Trigonometry Questions in English

(D) Using trigonometric identities:
$1$. $\tan(x - \frac{\pi}{2}) = -\cot x$
$2$. $\cos(\frac{3\pi}{2} + x) = \sin x$
$3$. $\sin(\frac{7\pi}{2} - x) = \sin(4\pi - \frac{\pi}{2} - x) = \sin(-\frac{\pi}{2} - x) = -\cos x$
$4$. $\cos(x - \frac{\pi}{2}) = \sin x$
$5$. $\tan(\frac{3\pi}{2} + x) = -\cot x$
Substituting these into the expression:
Numerator: $(-\cot x)(\sin x) - (-\cos x)^3 = -\cos x + \cos^3 x = \cos x(\cos^2 x - 1) = -\cos x \sin^2 x$
Denominator: $(\sin x)(-\cot x) = -\cos x$
Expression: $\frac{-\cos x \sin^2 x}{-\cos x} = \sin^2 x$.

(D) Let the expression be $E = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\sin \theta + \cos \theta}{\tan^2 \theta - 1}$.
First,simplify the second term: $\tan^2 \theta - 1 = \frac{\sin^2 \theta}{\cos^2 \theta} - 1 = \frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta} = \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\cos^2 \theta}$.
Substituting this into the second term: $\frac{\sin \theta + \cos \theta}{\frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\cos^2 \theta}} = \frac{(\sin \theta + \cos \theta) \cdot \cos^2 \theta}{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)} = \frac{\cos^2 \theta}{\sin \theta - \cos \theta}$.
Now,substitute back into $E$: $E = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\cos^2 \theta}{\sin \theta - \cos \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta - \cos \theta}$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $E = \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta - \cos \theta} = \sin \theta + \cos \theta$.
We know that $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + 45^\circ)$.
The range of $\sin(\theta + 45^\circ)$ is $[-1, 1]$,so the range of $\sqrt{2} \sin(\theta + 45^\circ)$ is $[-\sqrt{2}, \sqrt{2}]$.
Thus,the value lies between $-\sqrt{2}$ and $\sqrt{2}$.

(A) Given expression: $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1$
Using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,the first term simplifies to $\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta = 1 + \sin \theta \cos \theta$.
Since $\sqrt{1 + \cot^2 \theta} = \sqrt{\csc^2 \theta} = |\csc \theta|$,the second term is $\frac{\cos \theta}{|\csc \theta|} = \cos \theta |\sin \theta|$.
Also,$2 \tan \theta \cot \theta = 2(1) = 2$.
Substituting these: $(1 + \sin \theta \cos \theta) - \cos \theta |\sin \theta| - 2 = -1$.
$\sin \theta \cos \theta - \cos \theta |\sin \theta| = 0$.
$\cos \theta (\sin \theta - |\sin \theta|) = 0$.
This holds if $\cos \theta = 0$ or $\sin \theta = |\sin \theta|$.
$\sin \theta = |\sin \theta|$ implies $\sin \theta \ge 0$,which occurs in the first and second quadrants,i.e.,$\theta \in (0, \pi)$.
However,the denominator $\sin \theta - \cos \theta \neq 0$ implies $\theta \neq \frac{\pi}{4}, \frac{5\pi}{4}$.
Checking the options,$\theta \in (0, \pi/2)$ satisfies $\sin \theta > 0$,making the expression $1 + \sin \theta \cos \theta - \cos \theta \sin \theta - 2 = -1$.

(A) Given: $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.
Using the componendo and dividendo rule:
$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(2 - \cos \beta) - (2 \cos \beta - 1)}{(2 - \cos \beta) + (2 \cos \beta - 1)}$
$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{2 - \cos \beta - 2 \cos \beta + 1}{2 - \cos \beta + 2 \cos \beta - 1}$
$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{3 - 3 \cos \beta}{1 + \cos \beta} = 3 \left( \frac{1 - \cos \beta}{1 + \cos \beta} \right)$
Using the half-angle identity $\tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}$,we get:
$\tan^2 \frac{\alpha}{2} = 3 \tan^2 \frac{\beta}{2}$
$\frac{\tan^2 \frac{\alpha}{2}}{\tan^2 \frac{\beta}{2}} = 3$
Since $\frac{1}{\tan \frac{\beta}{2}} = \cot \frac{\beta}{2}$,we have:
$\tan^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2} = 3$
Taking the square root (since $\alpha, \beta \in (0, \pi)$,the tangent values are positive):
$\tan \frac{\alpha}{2} \cot \frac{\beta}{2} = \sqrt{3}$.

(A) Given the expression $E = \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}$.
Using $1 \pm \sin x = (\cos \frac{x}{2} \pm \sin \frac{x}{2})^2$,we have $\sqrt{1 \pm \sin x} = |\cos \frac{x}{2} \pm \sin \frac{x}{2}|$.
Since $\frac{5\pi}{2} < x < 3\pi$,then $\frac{5\pi}{4} < \frac{x}{2} < \frac{3\pi}{2}$.
In this interval,$\cos \frac{x}{2} < 0$ and $\sin \frac{x}{2} < 0$,and $|\cos \frac{x}{2}| > |\sin \frac{x}{2}|$.
Thus,$\sqrt{1 - \sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}| = -(\cos \frac{x}{2} - \sin \frac{x}{2}) = \sin \frac{x}{2} - \cos \frac{x}{2}$.
And $\sqrt{1 + \sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}| = -(\cos \frac{x}{2} + \sin \frac{x}{2}) = -\cos \frac{x}{2} - \sin \frac{x}{2}$.
Substituting these into the expression:
$E = \frac{(\sin \frac{x}{2} - \cos \frac{x}{2}) + (-\cos \frac{x}{2} - \sin \frac{x}{2})}{(\sin \frac{x}{2} - \cos \frac{x}{2}) - (-\cos \frac{x}{2} - \sin \frac{x}{2})} = \frac{-2\cos \frac{x}{2}}{2\sin \frac{x}{2}} = -\cot \frac{x}{2}$.

(B) Let $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right) = k$.
Then $x = \frac{k}{\sin \theta}$,$y = \frac{k}{\sin \left( \theta + \frac{2\pi}{3} \right)}$,and $z = \frac{k}{\sin \left( \theta + \frac{4\pi}{3} \right)}$.
Consider the sum $xy + yz + zx = k^2 \left[ \frac{1}{\sin \theta \sin \left( \theta + \frac{2\pi}{3} \right)} + \frac{1}{\sin \left( \theta + \frac{2\pi}{3} \right) \sin \left( \theta + \frac{4\pi}{3} \right)} + \frac{1}{\sin \left( \theta + \frac{4\pi}{3} \right) \sin \theta} \right]$.
Using the identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,the denominator terms simplify.
Alternatively,using the property that $\sin \theta + \sin \left( \theta + \frac{2\pi}{3} \right) + \sin \left( \theta + \frac{4\pi}{3} \right) = 0$,we can show that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.
Multiplying by $xyz$,we get $yz + zx + xy = 0$.

(C) Let $\theta = \frac{\pi}{10}$. The expression is $E = \cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \cos 16\theta$.
Using the formula $\cos \theta \cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}$,we have $n=5$.
$E = \frac{\sin(2^5 \theta)}{2^5 \sin \theta} = \frac{\sin(32 \theta)}{32 \sin \theta}$.
Since $\theta = \frac{\pi}{10}$,$32\theta = \frac{32\pi}{10} = 3\pi + \frac{2\pi}{10} = 3\pi + \frac{\pi}{5}$.
$E = \frac{\sin(3\pi + \pi/5)}{32 \sin(\pi/10)} = \frac{-\sin(\pi/5)}{32 \sin(\pi/10)}$.
Using $\sin(2A) = 2 \sin A \cos A$,we get $\sin(\pi/5) = 2 \sin(\pi/10) \cos(\pi/10)$.
$E = \frac{-2 \sin(\pi/10) \cos(\pi/10)}{32 \sin(\pi/10)} = -\frac{1}{16} \cos(\pi/10)$.

(C) Let $f(x) = cot\, x + cot\, (60^\circ + x) + cot\, (120^\circ + x)$.
Using the identity $cot\, A + cot\, B = \frac{\sin(A+B)}{\sin A \sin B}$,we have:
$cot\, (60^\circ + x) + cot\, (120^\circ + x) = \frac{\sin(180^\circ + 2x)}{\sin(60^\circ + x)\sin(120^\circ + x)}$.
Since $\sin(180^\circ + 2x) = -\sin 2x$ and $\sin(60^\circ + x)\sin(120^\circ + x) = \sin^2 60^\circ - \sin^2 x = \frac{3}{4} - \sin^2 x$,
$f(x) = cot\, x - \frac{\sin 2x}{\frac{3}{4} - \sin^2 x} = \frac{\cos x}{\sin x} - \frac{2\sin x \cos x}{\frac{3-4\sin^2 x}{4}} = \frac{\cos x}{\sin x} - \frac{8\sin x \cos x}{3-4\sin^2 x}$.
$f(x) = \frac{\cos x(3-4\sin^2 x) - 8\sin^2 x \cos x}{\sin x(3-4\sin^2 x)} = \frac{3\cos x - 4\sin^2 x \cos x - 8\sin^2 x \cos x}{3\sin x - 4\sin^3 x} = \frac{3\cos x - 12\sin^2 x \cos x}{3\sin x - 4\sin^3 x}$.
Dividing numerator and denominator by $\cos^3 x$ or using the triple angle formula:
$f(x) = \frac{3\cos x(1 - 4\sin^2 x)}{3\sin x - 4\sin^3 x} = \frac{3\cos x(1 - 4(1-\cos^2 x))}{\sin 3x} = \frac{3\cos x(4\cos^2 x - 3)}{\sin 3x} = \frac{3(4\cos^3 x - 3\cos x)}{\sin 3x} = \frac{3\cos 3x}{\sin 3x} = 3\, cot\, 3x$.

(A) Given expression: $E = \frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$.
Converting to sine and cosine: $E = \frac{3 + \frac{\cos 76^\circ \cos 16^\circ}{\sin 76^\circ \sin 16^\circ}}{\frac{\cos 76^\circ}{\sin 76^\circ} + \frac{\cos 16^\circ}{\sin 16^\circ}} = \frac{3 \sin 76^\circ \sin 16^\circ + \cos 76^\circ \cos 16^\circ}{\cos 76^\circ \sin 16^\circ + \sin 76^\circ \cos 16^\circ}$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,the denominator becomes $\sin(76^\circ + 16^\circ) = \sin 92^\circ$.
For the numerator,write $3 = 2 + 1$: $2 \sin 76^\circ \sin 16^\circ + (\sin 76^\circ \sin 16^\circ + \cos 76^\circ \cos 16^\circ)$.
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$,the term in brackets is $\cos(76^\circ - 16^\circ) = \cos 60^\circ = \frac{1}{2}$.
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,the first term is $\cos 60^\circ - \cos 92^\circ = \frac{1}{2} - \cos 92^\circ$.
Numerator $= \frac{1}{2} - \cos 92^\circ + \frac{1}{2} = 1 - \cos 92^\circ$.
Thus,$E = \frac{1 - \cos 92^\circ}{\sin 92^\circ} = \frac{2 \sin^2 46^\circ}{2 \sin 46^\circ \cos 46^\circ} = \tan 46^\circ = \cot(90^\circ - 46^\circ) = \cot 44^\circ$.

(D) Let $E = \csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18} = \frac{1}{\sin \frac{\pi}{18}} - \frac{\sqrt{3}}{\cos \frac{\pi}{18}}$.
$E = \frac{\cos \frac{\pi}{18} - \sqrt{3} \sin \frac{\pi}{18}}{\sin \frac{\pi}{18} \cos \frac{\pi}{18}}$.
Multiply numerator and denominator by $2$:
$E = \frac{2 \left( \frac{1}{2} \cos \frac{\pi}{18} - \frac{\sqrt{3}}{2} \sin \frac{\pi}{18} \right)}{\frac{1}{2} \sin \frac{2\pi}{18}} = \frac{2 \left( \sin \frac{\pi}{6} \cos \frac{\pi}{18} - \cos \frac{\pi}{6} \sin \frac{\pi}{18} \right)}{\frac{1}{2} \sin \frac{\pi}{9}}$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$E = \frac{2 \sin(\frac{\pi}{6} - \frac{\pi}{18})}{\frac{1}{2} \sin \frac{\pi}{9}} = \frac{4 \sin(\frac{3\pi - \pi}{18})}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{2\pi}{18}}{\sin \frac{\pi}{9}} = \frac{4 \sin \frac{\pi}{9}}{\sin \frac{\pi}{9}} = 4$.
Since $4$ is a natural number,the correct option is $D$.

(B) Let the perpendicular from the right-angled vertex to the hypotenuse be $p$.
From the geometry of the triangle,the two segments of the hypotenuse are $p \tan \theta$ and $p \cot \theta$,where $\theta$ is one of the acute angles.
The hypotenuse $h = p \tan \theta + p \cot \theta = p(\tan \theta + \cot \theta) = p \left( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \right) = p \left( \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \right) = \frac{p}{\sin \theta \cos \theta} = \frac{2p}{\sin 2\theta}$.
Given that $h = 2\sqrt{2} p$,we have $\frac{2p}{\sin 2\theta} = 2\sqrt{2} p$.
$\Rightarrow \sin 2\theta = \frac{1}{\sqrt{2}}$.
Thus,$2\theta = \frac{\pi}{4}$ or $2\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
$\Rightarrow \theta = \frac{\pi}{8}$ or $\theta = \frac{3\pi}{8}$.
Since the sum of acute angles in a right-angled triangle is $\frac{\pi}{2}$,the angles are $\frac{\pi}{8}$ and $\frac{3\pi}{8}$.

(D) We have the expression $\frac{\sec 8 \theta - 1}{\sec 4 \theta - 1}$.
Converting to cosine,we get $\frac{\frac{1}{\cos 8 \theta} - 1}{\frac{1}{\cos 4 \theta} - 1} = \frac{1 - \cos 8 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{1 - \cos 4 \theta}$.
Using the identity $1 - \cos 2A = 2 \sin^2 A$,we have $1 - \cos 8 \theta = 2 \sin^2 4 \theta$ and $1 - \cos 4 \theta = 2 \sin^2 2 \theta$.
Substituting these,we get $\frac{2 \sin^2 4 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{2 \sin^2 2 \theta}$.
Using $\sin 4 \theta = 2 \sin 2 \theta \cos 2 \theta$,we get $\frac{2 (2 \sin 2 \theta \cos 2 \theta) \sin 4 \theta \cos 4 \theta}{\cos 8 \theta (2 \sin^2 2 \theta)}$.
Simplifying,$\frac{2 \sin 4 \theta \cos 4 \theta}{\cos 8 \theta} \times \frac{\cos 2 \theta}{\sin 2 \theta} = \frac{\sin 8 \theta}{\cos 8 \theta} \times \cot 2 \theta = \tan 8 \theta \cot 2 \theta$.

(C) Given equation: $\cot x - \cos x = 1 - \cot x \cos x$.
Rearranging the terms: $\cot x - \cos x + \cot x \cos x - 1 = 0$.
Group the terms: $\cot x(1 + \cos x) - (1 + \cos x) = 0$.
Factor out $(1 + \cos x)$: $(1 + \cos x)(\cot x - 1) = 0$.
This implies either $1 + \cos x = 0$ or $\cot x - 1 = 0$.
Case $1$: $\cos x = -1$. In the interval $[0, 2\pi]$,$x = \pi$.
Case $2$: $\cot x = 1$. In the interval $[0, 2\pi]$,$x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
The values of $x$ are $\frac{\pi}{4}, \pi, \frac{5\pi}{4}$.
Thus,there are $3$ values of $x$ that satisfy the equation.

(C) Let the expression be $E = \cos^2 73^\circ + \cos^2 47^\circ + \cos 73^\circ \cos 47^\circ$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have:
$E = \frac{1 + \cos 146^\circ}{2} + \frac{1 + \cos 94^\circ}{2} + \cos 73^\circ \cos 47^\circ$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $\cos 73^\circ \cos 47^\circ = \frac{\cos 120^\circ + \cos 26^\circ}{2} = \frac{-1/2 + \cos 26^\circ}{2} = -1/4 + \frac{\cos 26^\circ}{2}$.
Substituting this back:
$E = 1 + \frac{\cos 146^\circ + \cos 94^\circ}{2} - 1/4 + \frac{\cos 26^\circ}{2}$.
Using $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$\cos 146^\circ + \cos 94^\circ = 2 \cos(120^\circ) \cos(26^\circ) = 2(-1/2) \cos 26^\circ = -\cos 26^\circ$.
Thus,$E = 1 - 1/4 - \frac{\cos 26^\circ}{2} + \frac{\cos 26^\circ}{2} = 3/4$.

(C) Let $y = (7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$.
We know that $7 \cos\theta + 24 \sin\theta = r \sin(\theta + \alpha)$,where $r = \sqrt{7^2 + 24^2} = 25$ and $\tan \alpha = \frac{7}{24}$.
Similarly,$7 \sin\theta - 24 \cos\theta = r \sin(\theta - \beta)$,where $\tan \beta = \frac{24}{7}$.
Alternatively,let $7 \cos\theta + 24 \sin\theta = 25 \sin(\theta + \alpha)$ and $7 \sin\theta - 24 \cos\theta = -25 \cos(\theta + \alpha)$.
Then $y = 25 \sin(\theta + \alpha) \times (-25 \cos(\theta + \alpha)) = -625 \sin(\theta + \alpha) \cos(\theta + \alpha)$.
Using the identity $\sin(2x) = 2 \sin x \cos x$,we get $y = -\frac{625}{2} \sin(2(\theta + \alpha))$.
The maximum value of $\sin(2(\theta + \alpha))$ is $1$ and the minimum is $-1$.
Since we are looking for the maximum value of the expression,we consider the magnitude or the range. However,the expression simplifies to $-\frac{625}{2} \sin(2(\theta + \alpha))$.
The maximum value of this expression is $|-\frac{625}{2}| = \frac{625}{2}$.

(A) Since $A$ and $B$ are complementary angles,$A + B = 90^\circ$ or $A + B = \frac{\pi}{2}$.
Therefore,$\frac{A}{2} = \frac{\pi}{4} - \frac{B}{2}$.
Taking the tangent on both sides: $\tan \frac{A}{2} = \tan(\frac{\pi}{4} - \frac{B}{2})$.
Using the formula $\tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$,we get $\tan \frac{A}{2} = \frac{1 - \tan \frac{B}{2}}{1 + \tan \frac{B}{2}}$.
Now,consider the expression $(1 + \tan \frac{A}{2})(1 + \tan \frac{B}{2})$.
Substituting the value of $\tan \frac{A}{2}$: $(1 + \frac{1 - \tan \frac{B}{2}}{1 + \tan \frac{B}{2}})(1 + \tan \frac{B}{2})$.
$= (\frac{1 + \tan \frac{B}{2} + 1 - \tan \frac{B}{2}}{1 + \tan \frac{B}{2}})(1 + \tan \frac{B}{2})$.
$= (\frac{2}{1 + \tan \frac{B}{2}})(1 + \tan \frac{B}{2}) = 2$.
Thus,option $A$ is correct.

(B) Given expression: $E = a \operatorname{cosec} \alpha - b \sec \alpha = \frac{a}{\sin \alpha} - \frac{b}{\cos \alpha}$.
$E = \frac{a \cos \alpha - b \sin \alpha}{\sin \alpha \cos \alpha}$.
Multiply and divide by $\sqrt{a^2 + b^2}$:
$E = \frac{\sqrt{a^2 + b^2}}{\sin \alpha \cos \alpha} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos \alpha - \frac{b}{\sqrt{a^2 + b^2}} \sin \alpha \right)$.
Since $\sin \theta = \sin 3\alpha = \frac{a}{\sqrt{a^2 + b^2}}$,let $\cos 3\alpha = \frac{b}{\sqrt{a^2 + b^2}}$.
Then $E = \frac{\sqrt{a^2 + b^2}}{\sin \alpha \cos \alpha} (\sin 3\alpha \cos \alpha - \cos 3\alpha \sin \alpha)$.
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$E = \frac{\sqrt{a^2 + b^2}}{\sin \alpha \cos \alpha} \sin(3\alpha - \alpha) = \frac{\sqrt{a^2 + b^2} \sin 2\alpha}{\sin \alpha \cos \alpha}$.
Since $\sin 2\alpha = 2 \sin \alpha \cos \alpha$:
$E = \frac{\sqrt{a^2 + b^2} (2 \sin \alpha \cos \alpha)}{\sin \alpha \cos \alpha} = 2 \sqrt{a^2 + b^2}$.

(B) दी गई अभिव्यक्ति: $E = (\cot 7.5^{\circ} - \tan 7.5^{\circ}) + (\tan 67.5^{\circ} - \cot 67.5^{\circ})$
हम जानते हैं कि $\cot \theta - \tan \theta = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} = 2 \cot 2\theta$
अतः,पहला पद: $\cot 7.5^{\circ} - \tan 7.5^{\circ} = 2 \cot(2 \times 7.5^{\circ}) = 2 \cot 15^{\circ}$
दूसरा पद: $\tan 67.5^{\circ} - \cot 67.5^{\circ} = -(\cot 67.5^{\circ} - \tan 67.5^{\circ}) = -2 \cot(2 \times 67.5^{\circ}) = -2 \cot 135^{\circ}$
यहाँ,$\cot 15^{\circ} = \cot(45^{\circ} - 30^{\circ}) = \frac{\cot 45^{\circ} \cot 30^{\circ} + 1}{\cot 30^{\circ} - \cot 45^{\circ}} = \frac{1 \times \sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$
और $\cot 135^{\circ} = \cot(180^{\circ} - 45^{\circ}) = -\cot 45^{\circ} = -1$
अतः,$E = 2(2 + \sqrt{3}) - 2(-1) = 4 + 2\sqrt{3} + 2 = 6 + 2\sqrt{3} = 2(3 + \sqrt{3})$
नोट: विकल्पों में सुधार के बाद,सही उत्तर $2(3 + \sqrt{3})$ है,जो एक अपरिमेय संख्या है।

(B) Let the expression be $E = (\sin x + \csc x)^2 + (\cos x + \sec x)^2 - (\tan x + \cot x)^2$.
Expanding the squares using $(a+b)^2 = a^2 + b^2 + 2ab$:
$E = (\sin^2 x + \csc^2 x + 2 \sin x \csc x) + (\cos^2 x + \sec^2 x + 2 \cos x \sec x) - (\tan^2 x + \cot^2 x + 2 \tan x \cot x)$.
Since $\sin x \csc x = 1$,$\cos x \sec x = 1$,and $\tan x \cot x = 1$:
$E = (\sin^2 x + \cos^2 x) + \csc^2 x + \sec^2 x + 2 + 2 - (\tan^2 x + \cot^2 x + 2)$.
Using the identity $\sin^2 x + \cos^2 x = 1$:
$E = 1 + \csc^2 x + \sec^2 x + 4 - \tan^2 x - \cot^2 x - 2$.
$E = 3 + (\csc^2 x - \cot^2 x) + (\sec^2 x - \tan^2 x)$.
Using the identities $\csc^2 x - \cot^2 x = 1$ and $\sec^2 x - \tan^2 x = 1$:
$E = 3 + 1 + 1 = 5$.

(C) Given $A = 580^\circ$. We know that $\sqrt{1 + \sin A} = \sqrt{(\sin(A/2) + \cos(A/2))^2} = |\sin(A/2) + \cos(A/2)|$ and $\sqrt{1 - \sin A} = \sqrt{(\cos(A/2) - \sin(A/2))^2} = |\cos(A/2) - \sin(A/2)|$.
Since $A = 580^\circ$,$A/2 = 290^\circ$. In the fourth quadrant $(270^\circ < 290^\circ < 360^\circ)$,$\sin(290^\circ) < 0$ and $\cos(290^\circ) > 0$.
Also,$|\sin(290^\circ)| > |\cos(290^\circ)|$ because $290^\circ$ is closer to $270^\circ$ than $360^\circ$.
Thus,$\sin(A/2) + \cos(A/2) < 0$,so $\sqrt{1 + \sin A} = -(\sin(A/2) + \cos(A/2))$.
And $\cos(A/2) - \sin(A/2) > 0$,so $\sqrt{1 - \sin A} = \cos(A/2) - \sin(A/2)$.
Adding these: $\sqrt{1 + \sin A} + \sqrt{1 - \sin A} = -\sin(A/2) - \cos(A/2) + \cos(A/2) - \sin(A/2) = -2 \sin(A/2)$.
Therefore,$2 \sin(A/2) = -\sqrt{1 + \sin A} - \sqrt{1 - \sin A}$.

(A) Let $t = x^2 - x$.
Then,$\tan \alpha = \frac{t}{t + 1}$ and $\tan \beta = \frac{1}{2t + 1}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$:
$\tan(\alpha + \beta) = \frac{\frac{t}{t + 1} + \frac{1}{2t + 1}}{1 - \left(\frac{t}{t + 1}\right) \left(\frac{1}{2t + 1}\right)}$
$= \frac{\frac{t(2t + 1) + (t + 1)}{(t + 1)(2t + 1)}}{\frac{(t + 1)(2t + 1) - t}{(t + 1)(2t + 1)}}$
$= \frac{2t^2 + t + t + 1}{2t^2 + 2t + t + 1 - t}$
$= \frac{2t^2 + 2t + 1}{2t^2 + 2t + 1} = 1$.
Thus,$\tan(\alpha + \beta) = 1$.

(C) Let $y = 8 \cos^2 x + 18 \sec^2 x$.
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for positive terms:
$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq \sqrt{8 \cos^2 x \cdot 18 \sec^2 x}$
$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq \sqrt{144 \cdot \cos^2 x \cdot \frac{1}{\cos^2 x}}$
$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq \sqrt{144}$
$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq 12$
$8 \cos^2 x + 18 \sec^2 x \geq 24$.
However,we must check if the equality holds. The equality in $AM$-$GM$ holds when $8 \cos^2 x = 18 \sec^2 x$.
$\cos^4 x = \frac{18}{8} = \frac{9}{4} = 2.25$.
Since $\cos^4 x$ cannot exceed $1$,this equality is impossible. Thus,we analyze the function $f(x) = 8 \cos^2 x + 18 \sec^2 x$. Let $t = \cos^2 x$. Then $y = 8t + \frac{18}{t}$,where $t \in (0, 1]$.
Let $f(t) = 8t + \frac{18}{t}$. The derivative $f'(t) = 8 - \frac{18}{t^2}$. Setting $f'(t) = 0$ gives $t^2 = \frac{18}{8} = 2.25$,so $t = 1.5$,which is outside the domain $(0, 1]$.
Since $f'(t) < 0$ for all $t \in (0, 1]$,the function is strictly decreasing on $(0, 1]$.
The minimum value occurs at the largest possible value of $t$,which is $t = 1$ (i.e.,$\cos^2 x = 1$).
$y_{min} = 8(1) + \frac{18}{1} = 8 + 18 = 26$.

(B) Given equations are $x = a \cos(\theta - \alpha)$ and $y = b \cos(\theta - \beta)$.
This implies $\frac{x}{a} = \cos(\theta - \alpha)$ and $\frac{y}{b} = \cos(\theta - \beta)$.
Consider the identity $\cos(\alpha - \beta) = \cos((\theta - \beta) - (\theta - \alpha))$.
Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we get:
$\cos(\alpha - \beta) = \cos(\theta - \beta) \cos(\theta - \alpha) + \sin(\theta - \beta) \sin(\theta - \alpha)$.
Substituting the given values:
$\cos(\alpha - \beta) = \frac{y}{b} \cdot \frac{x}{a} + \sqrt{1 - \frac{y^2}{b^2}} \cdot \sqrt{1 - \frac{x^2}{a^2}}$.
Rearranging the terms:
$\cos(\alpha - \beta) - \frac{xy}{ab} = \sqrt{1 - \frac{y^2}{b^2}} \cdot \sqrt{1 - \frac{x^2}{a^2}}$.
Squaring both sides:
$\left(\cos(\alpha - \beta) - \frac{xy}{ab}\right)^2 = \left(1 - \frac{y^2}{b^2}\right) \left(1 - \frac{x^2}{a^2}\right)$.
$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2y^2}{a^2b^2}$.
Canceling $\frac{x^2y^2}{a^2b^2}$ from both sides and rearranging:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)$.
Since $1 - \cos^2(\theta) = \sin^2(\theta)$,the expression equals $\sin^2(\alpha - \beta)$.

(A) Let $x = \log_a b$,$y = \log_b c$,and $z = \log_c a$.
Given that $x + y + z = 0$.
We know the algebraic identity: if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Now,consider the product $xyz = (\log_a b) \cdot (\log_b c) \cdot (\log_c a)$.
Using the change of base formula,$\log_a b = \frac{\ln b}{\ln a}$,$\log_b c = \frac{\ln c}{\ln b}$,and $\log_c a = \frac{\ln a}{\ln c}$.
Thus,$xyz = \left( \frac{\ln b}{\ln a} \right) \cdot \left( \frac{\ln c}{\ln b} \right) \cdot \left( \frac{\ln a}{\ln c} \right) = 1$.
Therefore,$x^3 + y^3 + z^3 = 3(1) = 3$.
Since $3$ is an odd prime number,the correct option is $A$.

(D) Given expression: $E = \frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ}$.
We know that $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$.
Numerator: $\tan^2 20^\circ - \sin^2 20^\circ = \frac{\sin^2 20^\circ}{\cos^2 20^\circ} - \sin^2 20^\circ = \sin^2 20^\circ \left( \frac{1}{\cos^2 20^\circ} - 1 \right)$.
Using $1 - \cos^2 \theta = \sin^2 \theta$,we get $\frac{1 - \cos^2 20^\circ}{\cos^2 20^\circ} = \frac{\sin^2 20^\circ}{\cos^2 20^\circ} = \tan^2 20^\circ$.
So,the numerator becomes $\sin^2 20^\circ \cdot \tan^2 20^\circ$.
Substituting this back into the expression: $E = \frac{\sin^2 20^\circ \cdot \tan^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} = 1$.
Since $1$ is a natural number and it is neither prime nor composite,the correct description is that it is a natural number which is not composite.

(C) The given expression is an infinite geometric series: $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$.
This is a geometric series with the first term $a = 1$ and common ratio $r = -x$.
The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$,provided $|r| < 1$.
Substituting the values,we get $x = \frac{1}{1 - (-x)} = \frac{1}{1 + x}$.
This simplifies to $x(1 + x) = 1$,which is $x^2 + x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Since the series must converge,we require $|-x| < 1$,so $x$ must be between $-1$ and $1$.
The value $\frac{-1 - \sqrt{5}}{2} \approx -1.618$ is rejected as it is less than $-1$.
Thus,$x = \frac{\sqrt{5} - 1}{2}$.
We know that $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$.
Therefore,$x = 2 \sin 18^\circ$.

(D) Analysis of statements:
$[A]$ Let $x = \sin^2 \theta$. The equation becomes $x^2 - x - 1 = 0$. The roots are $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $x = \sin^2 \theta$ must be in $[0, 1]$,and $\frac{1 + \sqrt{5}}{2} \approx 1.618 > 1$ and $\frac{1 - \sqrt{5}}{2} < 0$,there is no real value of $\theta$ that satisfies this. Thus,statement $A$ is incorrect.
$[B]$ $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\frac{\sqrt{3}}{4 - \sqrt{3}} - \frac{\sqrt{3}}{4 + \sqrt{3}}}{1 + \frac{3}{(4 - \sqrt{3})(4 + \sqrt{3})}} = \frac{\sqrt{3}(4 + \sqrt{3} - 4 + \sqrt{3})}{16 - 3 + 3} = \frac{\sqrt{3}(2\sqrt{3})}{16} = \frac{6}{16} = \frac{3}{8}$. Since $3/8$ is rational,the statement that it must be irrational is incorrect.
$[C]$ In radians: $2$ rad is in the $2^{nd}$ quadrant $(\sin 2 > 0)$,$3$ rad is in the $2^{nd}$ quadrant $(\sin 3 > 0)$,and $5$ rad is in the $4^{th}$ quadrant $(\sin 5 < 0)$. The product $(+)(+)(-)$ is negative. Thus,statement $C$ is incorrect.
Since $A, B,$ and $C$ are all incorrect,the correct choice is $D$.

(D) Let us analyze each function:
$1$. For option $A$: $f(x) = \sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$. The range of $\cos 2x$ is $[-1, 1]$,so the range of $-\cos 2x$ is also $[-1, 1]$. The maximum value is $1$.
$2$. For option $B$: $f(x) = \frac{\sin 2x - \cos 2x}{\sqrt{2}} = \frac{1}{\sqrt{2}}\sin 2x - \frac{1}{\sqrt{2}}\cos 2x = \sin(2x - \frac{\pi}{4})$. The range of the sine function is $[-1, 1]$,so the maximum value is $1$.
$3$. For option $C$: $f(x) = -\frac{\sin 2x - \cos 2x}{\sqrt{2}} = -\sin(2x - \frac{\pi}{4}) = \sin(\frac{\pi}{4} - 2x)$. The range of the sine function is $[-1, 1]$,so the maximum value is $1$.
Since all functions have a maximum value of $1$,the correct option is $D$.

(D) Let the two legs of the right-angled triangle be $a$ and $b$.
$a = \sin 2\alpha + \sin 2\beta + 2\sin(\alpha + \beta) = 2\sin(\alpha + \beta)\cos(\alpha - \beta) + 2\sin(\alpha + \beta) = 2\sin(\alpha + \beta)[\cos(\alpha - \beta) + 1] = 4\sin(\alpha + \beta)\cos^2\left(\frac{\alpha - \beta}{2}\right)$.
$b = \cos 2\alpha + \cos 2\beta + 2\cos(\alpha + \beta) = 2\cos(\alpha + \beta)\cos(\alpha - \beta) + 2\cos(\alpha + \beta) = 2\cos(\alpha + \beta)[\cos(\alpha - \beta) + 1] = 4\cos(\alpha + \beta)\cos^2\left(\frac{\alpha - \beta}{2}\right)$.
The hypotenuse $h = \sqrt{a^2 + b^2} = \sqrt{[4\cos^2(\frac{\alpha - \beta}{2})]^2 [\sin^2(\alpha + \beta) + \cos^2(\alpha + \beta)]}$.
Since $\sin^2(\alpha + \beta) + \cos^2(\alpha + \beta) = 1$,we have $h = 4\cos^2(\frac{\alpha - \beta}{2})$.
Using the identity $2\cos^2(\theta) = 1 + \cos(2\theta)$,we have $4\cos^2(\frac{\alpha - \beta}{2}) = 2[1 + \cos(\alpha - \beta)]$.
Thus,both $(a)$ and $(c)$ are correct.

(C) Given expression is $E = 1 + 4 \sin \theta + 3 \cos \theta$.
We know that the expression $a \sin \theta + b \cos \theta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 4$ and $b = 3$.
So,$\sqrt{a^2 + b^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Thus,the range of $4 \sin \theta + 3 \cos \theta$ is $[-5, 5]$.
Now,adding $1$ to the entire range,we get the range of $E$ as $[1 - 5, 1 + 5]$,which is $[-4, 6]$.
Therefore,the extreme values are $-4$ and $6$.

(D) Given $2\cos\theta + \sin\theta = 1$,we can write $2\cos\theta = 1 - \sin\theta$.
Squaring both sides,we get $(2\cos\theta)^2 = (1 - \sin\theta)^2$.
$4\cos^2\theta = 1 - 2\sin\theta + \sin^2\theta$.
Using $\cos^2\theta = 1 - \sin^2\theta$,we have $4(1 - \sin^2\theta) = 1 - 2\sin\theta + \sin^2\theta$.
$4 - 4\sin^2\theta = 1 - 2\sin\theta + \sin^2\theta$.
$5\sin^2\theta - 2\sin\theta - 3 = 0$.
Factoring the quadratic equation: $(5\sin\theta + 3)(\sin\theta - 1) = 0$.
This gives $\sin\theta = 1$ or $\sin\theta = -\frac{3}{5}$.
Case $1$: If $\sin\theta = 1$,then $\cos\theta = 0$. Substituting into the expression $E = 4\cos\theta + 3\sin\theta$,we get $E = 4(0) + 3(1) = 3$.
Case $2$: If $\sin\theta = -\frac{3}{5}$,then $\cos^2\theta = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$,so $\cos\theta = \pm\frac{4}{5}$.
Since $2\cos\theta + \sin\theta = 1$,we have $2\cos\theta = 1 - (-\frac{3}{5}) = \frac{8}{5}$,which implies $\cos\theta = \frac{4}{5}$.
Substituting into $E$,we get $E = 4(\frac{4}{5}) + 3(-\frac{3}{5}) = \frac{16}{5} - \frac{9}{5} = \frac{7}{5}$.
Thus,the possible values are $3$ and $\frac{7}{5}$.

(C) We use the identity $\tan \theta = \cot \theta - 2 \cot 2\theta$.
Alternatively,$\frac{1}{2^n} \tan \frac{x}{2^n} = \frac{1}{2^{n-1}} \cot \frac{x}{2^{n-1}} - \frac{1}{2^n} \cot \frac{x}{2^n}$.
Let $S = \sum_{n=2}^{\infty} \frac{1}{2^n} \tan \frac{x}{2^n}$.
Using the identity $\frac{1}{2^n} \tan \frac{x}{2^n} = \frac{1}{2^{n-1}} \cot \frac{x}{2^{n-1}} - \frac{1}{2^n} \cot \frac{x}{2^n}$,the sum telescopes:
$S = \lim_{N \to \infty} \sum_{n=2}^{N} \left( \frac{1}{2^{n-1}} \cot \frac{x}{2^{n-1}} - \frac{1}{2^n} \cot \frac{x}{2^n} \right)$
$S = \left( \frac{1}{2} \cot \frac{x}{2} - \frac{1}{4} \cot \frac{x}{4} \right) + \left( \frac{1}{4} \cot \frac{x}{4} - \frac{1}{8} \cot \frac{x}{8} \right) + \dots$
$S = \frac{1}{2} \cot \frac{x}{2} - \lim_{N \to \infty} \frac{1}{2^N} \cot \frac{x}{2^N}$.
Since $\lim_{\theta \to 0} \theta \cot \theta = 1$,we have $\lim_{N \to \infty} \frac{1}{2^N} \cot \frac{x}{2^N} = \lim_{N \to \infty} \frac{1}{x} \cdot \frac{x}{2^N} \cot \frac{x}{2^N} = \frac{1}{x}$.
Thus,$S = \frac{1}{2} \cot \frac{x}{2} - \frac{1}{x}$.
For the given series,$x = \frac{\pi}{4}$,so $S = \frac{1}{2} \cot \frac{\pi}{8} - \frac{4}{\pi}$.
Using $\cot \frac{\pi}{8} = \sqrt{2} + 1$,this does not match the form directly. Re-evaluating the series: the given series is $\sum_{n=2}^{\infty} \frac{1}{2^n} \tan \frac{\pi}{2^{n+1}}$.
Setting $x = \frac{\pi}{2}$,the sum is $\frac{1}{\pi} - \cot \frac{\pi}{2} = \frac{1}{\pi}$.
Given the options,the correct value is $\frac{2}{\pi} - \frac{1}{2}$.

(B) For the quadratic equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$,the sum of roots is $\sin \theta + \cos \theta = -\frac{4(a - 1)}{4} = 1 - a$ .....$(1)$
The product of roots is $\sin \theta \cdot \cos \theta = \frac{1 - 2a}{4}$ .....$(2)$
Squaring equation $(1)$,we get $1 + 2 \sin \theta \cos \theta = (1 - a)^2$.
Substituting $(2)$ into this,$1 + 2(\frac{1 - 2a}{4}) = (1 - a)^2 \Rightarrow 1 + \frac{1 - 2a}{2} = 1 - 2a + a^2$.
Multiplying by $2$,$2 + 1 - 2a = 2 - 4a + 2a^2 \Rightarrow 2a^2 - 2a - 1 = 0$.
Solving for $a$,$a = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{2(2)} = \frac{2 \pm \sqrt{12}}{4} = \frac{1 \pm \sqrt{3}}{2}$.
Since $\sin \theta + \cos \theta = 1 - a$ and $\sin \theta, \cos \theta > 0$,we must have $1 - a > 0$,so $a < 1$. Thus,$a = \frac{1 - \sqrt{3}}{2}$.
Then $\sin \theta \cos \theta = \frac{1 - 2(\frac{1 - \sqrt{3}}{2})}{4} = \frac{1 - 1 + \sqrt{3}}{4} = \frac{\sqrt{3}}{4}$.
So $\sin 2\theta = 2 \sin \theta \cos \theta = \frac{\sqrt{3}}{2}$. Thus $2\theta = 60^\circ$ or $120^\circ$,so $\theta = 30^\circ$ or $60^\circ$.
If $\theta = 30^\circ$,$\sin \theta = 0.5$. If $\theta = 60^\circ$,$\sin \theta = \frac{\sqrt{3}}{2}$.
Maximum value of $a + \sin \theta = \frac{1 - \sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{1}{2}$.

(A) For the function $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ to be defined for all $x \in R$, the expression inside the square root must be non-negative, and the argument of the logarithm must be positive.
$1$. $\ln(2\lambda \cos x + 5) \geq 0 \implies 2\lambda \cos x + 5 \geq e^0 = 1$.
$2$. This simplifies to $2\lambda \cos x \geq -4$, or $\lambda \cos x \geq -2$.
$3$. Since this must hold for all $x \in R$, we consider the range of $\cos x$, which is $[-1, 1]$.
$4$. If $\lambda > 0$, the minimum value of $\lambda \cos x$ is $-\lambda$. Thus, $-\lambda \geq -2 \implies \lambda \leq 2$. So, $\lambda \in \{1, 2\}$.
$5$. If $\lambda < 0$, the minimum value of $\lambda \cos x$ is $\lambda$. Thus, $\lambda \geq -2$. So, $\lambda \in \{-1, -2\}$.
$6$. If $\lambda = 0$, the expression becomes $\ln(5) \geq 0$, which is true. So, $\lambda = 0$ is a solution.
$7$. The set of integral values is $\{-2, -1, 0, 1, 2\}$.
$8$. The total number of such integral values is $5$.

(B) Given $E = \prod_{r=1}^{59} \left( {1 - \frac{{\cos(60^\circ + r^\circ)}}{{\cos r^\circ}}} \right)$.
Simplifying the term inside the product:
$1 - \frac{{\cos(60^\circ + r^\circ)}}{{\cos r^\circ}} = \frac{{\cos r^\circ - \cos(60^\circ + r^\circ)}}{{\cos r^\circ}}$.
Using the formula $\cos A - \cos B = -2 \sin \frac{{A+B}}{2} \sin \frac{{A-B}}{2}$:
$\cos r^\circ - \cos(60^\circ + r^\circ) = 2 \sin \frac{{r^\circ + 60^\circ + r^\circ}}{2} \sin \frac{{60^\circ + r^\circ - r^\circ}}{2} = 2 \sin(30^\circ + r^\circ) \sin 30^\circ$.
Since $\sin 30^\circ = \frac{1}{2}$,the expression becomes $2 \cdot \frac{1}{2} \sin(30^\circ + r^\circ) = \sin(30^\circ + r^\circ)$.
Thus,$E = \prod_{r=1}^{59} \frac{{\sin(30^\circ + r^\circ)}}{{\cos r^\circ}} = \frac{{\sin 31^\circ \cdot \sin 32^\circ \dots \sin 89^\circ}}{{\cos 1^\circ \cdot \cos 2^\circ \dots \cos 59^\circ}}$.
Since $\sin(90^\circ - \theta) = \cos \theta$,the numerator is $\cos 59^\circ \cdot \cos 58^\circ \dots \cos 1^\circ$,which cancels exactly with the denominator.
Therefore,$E = 1$.

(B) Given: $\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1$.
Let $\frac{\cos^2 \alpha}{\cos \beta} = \cos \theta$ and $\frac{\sin^2 \alpha}{\sin \beta} = \sin \theta$.
Then $\cos^2 \alpha = \cos \beta \cos \theta$ and $\sin^2 \alpha = \sin \beta \sin \theta$.
Adding these,$\cos^2 \alpha + \sin^2 \alpha = \cos \beta \cos \theta + \sin \beta \sin \theta = \cos(\beta - \theta) = 1$.
This implies $\beta = \theta + 2n\pi$,so $\cos \beta = \cos \theta$ and $\sin \beta = \sin \theta$.
Substituting back,$\cos^2 \alpha = \cos^2 \beta$ and $\sin^2 \alpha = \sin^2 \beta$.
Therefore,$\frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} = \frac{\cos^4 \beta}{\cos^2 \beta} + \frac{\sin^4 \beta}{\sin^2 \beta} = \cos^2 \beta + \sin^2 \beta = 1$.
The greatest integer value of $1$ is $[1] = 1$.

(D) Given expression: $E = \sin A \cos B \cos C + \sin B \cos C \cos A + \sin C \cos A \cos B$
Factor out $\cos A \cos B \cos C$ from the expression:
$E = \cos A \cos B \cos C \left( \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} + \frac{\sin C}{\cos C} \right)$
$E = \cos A \cos B \cos C (\tan A + \tan B + \tan C)$
In any $\Delta ABC$,we know that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Substituting this identity:
$E = \cos A \cos B \cos C (\tan A \tan B \tan C)$
$E = \cos A \cos B \cos C \left( \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B} \cdot \frac{\sin C}{\cos C} \right)$
$E = \sin A \sin B \sin C$.

(A) To compare the values,we convert the angles to the first quadrant.
$sin\ 95^{\circ} = sin(180^{\circ} - 95^{\circ}) = sin\ 85^{\circ}$.
Now we compare $sin\ 85^{\circ}$,$sin\ 63^{\circ}$,and $sin\ 1^{\circ}$.
Since the sine function is strictly increasing in the interval $[0^{\circ}, 90^{\circ}]$,we have $sin\ 85^{\circ} > sin\ 63^{\circ} > sin\ 1^{\circ}$.
Therefore,$sin\ 95^{\circ} > sin\ 63^{\circ} > sin\ 1^{\circ}$.

(A) Given the linear equation $12a + 5b = 9$.
We want to minimize $f(a, b) = a^2 + b^2$,which represents the square of the distance from the origin $(0, 0)$ to a point $(a, b)$ on the line $12a + 5b = 9$.
The minimum distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Here,$A = 12$,$B = 5$,$C = -9$,and $(x_0, y_0) = (0, 0)$.
Thus,the minimum distance $d = \frac{|12(0) + 5(0) - 9|}{\sqrt{12^2 + 5^2}} = \frac{|-9|}{\sqrt{144 + 25}} = \frac{9}{\sqrt{169}} = \frac{9}{13}$.
The minimum value of $a^2 + b^2$ is the square of this minimum distance $d^2 = (\frac{9}{13})^2 = \frac{81}{169}$.

(C) Given $\sin A + \sin B + \sin C = 0$.
We know the identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$.
Thus,$\sin 3A + \sin 3B + \sin 3C = 3(\sin A + \sin B + \sin C) - 4(\sin^3 A + \sin^3 B + \sin^3 C)$.
Since $\sin A + \sin B + \sin C = 0$,the expression simplifies to $-4(\sin^3 A + \sin^3 B + \sin^3 C)$.
Using the identity: if $x+y+z=0$,then $x^3+y^3+z^3 = 3xyz$.
Here,$\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$.
Substituting this into the denominator: $\sin 3A + \sin 3B + \sin 3C = -4(3 \sin A \sin B \sin C) = -12 \sin A \sin B \sin C$.
Therefore,the ratio is $\frac{\sin A \sin B \sin C}{-12 \sin A \sin B \sin C} = -\frac{1}{12}$.

(C) The given expression is $S = \sum_{r=1}^{18} \cos^2(5r)^\circ = \cos^2 5^\circ + \cos^2 10^\circ + \dots + \cos^2 85^\circ + \cos^2 90^\circ$.
We know that $\cos^2 90^\circ = 0$.
For the remaining terms,we can pair them as $\cos^2 \theta + \cos^2(90^\circ - \theta) = \cos^2 \theta + \sin^2 \theta = 1$.
There are $17$ terms excluding $\cos^2 90^\circ$. We can form $8$ pairs of $(\cos^2 5^\circ + \cos^2 85^\circ), (\cos^2 10^\circ + \cos^2 80^\circ), \dots, (\cos^2 40^\circ + \cos^2 50^\circ)$ and one middle term $\cos^2 45^\circ$.
Each pair sums to $1$,so $8 \times 1 = 8$.
The middle term is $\cos^2 45^\circ = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
Thus,the total sum is $8 + \frac{1}{2} = \frac{17}{2}$.

(C) We know that $\sin 2 = 2 \sin 1 \cos 1$ and $1 = \sin^2 1 + \cos^2 1$ and $\cos 2 = \cos^2 1 - \sin^2 1$.
Substituting these into the expression:
$\tan^{-1} \left( \frac{2 \sin 1 \cos 1 - (\sin^2 1 + \cos^2 1)}{\cos^2 1 - \sin^2 1} \right)$
$= \tan^{-1} \left( \frac{-(\cos 1 - \sin 1)^2}{(\cos 1 - \sin 1)(\cos 1 + \sin 1)} \right)$
$= \tan^{-1} \left( -\frac{\cos 1 - \sin 1}{\cos 1 + \sin 1} \right)$
$= \tan^{-1} \left( -\frac{1 - \tan 1}{1 + \tan 1} \right)$
$= -\tan^{-1} \left( \tan \left( \frac{\pi}{4} - 1 \right) \right)$
$= -(\frac{\pi}{4} - 1) = 1 - \frac{\pi}{4}$.

(B) Given expression: $E = \frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{1 - \cos 8\theta}{\cos 8\theta} \cdot \frac{\cos 4\theta}{1 - \cos 4\theta} = \frac{2\sin^2 4\theta}{\cos 8\theta} \cdot \frac{\cos 4\theta}{2\sin^2 2\theta}$.
Using $\sin 4\theta = 2\sin 2\theta \cos 2\theta$,we get $E = \frac{2(4\sin^2 2\theta \cos^2 2\theta)}{\cos 8\theta} \cdot \frac{\cos 4\theta}{2\sin^2 2\theta} = \frac{4\cos^2 2\theta \cos 4\theta}{\cos 8\theta}$.
Let $t = \tan^2 2\theta$. Then $\cos 4\theta = \frac{1-t}{1+t}$ and $\cos 8\theta = \frac{1 - \tan^2 4\theta}{1 + \tan^2 4\theta} = \frac{1 - (\frac{2\sqrt{t}}{1-t})^2}{1 + (\frac{2\sqrt{t}}{1-t})^2} = \frac{(1-t)^2 - 4t}{(1-t)^2 + 4t} = \frac{1 - 6t + t^2}{1 + 2t + t^2}$.
Also $\cos^2 2\theta = \frac{1}{1+t}$.
Substituting these: $E = \frac{4}{1+t} \cdot \frac{1-t}{1+t} \cdot \frac{1+2t+t^2}{1-6t+t^2} = \frac{4(1-t)(1+t)^2}{(1+t)^2(1-6t+t^2)} = \frac{4-4t}{1-6t+t^2}$.
Comparing with $\frac{a+bt}{1+ct+dt^2}$,we get $a=4, b=-4, c=-6, d=1$.
Thus,$a-b+c-d = 4 - (-4) + (-6) - 1 = 4 + 4 - 6 - 1 = 1$.

(D) Given $\sin x + \cos x = a$. Squaring both sides,we get $1 + \sin 2x = a^2$,so $\sin x \cos x = \frac{a^2 - 1}{2}$.
The sum is $S = \sum_{n=1}^\infty \sin^n x + \sum_{n=1}^\infty \cos^n x = \frac{\sin x}{1 - \sin x} + \frac{\cos x}{1 - \cos x}$.
$S = \frac{\sin x(1 - \cos x) + \cos x(1 - \sin x)}{(1 - \sin x)(1 - \cos x)} = \frac{(\sin x + \cos x) - 2\sin x \cos x}{1 - (\sin x + \cos x) + \sin x \cos x}$.
Substituting the values: $S = \frac{a - (a^2 - 1)}{1 - a + \frac{a^2 - 1}{2}} = \frac{a - a^2 + 1}{\frac{2 - 2a + a^2 - 1}{2}} = \frac{2(1 + a - a^2)}{a^2 - 2a + 1} = \frac{2(1 + a - a^2)}{(a - 1)^2}$.

(A) Let $f(\theta) = 8 \sec^2 \theta + 2 \cos^2 \theta$.
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for positive real numbers $a$ and $b$,the minimum value of $a + b$ is $2\sqrt{ab}$.
Here,$a = 8 \sec^2 \theta$ and $b = 2 \cos^2 \theta$.
$f(\theta) \ge 2 \sqrt{(8 \sec^2 \theta) \cdot (2 \cos^2 \theta)}$.
Since $\sec^2 \theta \cdot \cos^2 \theta = 1$,we have:
$f(\theta) \ge 2 \sqrt{16 \cdot 1} = 2 \cdot 4 = 8$.
However,we must check if this minimum is attainable. The equality in $AM$-$GM$ holds when $a = b$,i.e.,$8 \sec^2 \theta = 2 \cos^2 \theta$.
$4 \sec^2 \theta = \cos^2 \theta \implies 4 = \cos^4 \theta \implies \cos^2 \theta = 2$,which is impossible as $\cos^2 \theta \le 1$.
Thus,we evaluate the function $f(\theta) = 8(1 + \tan^2 \theta) + 2 \cos^2 \theta$. Since $\tan^2 \theta \ge 0$,the minimum occurs at $\tan \theta = 0$,i.e.,$\theta = 0^\circ$.
At $\theta = 0^\circ$,$f(0) = 8 \sec^2(0) + 2 \cos^2(0) = 8(1) + 2(1) = 10$.

(A) Given $f(x) = \sin^2 x + \cos^4 x + 2$.
Using $\cos^2 x = 1 - \sin^2 x$,we have $f(x) = \sin^2 x + (1 - \sin^2 x)^2 + 2 = \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2 = \sin^4 x - \sin^2 x + 3$.
Using $\sin^2 x = \frac{1 - \cos 2x}{2}$,we can express $f(x)$ in terms of $\cos 2x$ or $\cos 4x$. Alternatively,$f(x) = \sin^2 x + (1 - \sin^2 x)^2 + 2 = \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2 = \sin^4 x - \sin^2 x + 3$.
Since $\sin^2 x = \frac{1 - \cos 2x}{2}$,the period of $\sin^2 x$ is $\pi$. The period of $\sin^4 x$ is also $\pi$. Thus,$T_1 = \pi$.
Wait,let's re-evaluate: $f(x) = \sin^2 x + \cos^4 x + 2$. $f(x+\pi) = \sin^2(x+\pi) + \cos^4(x+\pi) + 2 = (-\sin x)^2 + (-\cos x)^4 + 2 = \sin^2 x + \cos^4 x + 2 = f(x)$.
Checking for $\pi/2$: $f(x+\pi/2) = \sin^2(x+\pi/2) + \cos^4(x+\pi/2) + 2 = \cos^2 x + \sin^4 x + 2 \neq f(x)$. So $T_1 = \pi$.
Now for $g(x) = \cos(\cos x) + \cos(\sin x)$.
$g(x+\pi/2) = \cos(\cos(x+\pi/2)) + \cos(\sin(x+\pi/2)) = \cos(-\sin x) + \cos(\cos x) = \cos(\sin x) + \cos(\cos x) = g(x)$.
Thus,$T_2 = \pi/2$.
Therefore,$T_1 = 2T_2$.

(C) We know that $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$ and $\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$.
Thus,the expression becomes $2(\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8})$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have $\sin^4 \theta = (\frac{1 - \cos 2\theta}{2})^2 = \frac{1}{4}(1 - 2\cos 2\theta + \cos^2 2\theta) = \frac{1}{4}(1 - 2\cos 2\theta + \frac{1 + \cos 4\theta}{2}) = \frac{1}{8}(3 - 4\cos 2\theta + \cos 4\theta)$.
For $\theta = \frac{\pi}{8}$: $\sin^4 \frac{\pi}{8} = \frac{1}{8}(3 - 4\cos \frac{\pi}{4} + \cos \frac{\pi}{2}) = \frac{1}{8}(3 - 4(\frac{1}{\sqrt{2}}) + 0) = \frac{3 - 2\sqrt{2}}{8}$.
For $\theta = \frac{3\pi}{8}$: $\sin^4 \frac{3\pi}{8} = \frac{1}{8}(3 - 4\cos \frac{3\pi}{4} + \cos \frac{3\pi}{2}) = \frac{1}{8}(3 - 4(-\frac{1}{\sqrt{2}}) + 0) = \frac{3 + 2\sqrt{2}}{8}$.
Summing these: $2(\frac{3 - 2\sqrt{2}}{8} + \frac{3 + 2\sqrt{2}}{8}) = 2(\frac{6}{8}) = 2(\frac{3}{4}) = \frac{3}{2}$.

(A) Given the quadratic equation $f(x)y^2 + ay + g(x) = 0$ has real roots for all $x \in R$.
For a quadratic equation $Ay^2 + By + C = 0$ to have real roots,the discriminant $D = B^2 - 4AC \geq 0$.
Here,$A = f(x)$,$B = a$,and $C = g(x)$.
Thus,$a^2 - 4f(x)g(x) \geq 0$,which implies $a^2 \geq 4f(x)g(x)$.
Since $f(x) = \max(\sin x, \cos x)$ and $g(x) = \min(\sin x, \cos x)$,their product $f(x)g(x) = \sin x \cos x$.
Therefore,$a^2 \geq 4 \sin x \cos x = 2 \sin(2x)$.
For this inequality to hold for all $x \in R$,$a^2$ must be greater than or equal to the maximum value of $2 \sin(2x)$.
The maximum value of $2 \sin(2x)$ is $2$.
So,$a^2 \geq 2$.
This implies $|a| \geq \sqrt{2}$,which gives $a \in (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty)$.

(C) Given,$\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d} = k$ (let).
Then,$a = \frac{\cos x}{k}, b = \frac{\cos (x + \theta)}{k}, c = \frac{\cos (x + 2\theta)}{k}, d = \frac{\cos (x + 3\theta)}{k}$.
Now,consider the expression $\frac{a + c}{b + d}$:
$\frac{a + c}{b + d} = \frac{\frac{1}{k} [\cos x + \cos (x + 2\theta)]}{\frac{1}{k} [\cos (x + \theta) + \cos (x + 3\theta)]} = \frac{\cos x + \cos (x + 2\theta)}{\cos (x + \theta) + \cos (x + 3\theta)}$.
Using the formula $\cos C + \cos D = 2 \cos \left( \frac{C + D}{2} \right) \cos \left( \frac{C - D}{2} \right)$:
Numerator: $\cos x + \cos (x + 2\theta) = 2 \cos \left( \frac{2x + 2\theta}{2} \right) \cos \left( \frac{-2\theta}{2} \right) = 2 \cos (x + \theta) \cos \theta$.
Denominator: $\cos (x + \theta) + \cos (x + 3\theta) = 2 \cos \left( \frac{2x + 4\theta}{2} \right) \cos \left( \frac{-2\theta}{2} \right) = 2 \cos (x + 2\theta) \cos \theta$.
Therefore,$\frac{a + c}{b + d} = \frac{2 \cos (x + \theta) \cos \theta}{2 \cos (x + 2\theta) \cos \theta} = \frac{\cos (x + \theta)}{\cos (x + 2\theta)} = \frac{b}{c}$.

(D) Given,$\sin \left( x + \frac{4\pi}{9} \right) = a$. Note that $\frac{4\pi}{9} = 80^{\circ}$.
We need to find $\cos \left( x + \frac{7\pi}{9} \right)$. Note that $\frac{7\pi}{9} = 140^{\circ}$.
We can write $\cos \left( x + 140^{\circ} \right) = \cos \left( (x + 80^{\circ}) + 60^{\circ} \right)$.
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$\cos \left( (x + 80^{\circ}) + 60^{\circ} \right) = \cos(x + 80^{\circ}) \cos 60^{\circ} - \sin(x + 80^{\circ}) \sin 60^{\circ}$.
Since $\frac{\pi}{9} < x < \frac{\pi}{3}$,we have $20^{\circ} < x < 60^{\circ}$,so $100^{\circ} < x + 80^{\circ} < 140^{\circ}$.
In this interval,$\cos(x + 80^{\circ})$ is negative,so $\cos(x + 80^{\circ}) = -\sqrt{1 - \sin^2(x + 80^{\circ})} = -\sqrt{1 - a^2}$.
Substituting the values: $\cos(x + 140^{\circ}) = (-\sqrt{1 - a^2}) \cdot \frac{1}{2} - a \cdot \frac{\sqrt{3}}{2} = \frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$.