{\circ}) \sin(50^{\circ}) \sin(70^{\circ}) = \frac{1}{4} \sin(30^{\circ}) = \frac{1}{8}$.
Substituting this into the expression:
$S = \sin(10^{\circ}) \cdot \sin(20^{\circ}) \sin(40^{\circ}) \cdot [\sin(10^{\circ}) \sin(50^{\circ}) \sin(70^{\circ})]$
$S = \sin(10^{\circ}) \cdot \sin(20^{\circ}) \sin(40^{\circ}) \cdot \frac{1}{8}$.
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$S = \frac{1}{8} \sin(10^{\circ}) \cdot \frac{1}{2} [\cos(20^{\circ}) - \cos(60^{\circ})] = \frac{1}{16} \sin(10^{\circ}) [\cos(20^{\circ}) - \frac{1}{2}]$
$S = \frac{1}{16} \sin(10^{\circ}) \cos(20^{\circ}) - \frac{1}{32} \sin(10^{\circ})$.
Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$S = \frac{1}{32} [\sin(30^{\circ}) + \sin(-10^{\circ})] - \frac{1}{32} \sin(10^{\circ})$
$S = \frac{1}{32} [\frac{1}{2} - \sin(10^{\circ})] - \frac{1}{32} \sin(10^{\circ}) = \frac{1}{64} - \frac{1}{16} \sin(10^{\circ})$.
Comparing with $\alpha - \frac{1}{16} \sin(10^{\circ})$,we get $\alpha = \frac{1}{64}$.
Thus,$16 + \alpha^{-1} = 16 + 64 = 80$.

(A) Given equation: $3 \cos^2 2\theta + 6 \cos 2\theta - 10 \cos^2 \theta + 5 = 0$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$3 \cos^2 2\theta + 6 \cos 2\theta - 10(\frac{1 + \cos 2\theta}{2}) + 5 = 0$
$3 \cos^2 2\theta + 6 \cos 2\theta - 5 - 5 \cos 2\theta + 5 = 0$
$3 \cos^2 2\theta + \cos 2\theta = 0$
$\cos 2\theta(3 \cos 2\theta + 1) = 0$
This gives two cases:
Case $1$: $\cos 2\theta = 0$
For $\theta \in [-4\pi, 4\pi]$,$2\theta \in [-8\pi, 8\pi]$.
$\cos 2\theta = 0 \implies 2\theta = (2n+1)\frac{\pi}{2}$,where $n \in \{-8, -7, \dots, 7\}$.
There are $16$ values for $2\theta$ in the interval $[-8\pi, 8\pi]$,so there are $16$ values for $\theta$.
Case $2$: $\cos 2\theta = -\frac{1}{3}$
Since $-1 < -\frac{1}{3} < 1$,there are $2$ solutions for $2\theta$ in each interval of length $2\pi$.
In the interval $[-8\pi, 8\pi]$ (length $16\pi$),there are $8 \times 2 = 16$ solutions.
Total number of elements $= 16 + 16 = 32$.

(B) Let $S = 2 \sin \frac{\pi}{22} \sin \frac{3 \pi}{22} \sin \frac{5 \pi}{22} \sin \frac{7 \pi}{22} \sin \frac{9 \pi}{22}$.
Using $\sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$,we have:
$\sin \frac{\pi}{22} = \cos \frac{10 \pi}{22} = \cos \frac{5 \pi}{11}$
$\sin \frac{3 \pi}{22} = \cos \frac{8 \pi}{22} = \cos \frac{4 \pi}{11}$
$\sin \frac{5 \pi}{22} = \cos \frac{6 \pi}{22} = \cos \frac{3 \pi}{11}$
$\sin \frac{7 \pi}{22} = \cos \frac{4 \pi}{22} = \cos \frac{2 \pi}{11}$
$\sin \frac{9 \pi}{22} = \cos \frac{2 \pi}{22} = \cos \frac{\pi}{11}$
Thus,$S = 2 \cos \frac{\pi}{11} \cos \frac{2 \pi}{11} \cos \frac{3 \pi}{11} \cos \frac{4 \pi}{11} \cos \frac{5 \pi}{11}$.
Using the formula $\prod_{k=1}^{n} \cos \frac{k \pi}{2n+1} = \frac{1}{2^n}$,for $n=5$,we have $\prod_{k=1}^{5} \cos \frac{k \pi}{11} = \frac{1}{2^5} = \frac{1}{32}$.
Therefore,$S = 2 \times \frac{1}{32} = \frac{1}{16}$.

(C) Let $y = 8^{2 \sin^2 \theta}$. Since $\cos^2 \theta = 1 - \sin^2 \theta$,the equation becomes $y + 8^{2(1 - \sin^2 \theta)} = 16$,which is $y + \frac{64}{y} = 16$.
Multiplying by $y$,we get $y^2 - 16y + 64 = 0$,so $(y - 8)^2 = 0$,which gives $y = 8$.
Thus,$8^{2 \sin^2 \theta} = 8^1$,implying $2 \sin^2 \theta = 1$,or $\sin^2 \theta = \frac{1}{2}$.
This means $\sin \theta = \pm \frac{1}{\sqrt{2}}$,so $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. Thus,$n(S) = 4$.
For each $\theta \in S$,$2\theta \in \{\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\}$.
Then $\frac{\pi}{4} + 2\theta \in \{\frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}, \frac{15\pi}{4}\}$.
The expression is $\sum \frac{1}{\cos(\frac{\pi}{4} + 2\theta) \sin(\frac{\pi}{4} + 2\theta)} = \sum \frac{2}{\sin(\frac{\pi}{2} + 4\theta)} = \sum \frac{2}{\cos(4\theta)}$.
For all $\theta \in S$,$4\theta$ is an odd multiple of $\pi$,so $\cos(4\theta) = -1$.
Thus,the sum is $\sum_{\theta \in S} (-2) = 4 \times (-2) = -8$.
Finally,$n(S) + \text{sum} = 4 - 8 = -4$.

(C) Let $\alpha = \theta + (m-1) \frac{\pi}{6}$ and $\beta = \theta + m \frac{\pi}{6}$.
Then $\beta - \alpha = \frac{\pi}{6}$.
The sum is given by $\sum_{m=1}^{9} \sec \alpha \sec \beta = \sum_{m=1}^{9} \frac{1}{\cos \alpha \cos \beta}$.
Multiplying and dividing by $\sin(\beta - \alpha) = \sin(\frac{\pi}{6}) = \frac{1}{2}$,we get:
$2 \sum_{m=1}^{9} \frac{\sin(\beta - \alpha)}{\cos \alpha \cos \beta} = 2 \sum_{m=1}^{9} (\tan \beta - \tan \alpha)$.
This is a telescoping sum: $2 \left( \tan \left( \theta + \frac{9\pi}{6} \right) - \tan \theta \right) = 2 (\tan(\theta + \frac{3\pi}{2}) - \tan \theta) = 2(-\cot \theta - \tan \theta)$.
Given $2(-\cot \theta - \tan \theta) = -\frac{8}{\sqrt{3}}$,we have $\tan \theta + \cot \theta = \frac{4}{\sqrt{3}}$.
Let $x = \tan \theta$,then $x + \frac{1}{x} = \frac{4}{\sqrt{3}} \implies \sqrt{3}x^2 - 4x + \sqrt{3} = 0$.
Solving the quadratic: $(\sqrt{3}x - 1)(x - \sqrt{3}) = 0$,so $\tan \theta = \frac{1}{\sqrt{3}}$ or $\tan \theta = \sqrt{3}$.
Thus,$\theta = \frac{\pi}{6}$ or $\theta = \frac{\pi}{3}$.
$S = \left\{ \frac{\pi}{6}, \frac{\pi}{3} \right\}$,so $\sum_{\theta \in S} \theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.

(A) Given equation: $2 \cos \left(\frac{x^{2}+x}{6}\right) = 4^{x} + 4^{-x}$
We know that the range of the cosine function is $[-1, 1]$,so $2 \cos \left(\frac{x^{2}+x}{6}\right) \leq 2$.
For the $R$.$H$.$S$.,by the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,we have $\frac{4^{x} + 4^{-x}}{2} \geq \sqrt{4^{x} \cdot 4^{-x}} = 1$,which implies $4^{x} + 4^{-x} \geq 2$.
For the equation to hold,both sides must be equal to $2$.
$2 \cos \left(\frac{x^{2}+x}{6}\right) = 2 \implies \cos \left(\frac{x^{2}+x}{6}\right) = 1$
$4^{x} + 4^{-x} = 2 \implies (2^{x} - 2^{-x})^{2} + 2 = 2 \implies 2^{x} = 2^{-x} \implies x = 0$.
Substituting $x = 0$ into the cosine part: $\cos \left(\frac{0^{2}+0}{6}\right) = \cos(0) = 1$. This satisfies the equation.
Thus,there is only $1$ element in the set $S$.

(B) Given the trigonometric relation:
$(\frac{1}{3} \sin \theta + \frac{2}{3} \cos \theta)^2 = \frac{1}{3} \sin^2 \theta + \frac{2}{3} \cos^2 \theta$
Expanding the left side:
$\frac{1}{9} \sin^2 \theta + \frac{4}{9} \cos^2 \theta + \frac{4}{9} \sin \theta \cos \theta = \frac{1}{3} \sin^2 \theta + \frac{2}{3} \cos^2 \theta$
Multiplying by $9$:
$\sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 3 \sin^2 \theta + 6 \cos^2 \theta$
Rearranging terms:
$2 \sin^2 \theta + 2 \cos^2 \theta - 4 \sin \theta \cos \theta = 0$
Dividing by $2$:
$\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = 0$
$(\sin \theta - \cos \theta)^2 = 0$
$\sin \theta = \cos \theta \Rightarrow \tan \theta = 1$
For $\theta \in [0, \pi]$,$\tan \theta = 1$ implies $\theta = \frac{\pi}{4}$.
Thus,$A \cap [0, \pi] = \{\frac{\pi}{4}\}$,which has exactly one point.

(D) We are given the set $S = \{x \in R : \cos(x) + \cos(\sqrt{2}x) < 2\}$.
We know that the maximum value of the cosine function is $1$,which occurs when the argument is an integer multiple of $2\pi$.
For $\cos(x) + \cos(\sqrt{2}x) = 2$,both $\cos(x)$ and $\cos(\sqrt{2}x)$ must simultaneously be equal to $1$.
This implies $x = 2n\pi$ and $\sqrt{2}x = 2m\pi$ for some integers $n, m$.
If $x = 0$,then $n = 0$ and $m = 0$,which satisfies the condition $\cos(0) + \cos(0) = 1 + 1 = 2$.
If $x \neq 0$,then $\frac{\sqrt{2}x}{x} = \frac{2m\pi}{2n\pi} \implies \sqrt{2} = \frac{m}{n}$,which is impossible since $\sqrt{2}$ is irrational.
Thus,the sum $\cos(x) + \cos(\sqrt{2}x)$ is equal to $2$ if and only if $x = 0$,and it is strictly less than $2$ for all $x \in R \setminus \{0\}$.
Therefore,$S = R \setminus \{0\}$.

(B) Given the equation: $\cos^2(x \sin(2x)) + \frac{1}{1+x^2} = \cos^2 x + \sec^2 x$.
Consider the Left Hand Side $(LHS)$: $f(x) = \cos^2(x \sin(2x)) + \frac{1}{1+x^2}$.
Since $\cos^2(\theta) \leq 1$ and $\frac{1}{1+x^2} \leq 1$ for all real $x$,we have $f(x) \leq 1 + 1 = 2$.
Consider the Right Hand Side $(RHS)$: $g(x) = \cos^2 x + \sec^2 x$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,$\cos^2 x + \sec^2 x \geq 2 \sqrt{\cos^2 x \cdot \sec^2 x} = 2 \sqrt{1} = 2$.
For the equation to hold,we must have $f(x) = 2$ and $g(x) = 2$.
$g(x) = 2$ occurs when $\cos^2 x = \sec^2 x$,which implies $\cos^4 x = 1$,so $\cos x = \pm 1$,meaning $x = n\pi$ for $n \in \mathbb{Z}$.
If $x = 0$,$f(0) = \cos^2(0) + \frac{1}{1+0} = 1 + 1 = 2$ and $g(0) = \cos^2(0) + \sec^2(0) = 1 + 1 = 2$.
If $x = n\pi$ where $n \neq 0$,then $\frac{1}{1+x^2} < 1$,so $f(x) < 2$,which contradicts $f(x) = 2$.
Thus,the only real solution is $x = 0$.

(A) Given the parametric equations: $x(\theta) = |\cos 4\theta| \cos \theta$ and $y(\theta) = |\cos 4\theta| \sin \theta$.
We can express this in polar form by noting that $r^2 = x^2 + y^2 = |\cos 4\theta|^2 (\cos^2 \theta + \sin^2 \theta) = \cos^2 4\theta$. Thus,$r = |\cos 4\theta|$.
The curve $r = |\cos 4\theta|$ is a rose curve with $8$ petals because the coefficient of $\theta$ is $4$ (even,so $2n = 8$ petals).
Evaluating points:

$\theta$	$x(\theta)$	$y(\theta)$
$0$	$1$	$0$
$45^{\circ}$	$0$	$0$
$90^{\circ}$	$0$	$-1$
$180^{\circ}$	$-1$	$0$

The graph shows $8$ petals symmetric about the axes,which matches the first option.

(C) Let $S = \sum_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$.
Multiply and divide by $\sin 1^{\circ}$:
$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} \frac{\sin((k+1)^{\circ} - k^{\circ})}{\cos k^{\circ} \cos (k+1)^{\circ}}$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} (\tan (k+1)^{\circ} - \tan k^{\circ})$.
This is a telescoping sum:
$S = \frac{1}{\sin 1^{\circ}} [(\tan 1^{\circ} - \tan 0^{\circ}) + (\tan 2^{\circ} - \tan 1^{\circ}) + \dots + (\tan 45^{\circ} - \tan 44^{\circ})]$.
$S = \frac{1}{\sin 1^{\circ}} (\tan 45^{\circ} - \tan 0^{\circ}) = \frac{1}{\sin 1^{\circ}} (1 - 0) = \frac{1}{\sin 1^{\circ}}$.
Since $\sin 1^{\circ} \approx 0.01745$,we have $S \approx \frac{1}{0.01745} \approx 57.299$.
The integer part of $57.299$ is $57$.

(B) Given the equation: $\frac{\sin (\lambda \alpha) \cos (\lambda \alpha)}{\sin \alpha \cos \alpha} = \lambda - 1$.
Multiply both sides by $2$ to simplify the numerator and denominator:
$\frac{2 \sin (\lambda \alpha) \cos (\lambda \alpha)}{2 \sin \alpha \cos \alpha} = \lambda - 1$
$\Rightarrow \frac{\sin (2 \lambda \alpha)}{\sin (2 \alpha)} = \lambda - 1$
$\Rightarrow \sin (2 \lambda \alpha) = (\lambda - 1) \sin (2 \alpha)$.
For this to hold for all $\alpha$,we compare the derivatives with respect to $\alpha$ at $\alpha = 0$:
$2 \lambda \cos (2 \lambda \alpha) = 2 (\lambda - 1) \cos (2 \alpha)$.
At $\alpha = 0$,$2 \lambda = 2 (\lambda - 1) \Rightarrow \lambda = \lambda - 1$,which is impossible.
However,checking specific values:
If $\lambda = 1$,$\sin (2 \alpha) = 0 \cdot \sin (2 \alpha) = 0$,which is not true for all $\alpha$.
If $\lambda = 0$,$\sin (0) = -1 \cdot \sin (2 \alpha)$,not true.
Re-evaluating the original expression: $\frac{\sin (2 \lambda \alpha)}{2 \sin \alpha \cos \alpha} = \lambda - 1$ $\Rightarrow \sin (2 \lambda \alpha) = 2(\lambda - 1) \sin \alpha \cos \alpha = (\lambda - 1) \sin (2 \alpha)$.
This holds if $2 \lambda = 2$ and $\lambda - 1 = 1$ (i.e.,$\lambda = 2$) or if $\lambda - 1 = 0$ and $\sin (2 \lambda \alpha) = 0$ (i.e.,$\lambda = 1$).
Thus,$\lambda = 1$ and $\lambda = 2$ are the solutions.

(D) Given the function $f(x) = \frac{\sin(x-a) + \sin(x+a)}{\cos(x-a) - \cos(x+a)}$.
Using the trigonometric identities $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$ and $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$,we get:
$f(x) = \frac{2\sin x \cos a}{2\sin x \sin a}$
Assuming $\sin x \neq 0$,we can simplify the expression:
$f(x) = \frac{\cos a}{\sin a} = \cot a$
Since $a$ is a constant,$\cot a$ is also a constant.
Therefore,$f(x)$ is a constant function.

(D) We have,$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$
$= (\tan 81^{\circ} + \tan 9^{\circ}) - (\tan 63^{\circ} + \tan 27^{\circ})$
$= (\cot 9^{\circ} + \tan 9^{\circ}) - (\cot 27^{\circ} + \tan 27^{\circ})$
$= \left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}} + \frac{\sin 9^{\circ}}{\cos 9^{\circ}}\right) - \left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}} + \frac{\sin 27^{\circ}}{\cos 27^{\circ}}\right)$
$= \left(\frac{\cos^2 9^{\circ} + \sin^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}\right) - \left(\frac{\cos^2 27^{\circ} + \sin^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}\right)$
$= \left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}\right) - \left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}\right)$
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
$= 2 \left(\frac{1}{\sin 18^{\circ}} - \frac{1}{\cos 36^{\circ}}\right)$
$= 2 \left(\frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1}\right)$
$= 8 \left(\frac{\sqrt{5} + 1 - (\sqrt{5} - 1)}{5 - 1}\right)$
$= 8 \left(\frac{2}{4}\right) = 4$

(C) Let $P = (1+\tan 1^{\circ})(1+\tan 2^{\circ}) \dots (1+\tan 44^{\circ})(1+\tan 45^{\circ})$.
We use the identity $(1+\tan \theta)(1+\tan(45^{\circ}-\theta)) = 1 + \tan \theta + \tan(45^{\circ}-\theta) + \tan \theta \tan(45^{\circ}-\theta)$.
Since $\tan(45^{\circ}-\theta) = \frac{1-\tan \theta}{1+\tan \theta}$,the expression becomes $1 + \tan \theta + \frac{1-\tan \theta}{1+\tan \theta} + \tan \theta \left(\frac{1-\tan \theta}{1+\tan \theta}\right) = 1 + \frac{(1+\tan \theta)^2 + (1-\tan \theta) + \tan \theta - \tan^2 \theta}{1+\tan \theta} = 1 + \frac{1+2\tan \theta + \tan^2 \theta + 1 - \tan^2 \theta}{1+\tan \theta} = 1 + \frac{2+2\tan \theta}{1+\tan \theta} = 1 + 2 = 3$ (Wait,the identity is $(1+\tan \theta)(1+\tan(45^{\circ}-\theta)) = 2$).
Pairing terms: $(1+\tan 1^{\circ})(1+\tan 44^{\circ}) = 2$,$(1+\tan 2^{\circ})(1+\tan 43^{\circ}) = 2$,...,$(1+\tan 22^{\circ})(1+\tan 23^{\circ}) = 2$.
There are $22$ such pairs,so the product of these is $2^{22}$.
Finally,we have the term $(1+\tan 45^{\circ}) = 1+1 = 2$.
Thus,the total product is $2^{22} \times 2 = 2^{23}$.

(A) Given the equation: $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$.
Rearranging the terms,we get: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\beta-\alpha)$.
Since $\sin^2(\beta-\alpha) = \sin^2(\alpha-\beta)$,the equation becomes: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)$.
Using the identity $\cos^2\theta+\sin^2\theta=1$,we have: $\operatorname{cosec}^2(\alpha+\beta)+\sin^2(2\alpha-\beta)=1$.
Since $\operatorname{cosec}^2(\alpha+\beta) \geq 1$ and $\sin^2(2\alpha-\beta) \geq 0$,the sum can be $1$ only if $\operatorname{cosec}^2(\alpha+\beta)=1$ and $\sin^2(2\alpha-\beta)=0$.
This implies $\alpha+\beta = \frac{\pi}{2}$ and $2\alpha-\beta = 0$.
Adding the two equations: $3\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{6}$.
Substituting $\alpha$ back: $\beta = 2\alpha = \frac{\pi}{3}$.
Thus,$\sin(\alpha-\beta) = \sin(\frac{\pi}{6}-\frac{\pi}{3}) = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$.

(A) Given: $\sin x + \sin y = \frac{7}{5}$ $(i)$
And: $\cos x + \cos y = \frac{1}{5}$ $(ii)$
Using sum-to-product formulas:
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{7}{5}$ $(iii)$
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{1}{5}$ $(iv)$
Dividing $(iii)$ by $(iv)$:
$\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{7/5}{1/5}$
$\tan \left(\frac{x+y}{2}\right) = 7$
Using the identity $\sin(x+y) = \frac{2 \tan \left(\frac{x+y}{2}\right)}{1 + \tan^2 \left(\frac{x+y}{2}\right)}$:
$\sin(x+y) = \frac{2(7)}{1 + 7^2} = \frac{14}{1 + 49} = \frac{14}{50} = \frac{7}{25}$

(C) Let $a, b, c$ be the sides of a triangle.
Since $a^2+b^2 \geq 2ab$,$b^2+c^2 \geq 2bc$,and $c^2+a^2 \geq 2ac$,adding these inequalities gives $2(a^2+b^2+c^2) \geq 2(ab+bc+ca)$,which implies $\frac{a^2+b^2+c^2}{ab+bc+ca} \geq 1$. Thus,$t \geq 1$.
For a triangle,the triangle inequality states $a+b > c$,$b+c > a$,and $c+a > b$.
Squaring $a+b > c$ gives $a^2+b^2+2ab > c^2$,or $a^2+b^2-c^2 > -2ab$. This does not directly bound $t$.
However,consider $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$ for distinct sides,which leads to $a^2+b^2+c^2 > ab+bc+ca$,so $t > 1$.
Using the triangle inequality $a < b+c$,we have $a^2 < a(b+c) = ab+ac$.
Similarly,$b^2 < ab+bc$ and $c^2 < ac+bc$.
Adding these,$a^2+b^2+c^2 < 2(ab+bc+ca)$,which implies $t < 2$.
Combining these,the set of values is $1 \leq t < 2$ (equality holds for equilateral triangles where $a=b=c$).
Therefore,the set is $\{x \in \mathbb{R} \mid 1 \leq x < 2\}$.

(D) The area of a regular $n$-sided polygon inscribed in a circle of radius $r=1$ is given by $A_n = n \times \frac{1}{2} \times r^2 \times \sin\left(\frac{2\pi}{n}\right) = \frac{n}{2} \sin\left(\frac{2\pi}{n}\right)$.
For a pentagon $(n=5)$,$X = \frac{5}{2} \sin\left(\frac{2\pi}{5}\right)$.
For a hexagon $(n=6)$,$Y = \frac{6}{2} \sin\left(\frac{2\pi}{6}\right)$.
For a heptagon $(n=7)$,$Z = \frac{7}{2} \sin\left(\frac{2\pi}{7}\right)$.
Dividing by $n$,we get $\frac{X}{5} = \frac{1}{2} \sin\left(\frac{2\pi}{5}\right)$,$\frac{Y}{6} = \frac{1}{2} \sin\left(\frac{2\pi}{6}\right)$,and $\frac{Z}{7} = \frac{1}{2} \sin\left(\frac{2\pi}{7}\right)$.
Since the function $f(\theta) = \sin(\theta)$ is decreasing for $\theta \in (0, \pi/2)$,and $\frac{2\pi}{5} > \frac{2\pi}{6} > \frac{2\pi}{7}$,it follows that $\sin\left(\frac{2\pi}{5}\right) > \sin\left(\frac{2\pi}{6}\right) > \sin\left(\frac{2\pi}{7}\right)$.
Thus,$\frac{X}{5} > \frac{Y}{6} > \frac{Z}{7}$.
As $n$ increases,the area of the inscribed polygon approaches the area of the circle $(\pi r^2 = \pi)$,so $X < Y < Z$.

(A) The angle $\theta$ between the hour hand and the minute hand at $H:M$ is given by the formula $\theta = |30H - 5.5M|$.
For $H = 6$ and $M = 15$:
$\theta = |30(6) - 5.5(15)| = |180 - 82.5| = 97.5^{\circ}$.
Let the two angles be $\alpha$ and $\beta$. We have $\alpha = 97.5^{\circ}$ and $\alpha + \beta = 360^{\circ}$.
Then $\beta = 360^{\circ} - 97.5^{\circ} = 262.5^{\circ}$.
The difference between these two angles is $\beta - \alpha = 262.5^{\circ} - 97.5^{\circ} = 165^{\circ}$.