Trigonometry Questions in English

(A) In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^\circ$.
Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.
From $A + C = 180^\circ$,we have $A = 180^\circ - C$.
Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.
Similarly,from $B + D = 180^\circ$,we have $B = 180^\circ - D$.
Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.
Now,consider the expression $\cos A - \cos B + \cos C - \cos D$.
Rearranging the terms,we get $(\cos A + \cos C) - (\cos B + \cos D)$.
Substituting the values derived above,we get $0 - 0 = 0$.

(B) In $\Delta ABC,$ we have $A + B + C = 180^\circ.$
We need to evaluate $\sin A + \sin B + \sin C.$
Using the sum-to-product formula $\sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2},$ we get:
$\sin A + \sin B + \sin C = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$
Since $A+B = 180^\circ - C,$ then $\frac{A+B}{2} = 90^\circ - \frac{C}{2}.$
Thus,$\sin \frac{A+B}{2} = \cos \frac{C}{2}.$
Substituting this into the expression:
$= 2\cos \frac{C}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$
$= 2\cos \frac{C}{2} \left( \cos \frac{A-B}{2} + \sin \frac{C}{2} \right)$
Since $\sin \frac{C}{2} = \cos \frac{A+B}{2},$ we have:
$= 2\cos \frac{C}{2} \left( \cos \frac{A-B}{2} + \cos \frac{A+B}{2} \right)$
Using the formula $\cos(x-y) + \cos(x+y) = 2\cos x \cos y,$ we get:
$= 2\cos \frac{C}{2} \left( 2\cos \frac{A}{2} \cos \frac{B}{2} \right) = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$
Therefore,the correct option is $B.$

(A) In $\Delta ABC$,we have $A + B + C = 180^\circ$.
Consider the expression $\sin 2A + \sin 2B + \sin 2C$.
Using the sum-to-product formula $\sin X + \sin Y = 2\sin(\frac{X+Y}{2})\cos(\frac{X-Y}{2})$,we get:
$\sin 2A + \sin 2B = 2\sin(A+B)\cos(A-B)$.
Since $A+B = 180^\circ - C$,then $\sin(A+B) = \sin(180^\circ - C) = \sin C$.
So,$\sin 2A + \sin 2B = 2\sin C \cos(A-B)$.
Now,$\sin 2C = 2\sin C \cos C$.
Since $C = 180^\circ - (A+B)$,then $\cos C = \cos(180^\circ - (A+B)) = -\cos(A+B)$.
So,$\sin 2C = -2\sin C \cos(A+B)$.
Substituting these back into the expression:
$\sin 2A + \sin 2B + \sin 2C = 2\sin C \cos(A-B) - 2\sin C \cos(A+B)$
$= 2\sin C [\cos(A-B) - \cos(A+B)]$.
Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$,we get:
$= 2\sin C [2\sin A \sin B] = 4\sin A \sin B \sin C$.

(B) Given: $x + y + z = 180^o \implies x + y = 180^o - z$.
Consider the expression: $\cos 2x + \cos 2y - \cos 2z$.
Using the formula $\cos C + \cos D = 2\cos(\frac{C+D}{2})\cos(\frac{C-D}{2})$:
$\cos 2x + \cos 2y = 2\cos(x+y)\cos(x-y)$.
Using the identity $\cos 2z = 2\cos^2 z - 1$:
Expression $= 2\cos(x+y)\cos(x-y) - (2\cos^2 z - 1)$.
Since $x+y = 180^o - z$,then $\cos(x+y) = \cos(180^o - z) = -\cos z$.
Substituting this into the expression:
$= 2(-\cos z)\cos(x-y) - 2\cos^2 z + 1$
$= 1 - 2\cos z [\cos(x-y) + \cos z]$.
Since $z = 180^o - (x+y)$,then $\cos z = \cos(180^o - (x+y)) = -\cos(x+y)$.
$= 1 - 2\cos z [\cos(x-y) - \cos(x+y)]$.
Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:
$= 1 - 2\cos z [2\sin x \sin y]$
$= 1 - 4\sin x \sin y \cos z$.

(A) Given that $\alpha + \beta + \gamma = 2\pi$.
Dividing by $2$,we get $\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$.
Taking the tangent of both sides,we have $\tan \left( \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} \right) = \tan \pi = 0$.
Using the formula $\tan(A+B+C) = \frac{\sum \tan A - \prod \tan A}{1 - \sum \tan A \tan B}$,we get:
$\frac{\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} - \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}}{1 - (\tan \frac{\alpha}{2} \tan \frac{\beta}{2} + \tan \frac{\beta}{2} \tan \frac{\gamma}{2} + \tan \frac{\gamma}{2} \tan \frac{\alpha}{2})} = 0$.
Since the fraction is $0$,the numerator must be $0$:
$\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} - \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2} = 0$.
Therefore,$\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} = \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$.

(C) Given $A + B + C = \pi$,so $A + B = \pi - C$.
Using the sum-to-product formula: $\cos 2A + \cos 2B = 2 \cos(A + B) \cos(A - B)$.
Also,$\cos 2C = 2 \cos^2 C - 1$.
Substituting these into the expression:
$L.H.S. = 2 \cos(A + B) \cos(A - B) + 2 \cos^2 C - 1$.
Since $\cos(A + B) = \cos(\pi - C) = - \cos C$,we have:
$L.H.S. = 2(-\cos C) \cos(A - B) + 2 \cos^2 C - 1$.
$L.H.S. = - 1 - 2 \cos C [\cos(A - B) - \cos C]$.
Since $\cos C = - \cos(A + B)$,we substitute:
$L.H.S. = - 1 - 2 \cos C [\cos(A - B) + \cos(A + B)]$.
Using $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$:
$L.H.S. = - 1 - 2 \cos C [2 \cos A \cos B] = - 1 - 4 \cos A \cos B \cos C$.

(B) Given $A + B + C = 180^\circ.$
Numerator $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\sin \frac{B}{2}\cos \frac{B}{2})(2\sin \frac{C}{2}\cos \frac{C}{2}) = 32\sin \frac{A}{2}\cos \frac{A}{2}\sin \frac{B}{2}\cos \frac{B}{2}\sin \frac{C}{2}\cos \frac{C}{2}.$
Denominator $(D^r)$ = $\cos A + \cos B + \cos C - 1 = -4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$
Wait,using the identity $\cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},$ we have $D^r = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$
Now,$N^r = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\sin \frac{B}{2}\cos \frac{B}{2})(2\sin \frac{C}{2}\cos \frac{C}{2}) = 32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$
Therefore,$\frac{N^r}{D^r} = \frac{32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} = 8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$

(B) In a triangle $ABC$,$A + B + C = \pi$,so $C = \pi - (A + B)$.
We know that $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$.
Proof:
$\sin^2 A + \sin^2 B + \sin^2 C = \sin^2 A + \sin^2 B + \sin^2 (A + B)$
$= \sin^2 A + \sin^2 B + (\sin A \cos B + \cos A \sin B)^2$
$= \sin^2 A + \sin^2 B + \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2 \sin A \cos B \cos A \sin B$
$= \sin^2 A(1 + \cos^2 B) + \sin^2 B(1 + \cos^2 A) + 2 \sin A \cos A \sin B \cos B$
Alternatively,using the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$ for any triangle $ABC$:
$\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = (2 + 2\cos A \cos B \cos C) - 2\cos A \cos B \cos C = 2$.
Thus,the correct option is $B$.

(C) In any triangle $ABC$,the identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ holds true.
Given $\tan A + \tan B + \tan C = 6$ and $\tan A \tan B = 2$.
Substituting these into the identity: $6 = 2 \times \tan C$.
Therefore,$\tan C = \frac{6}{2} = 3$.
Now,we have $\tan A + \tan B = 6 - \tan C = 6 - 3 = 3$.
We are given $\tan A \tan B = 2$.
Let $\tan A$ and $\tan B$ be the roots of the quadratic equation $x^2 - (\tan A + \tan B)x + \tan A \tan B = 0$.
Substituting the values: $x^2 - 3x + 2 = 0$.
Factoring the equation: $(x - 1)(x - 2) = 0$.
Thus,the roots are $x = 1$ and $x = 2$.
So,the set of values ${\tan A, \tan B, \tan C}$ is ${1, 2, 3}$.
This corresponds to both $(a)$ and $(b)$.

(B) Let $x = \tan \frac{A}{2}$,$y = \tan \frac{B}{2}$,and $z = \tan \frac{C}{2}$.
Since $A + B + C = \pi$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$.
Using the identity for $\tan(X+Y+Z)$,we know that if $X+Y+Z = \frac{\pi}{2}$,then $xy + yz + zx = 1$.
We know the algebraic inequality $(x-y)^2 + (y-z)^2 + (z-x)^2 \ge 0$.
Expanding this,we get $2(x^2 + y^2 + z^2) - 2(xy + yz + zx) \ge 0$.
Dividing by $2$,we get $x^2 + y^2 + z^2 \ge xy + yz + zx$.
Substituting the value $xy + yz + zx = 1$,we get $x^2 + y^2 + z^2 \ge 1$.
Thus,$\tan^2 \frac{A}{2} + \tan^2 \frac{B}{2} + \tan^2 \frac{C}{2} \ge 1$.

(C) In any triangle $ABC$,where $A + B + C = 180^o$,the identity for the sum of tangents is given by $\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$.
Dividing both sides by the product $\tan A \cdot \tan B \cdot \tan C$,we get:
$\frac{\tan A + \tan B + \tan C}{\tan A \cdot \tan B \cdot \tan C} = 1$.

(D) Given expression: $\sin 2A + \sin 2B - \sin 2C$
Using the formula $\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$:
$= 2 \sin(A+B) \cos(A-B) - \sin 2C$
Since $A+B+C = \pi$,we have $A+B = \pi - C$,so $\sin(A+B) = \sin C$ and $\cos(A+B) = -\cos C$.
$= 2 \sin C \cos(A-B) - 2 \sin C \cos C$
$= 2 \sin C [\cos(A-B) - \cos C]$
$= 2 \sin C [\cos(A-B) + \cos(A+B)]$
Using the formula $\cos(X-Y) + \cos(X+Y) = 2 \cos X \cos Y$:
$= 2 \sin C [2 \cos A \cos B]$
$= 4 \cos A \cos B \sin C$.
Thus,the correct option is $(d)$.

(C) We know that in a triangle $ABC$,$A+B+C = 180^\circ$,so $\frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2}$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have:
${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2} = \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \sin^2 \frac{C}{2}$
$= 1 - \frac{1}{2}(\cos A + \cos B) + \sin^2 \frac{C}{2}$
$= 1 - \cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) + \sin^2 \frac{C}{2}$
Since $\cos(\frac{A+B}{2}) = \sin \frac{C}{2}$,we get:
$= 1 - \sin \frac{C}{2} \cos(\frac{A-B}{2}) + \sin^2 \frac{C}{2}$
$= 1 - \sin \frac{C}{2} [\cos(\frac{A-B}{2}) - \sin \frac{C}{2}]$
$= 1 - \sin \frac{C}{2} [\cos(\frac{A-B}{2}) - \cos(\frac{A+B}{2})]$
Using $\cos X - \cos Y = 2\sin(\frac{X+Y}{2})\sin(\frac{Y-X}{2})$,we get:
$= 1 - \sin \frac{C}{2} [2\sin \frac{A}{2} \sin \frac{B}{2}]$
$= 1 - 2\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

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

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

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

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

(B) Given $A + B + C = 180^\circ$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$.
Thus,$\frac{A}{2} = 90^\circ - (\frac{B}{2} + \frac{C}{2})$.
Taking tangent on both sides,$\tan \frac{A}{2} = \tan(90^\circ - (\frac{B}{2} + \frac{C}{2})) = \cot(\frac{B}{2} + \frac{C}{2})$.
Using the identity $\cot \theta = \frac{1}{\tan \theta}$,we get $\tan \frac{A}{2} = \frac{1}{\tan(\frac{B}{2} + \frac{C}{2})}$.
Therefore,$\tan \frac{A}{2} \tan(\frac{B}{2} + \frac{C}{2}) = 1$.
Expanding the tangent sum formula: $\tan \frac{A}{2} \left( \frac{\tan \frac{B}{2} + \tan \frac{C}{2}}{1 - \tan \frac{B}{2} \tan \frac{C}{2}} \right) = 1$.
$\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{A}{2} \tan \frac{C}{2} = 1 - \tan \frac{B}{2} \tan \frac{C}{2}$.
Rearranging the terms,we get $\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1$.

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

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

(D) Given $A + B + C = \frac{3\pi}{2}$.
We need to evaluate $\cos 2A + \cos 2B + \cos 2C$.
Using the formula $\cos X + \cos Y = 2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2}$,we get:
$\cos 2A + \cos 2B = 2\cos(A+B)\cos(A-B)$.
Since $A+B = \frac{3\pi}{2} - C$,then $\cos(A+B) = \cos(\frac{3\pi}{2} - C) = -\sin C$.
So,$\cos 2A + \cos 2B = -2\sin C \cos(A-B)$.
Now,$\cos 2C = 1 - 2\sin^2 C$.
Substituting these into the expression:
$-2\sin C \cos(A-B) + 1 - 2\sin^2 C = 1 - 2\sin C(\cos(A-B) + \sin C)$.
Since $\sin C = \sin(\frac{3\pi}{2} - (A+B)) = -\cos(A+B)$,
$= 1 - 2\sin C(\cos(A-B) - \cos(A+B))$.
Using $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$,we get:
$= 1 - 2\sin C(2\sin A \sin B) = 1 - 4\sin A \sin B \sin C$.

(D) The given function is $f(x) = \sin x + \cos x$.
We can write this in the form $a \sin x + b \cos x$,where $a = 1$ and $b = 1$.
The maximum value of the function $f(x) = a \sin x + b \cos x$ is given by the formula $\sqrt{a^2 + b^2}$.
Substituting the values,we get $\sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,the maximum value is $\sqrt{2}$.

(B) The function is given by $f(x) = \sqrt{3} \sin x + \cos x$.
To find the maximum distance from the $x$-axis,we need to find the maximum absolute value of the function,which is the amplitude of the expression $a \sin x + b \cos x$.
The amplitude is given by $\sqrt{a^2 + b^2}$.
Here,$a = \sqrt{3}$ and $b = 1$.
Maximum distance $= \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
Thus,the maximum distance of a point from the $x$-axis is $2$.

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

(C) The length of the wire is equal to the circumference of the circular wire with diameter $d_1 = 10\,cm$.
Length of wire $(l)$ = $\pi \times d_1 = 10\pi\,cm$.
The wire is placed along the circumference of a circle with diameter $d_2 = 1\,m = 100\,cm$.
The radius of this circle $(r)$ = $\frac{d_2}{2} = \frac{100}{2} = 50\,cm$.
The angle $( heta)$ subtended by an arc of length $l$ at the centre of a circle of radius $r$ is given by $\theta = \frac{l}{r}$.
Substituting the values,$\theta = \frac{10\pi}{50} = \frac{\pi}{5}\,\text{radian}$.

(D) The given expression is $S = \sin^2 5^\circ + \sin^2 10^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$.
This series contains $18$ terms (from $5^\circ$ to $90^\circ$ in steps of $5^\circ$).
We know that $\sin^2 \theta + \sin^2(90^\circ - \theta) = \sin^2 \theta + \cos^2 \theta = 1$.
Pairing the terms: $(\sin^2 5^\circ + \sin^2 85^\circ) + (\sin^2 10^\circ + \sin^2 80^\circ) + \dots + (\sin^2 40^\circ + \sin^2 50^\circ) + \sin^2 45^\circ + \sin^2 90^\circ$.
There are $8$ such pairs,each summing to $1$.
Thus,$S = 8(1) + \sin^2 45^\circ + \sin^2 90^\circ$.
Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$,$\sin^2 45^\circ = \frac{1}{2}$.
Since $\sin 90^\circ = 1$,$\sin^2 90^\circ = 1$.
Therefore,$S = 8 + \frac{1}{2} + 1 = 9\frac{1}{2}$.

(C) Given expression: $\sqrt{\csc^2 \alpha + 2\cot \alpha}$
Using the identity $\csc^2 \alpha = 1 + \cot^2 \alpha$,we get:
$= \sqrt{1 + \cot^2 \alpha + 2\cot \alpha}$
$= \sqrt{(1 + \cot \alpha)^2}$
$= |1 + \cot \alpha|$
Given the interval $\frac{3\pi}{4} < \alpha < \pi$,the value of $\cot \alpha$ lies in the range $(-\infty, -1)$.
Therefore,$\cot \alpha < -1$,which implies $1 + \cot \alpha < 0$.
Since the expression inside the absolute value is negative,$|1 + \cot \alpha| = -(1 + \cot \alpha) = -1 - \cot \alpha$.

(C) Given equations are:
$m = a \cos^3 \alpha + 3a \cos \alpha \sin^2 \alpha$
$n = a \sin^3 \alpha + 3a \cos^2 \alpha \sin \alpha$
Adding the two equations:
$m + n = a(\cos^3 \alpha + 3 \cos^2 \alpha \sin \alpha + 3 \cos \alpha \sin^2 \alpha + \sin^3 \alpha)$
$m + n = a(\cos \alpha + \sin \alpha)^3$
Subtracting the two equations:
$m - n = a(\cos^3 \alpha - 3 \cos^2 \alpha \sin \alpha + 3 \cos \alpha \sin^2 \alpha - \sin^3 \alpha)$
$m - n = a(\cos \alpha - \sin \alpha)^3$
Now,calculating $(m + n)^{2/3} + (m - n)^{2/3}$:
$= [a(\cos \alpha + \sin \alpha)^3]^{2/3} + [a(\cos \alpha - \sin \alpha)^3]^{2/3}$
$= a^{2/3} [(\cos \alpha + \sin \alpha)^2 + (\cos \alpha - \sin \alpha)^2]$
$= a^{2/3} [(\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha) + (\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha)]$
$= a^{2/3} [1 + 1] = 2a^{2/3}$.

(C) Given: $\cos(\theta - \alpha) = a$ and $\sin(\theta - \beta) = b$.
Let $x = \theta - \alpha$ and $y = \theta - \beta$. Then $\cos x = a$ and $\sin y = b$.
Note that $x - y = (\theta - \alpha) - (\theta - \beta) = \beta - \alpha$,so $\alpha - \beta = y - x$.
We need to find $\cos^2(\alpha - \beta) + 2ab\sin(\alpha - \beta) = \cos^2(y - x) + 2ab\sin(y - x)$.
Using $\cos(y - x) = \cos y \cos x + \sin y \sin x = \cos y \cdot a + b \cdot \sin x$.
Since $\sin x = \sqrt{1 - a^2}$ and $\cos y = \sqrt{1 - b^2}$,we have $\cos(y - x) = a\sqrt{1 - b^2} + b\sqrt{1 - a^2}$.
Using $\sin(y - x) = \sin y \cos x - \cos y \sin x = ba - \sqrt{1 - b^2}\sqrt{1 - a^2}$.
Substituting these into the expression:
$\cos^2(y - x) + 2ab\sin(y - x) = (a\sqrt{1 - b^2} + b\sqrt{1 - a^2})^2 + 2ab(ab - \sqrt{1 - b^2}\sqrt{1 - a^2})$
$= a^2(1 - b^2) + b^2(1 - a^2) + 2ab\sqrt{1 - b^2}\sqrt{1 - a^2} + 2a^2b^2 - 2ab\sqrt{1 - b^2}\sqrt{1 - a^2}$
$= a^2 - a^2b^2 + b^2 - a^2b^2 + 2a^2b^2 = a^2 + b^2$.

(B) Given $\sin A = n \sin B,$ we have $\frac{n}{1} = \frac{\sin A}{\sin B}.$
Applying the componendo and dividendo rule,we get $\frac{n - 1}{n + 1} = \frac{\sin A - \sin B}{\sin A + \sin B}.$
Using the sum-to-product formulas $\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$ and $\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2},$
$\frac{n - 1}{n + 1} = \frac{2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}}{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}} = \cot \frac{A + B}{2} \tan \frac{A - B}{2}.$
Multiplying both sides by $\tan \frac{A + B}{2},$ we get $\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = \tan \frac{A - B}{2} \cot \frac{A + B}{2} \tan \frac{A + B}{2} = \tan \frac{A - B}{2}.$

(B) Given that $x + \frac{1}{x} = 2\cos \theta$.
We use the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.
Substituting $a = x$ and $b = \frac{1}{x}$,we get:
${x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x \cdot \frac{1}{x}} \right)\left( {x + \frac{1}{x}} \right)$
$= {(2\cos \theta )^3} - 3(1)(2\cos \theta )$
$= 8\cos^3 \theta - 6\cos \theta$
$= 2(4\cos^3 \theta - 3\cos \theta)$
Using the trigonometric identity $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$,we obtain:
$= 2\cos 3\theta$.

(A) Given the equation: $\sin x + \csc x = 2$.
Since $\csc x = \frac{1}{\sin x}$,we can write the equation as $\sin x + \frac{1}{\sin x} = 2$.
Multiplying by $\sin x$,we get $\sin^2 x + 1 = 2 \sin x$,which simplifies to $\sin^2 x - 2 \sin x + 1 = 0$.
This is a perfect square: $(\sin x - 1)^2 = 0$.
Therefore,$\sin x = 1$.
Since $\sin x = 1$,then $\csc x = \frac{1}{1} = 1$.
Substituting these values into the expression $\sin^n x + \csc^n x$,we get $1^n + 1^n = 1 + 1 = 2$.

(A) Given $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.
Dividing numerator and denominator by $\cos \alpha$,we get $\tan \theta = \frac{\tan \alpha - 1}{\tan \alpha + 1} = \tan(\alpha - 45^\circ)$.
Thus,$\theta = \alpha - 45^\circ$,which implies $\alpha = \theta + 45^\circ$.
Now,$\sin \alpha + \cos \alpha = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \alpha + \frac{1}{\sqrt{2}} \cos \alpha \right) = \sqrt{2} \sin(\alpha + 45^\circ) = \sqrt{2} \sin(\theta + 45^\circ + 45^\circ) = \sqrt{2} \sin(\theta + 90^\circ) = \sqrt{2} \cos \theta$.
Similarly,$\sin \alpha - \cos \alpha = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \alpha - \frac{1}{\sqrt{2}} \cos \alpha \right) = \sqrt{2} \sin(\alpha - 45^\circ) = \sqrt{2} \sin(\theta + 45^\circ - 45^\circ) = \sqrt{2} \sin \theta$.

(B) Given the equation: $\cos^6 \alpha + \sin^6 \alpha + K \sin^2 2\alpha = 1$.
We know that $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.
Let $a = \cos^2 \alpha$ and $b = \sin^2 \alpha$. Then $a + b = \cos^2 \alpha + \sin^2 \alpha = 1$.
So,$\cos^6 \alpha + \sin^6 \alpha = (\cos^2 \alpha)^3 + (\sin^2 \alpha)^3 = (\cos^2 \alpha + \sin^2 \alpha)^3 - 3 \cos^2 \alpha \sin^2 \alpha (\cos^2 \alpha + \sin^2 \alpha)$.
$= 1^3 - 3 \cos^2 \alpha \sin^2 \alpha (1) = 1 - 3 \cos^2 \alpha \sin^2 \alpha$.
Substitute this into the original equation:
$1 - 3 \cos^2 \alpha \sin^2 \alpha + K \sin^2 2\alpha = 1$.
$-3 \cos^2 \alpha \sin^2 \alpha + K (2 \sin \alpha \cos \alpha)^2 = 0$.
$-3 \cos^2 \alpha \sin^2 \alpha + K (4 \sin^2 \alpha \cos^2 \alpha) = 0$.
Divide by $\sin^2 \alpha \cos^2 \alpha$ (assuming $\sin \alpha \cos \alpha \neq 0$):
$-3 + 4K = 0$.
$4K = 3$,which gives $K = \frac{3}{4}$.

(C) We use the identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta$.
Given expression: $E = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$.
Substitute $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
$E = \frac{\sqrt{3}}{2} (\sin 20^\circ \sin 40^\circ \sin 80^\circ)$.
Using the identity with $\theta = 20^\circ$:
$\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \sin(3 \times 20^\circ) = \frac{1}{4} \sin 60^\circ$.
Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,we have:
$\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.
Now substitute this back into the expression for $E$:
$E = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{8} = \frac{3}{16}$.

(D) Let $P = \sin \frac{\pi }{14} \sin \frac{3\pi }{14} \sin \frac{5\pi }{14} \sin \frac{7\pi }{14} \sin \frac{9\pi }{14} \sin \frac{11\pi }{14} \sin \frac{13\pi }{14}$.
Using the property $\sin(\pi - \theta) = \sin \theta$,we have:
$\sin \frac{13\pi }{14} = \sin \frac{\pi }{14}$,$\sin \frac{11\pi }{14} = \sin \frac{3\pi }{14}$,and $\sin \frac{9\pi }{14} = \sin \frac{5\pi }{14}$.
Also,$\sin \frac{7\pi }{14} = \sin \frac{\pi }{2} = 1$.
Substituting these,we get:
$P = \left( \sin \frac{\pi }{14} \sin \frac{3\pi }{14} \sin \frac{5\pi }{14} \right)^2 \times 1$.
Using the identity $\prod_{k=1}^{n} \sin \frac{k\pi}{2n+1} = \frac{\sqrt{2n+1}}{2^n}$,for $n=7$,$2n+1=15$ (not applicable directly).
Alternatively,using $\sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin \frac{5\pi}{14} = \frac{1}{8}$,we get:
$P = \left( \frac{1}{8} \right)^2 \times 1 = \frac{1}{64}$.

(C) We use the identity $\cot \theta - \tan \theta = 2\cot 2\theta$,which implies $\tan \theta = \cot \theta - 2\cot 2\theta$.
Applying this to the last term: $8\cot 8\alpha = 4(2\cot 8\alpha) = 4(\cot 4\alpha - \tan 4\alpha)$.
Substituting this into the expression:
$= \tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 4(\cot 4\alpha - \tan 4\alpha)$
$= \tan \alpha + 2\tan 2\alpha + 4\cot 4\alpha$
Now,apply the identity again for $4\cot 4\alpha = 2(2\cot 4\alpha) = 2(\cot 2\alpha - \tan 2\alpha)$:
$= \tan \alpha + 2\tan 2\alpha + 2(\cot 2\alpha - \tan 2\alpha)$
$= \tan \alpha + 2\cot 2\alpha$
Finally,apply the identity for $2\cot 2\alpha = \cot \alpha - \tan \alpha$:
$= \tan \alpha + (\cot \alpha - \tan \alpha)$
$= \cot \alpha$.

(C) $\sqrt{3} \csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$
$= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$
Multiply numerator and denominator by $2$:
$= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ \right)}{\frac{1}{2} (2 \sin 20^\circ \cos 20^\circ)}$
$= \frac{2 (\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)}{\frac{1}{2} \sin 40^\circ}$
$= \frac{2 \sin(60^\circ - 20^\circ)}{\frac{1}{2} \sin 40^\circ} = \frac{2 \sin 40^\circ}{\frac{1}{2} \sin 40^\circ} = 4$.

(C) Given expression: $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $
Using the identity $1 + \cos 2\theta = 2 \cos^2 \theta$,we have $1 + \cos 56^\circ = 2 \cos^2 28^\circ$.
Using the identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$,we have $\cos 58^\circ - \cos 66^\circ = -2 \sin \frac{58^\circ + 66^\circ}{2} \sin \frac{58^\circ - 66^\circ}{2} = -2 \sin 62^\circ \sin(-4^\circ) = 2 \sin 62^\circ \sin 4^\circ$.
Since $\sin 62^\circ = \cos 28^\circ$,the expression becomes:
$2 \cos^2 28^\circ + 2 \cos 28^\circ \sin 4^\circ = 2 \cos 28^\circ (\cos 28^\circ + \sin 4^\circ)$.
Since $\sin 4^\circ = \cos 86^\circ$,we have:
$2 \cos 28^\circ (\cos 28^\circ + \cos 86^\circ) = 2 \cos 28^\circ [2 \cos \frac{28^\circ + 86^\circ}{2} \cos \frac{86^\circ - 28^\circ}{2}]$
$= 4 \cos 28^\circ \cos 57^\circ \cos 29^\circ$.
Since $\cos 57^\circ = \sin 33^\circ$,the final result is $4 \cos 28^\circ \cos 29^\circ \sin 33^\circ$.

(B) Given: $x = \sin 130^\circ \cos 80^\circ$ and $y = \sin 80^\circ \cos 130^\circ$.
Using the identity $\sin(180^\circ - \theta) = \sin \theta$ and $\cos(180^\circ - \theta) = -\cos \theta$:
$x = \sin(180^\circ - 50^\circ) \cos 80^\circ = \sin 50^\circ \cos 80^\circ$. Since $50^\circ$ and $80^\circ$ are in the first quadrant,$\sin 50^\circ > 0$ and $\cos 80^\circ > 0$,so $x > 0$.
$y = \sin 80^\circ \cos(180^\circ - 50^\circ) = \sin 80^\circ (-\cos 50^\circ)$. Since $\sin 80^\circ > 0$ and $\cos 50^\circ > 0$,$y < 0$.
Now,$xy = (\sin 130^\circ \cos 80^\circ)(\sin 80^\circ \cos 130^\circ) = (\sin 130^\circ \cos 130^\circ)(\sin 80^\circ \cos 80^\circ)$.
Using $2 \sin \theta \cos \theta = \sin 2\theta$,we get $xy = (\frac{1}{2} \sin 260^\circ)(\frac{1}{2} \sin 160^\circ) = \frac{1}{4} \sin(270^\circ - 10^\circ) \sin(180^\circ - 20^\circ) = \frac{1}{4} (-\cos 10^\circ)(\sin 20^\circ)$.
Since $\cos 10^\circ > 0$ and $\sin 20^\circ > 0$,$xy < 0$.
Since $xy$ is a small negative value,$z = 1 + xy$ implies $0 < z < 1$.

(A) We have $\sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma)$.
Using the expansion $\sin(\alpha + \beta + \gamma) = \sin \alpha \cos \beta \cos \gamma + \cos \alpha \sin \beta \cos \gamma + \cos \alpha \cos \beta \sin \gamma - \sin \alpha \sin \beta \sin \gamma$.
Substituting this into the expression:
$= \sin \alpha(1 - \cos \beta \cos \gamma) + \sin \beta(1 - \cos \alpha \cos \gamma) + \sin \gamma(1 - \cos \alpha \cos \beta) + \sin \alpha \sin \beta \sin \gamma$.
Since $\alpha, \beta, \gamma \in (0, \pi/2)$,we have $\cos \beta \cos \gamma < 1$,$\cos \alpha \cos \gamma < 1$,and $\cos \alpha \cos \beta < 1$.
Thus,each term is positive,so $\sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma) > 0$.
Therefore,$\sin \alpha + \sin \beta + \sin \gamma > \sin(\alpha + \beta + \gamma)$.
This implies $\frac{\sin(\alpha + \beta + \gamma)}{\sin \alpha + \sin \beta + \sin \gamma} < 1$.

(B) Given the equation: $a \cos 2\theta + b \sin 2\theta = c$.
Using the half-angle identities $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we substitute these into the equation:
$a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = c$.
Multiplying both sides by $(1 + \tan^2 \theta)$:
$a(1 - \tan^2 \theta) + 2b \tan \theta = c(1 + \tan^2 \theta)$.
Rearranging the terms to form a quadratic equation in $\tan \theta$:
$a - a \tan^2 \theta + 2b \tan \theta = c + c \tan^2 \theta$.
$(a + c) \tan^2 \theta - 2b \tan \theta + (c - a) = 0$.
Since $\alpha$ and $\beta$ are the solutions for $\theta$,$\tan \alpha$ and $\tan \beta$ are the roots of this quadratic equation.
Using the sum of roots formula for a quadratic equation $Ax^2 + Bx + C = 0$,where the sum of roots is $-B/A$:
$\tan \alpha + \tan \beta = - \frac{-2b}{a + c} = \frac{2b}{c + a}$.

(B) Given expressions are $y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$.
Adding $y$ and $z$:
$y + z = a(\cos^2 x + \sin^2 x) + c(\sin^2 x + \cos^2 x) = a(1) + c(1) = a + c$.
Thus,$y + z = a + c$ is the correct relation.
Subtracting $z$ from $y$:
$y - z = a(\cos^2 x - \sin^2 x) + 4b \sin x \cos x - c(\cos^2 x - \sin^2 x) = (a - c) \cos 2x + 2b \sin 2x$.
Using $\tan x = \frac{2b}{a - c}$,we substitute $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$ and $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$.
$y - z = (a - c) \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) + 2b \left( \frac{2 \tan x}{1 + \tan^2 x} \right) = \frac{(a - c)(1 - \tan^2 x) + 4b \tan x}{1 + \tan^2 x}$.
Substituting $\tan x = \frac{2b}{a - c}$:
$y - z = \frac{(a - c)(1 - \frac{4b^2}{(a - c)^2}) + 4b(\frac{2b}{a - c})}{1 + \frac{4b^2}{(a - c)^2}} = \frac{(a - c) - \frac{4b^2}{a - c} + \frac{8b^2}{a - c}}{\frac{(a - c)^2 + 4b^2}{(a - c)^2}} = \frac{(a - c) + \frac{4b^2}{a - c}}{\frac{(a - c)^2 + 4b^2}{(a - c)^2}} = \frac{\frac{(a - c)^2 + 4b^2}{a - c}}{\frac{(a - c)^2 + 4b^2}{(a - c)^2}} = a - c$.

(B) Given,$\csc \theta = \frac{p + q}{p - q}$.
This implies $\frac{1}{\sin \theta} = \frac{p + q}{p - q}$.
Applying the componendo and dividendo rule:
$\frac{1 + \sin \theta}{1 - \sin \theta} = \frac{(p + q) + (p - q)}{(p + q) - (p - q)} = \frac{2p}{2q} = \frac{p}{q}$.
We know that $1 + \sin \theta = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2$ and $1 - \sin \theta = (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2$.
So,$\left( \frac{\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} - \sin \frac{\theta}{2}} \right)^2 = \frac{p}{q}$.
Dividing the numerator and denominator by $\cos \frac{\theta}{2}$,we get:
$\left( \frac{1 + \tan \frac{\theta}{2}}{1 - \tan \frac{\theta}{2}} \right)^2 = \frac{p}{q}$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = \frac{\pi}{4}$ and $B = \frac{\theta}{2}$:
$\tan^2 \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = \frac{p}{q}$.
Taking the reciprocal:
$\cot^2 \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = \frac{q}{p}$.
Therefore,$\cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = \sqrt{\frac{q}{p}}$.

(B) Given: $a \sin^2 x + b \cos^2 x = c$
Divide by $\cos^2 x$: $a \tan^2 x + b = c \sec^2 x = c(1 + \tan^2 x)$
$a \tan^2 x + b = c + c \tan^2 x \Rightarrow (a - c) \tan^2 x = c - b \Rightarrow \tan^2 x = \frac{b - c}{a - c} = \frac{b - c}{-(c - a)} = \frac{b - c}{c - a}$
Given: $b \sin^2 y + a \cos^2 y = d$
Divide by $\cos^2 y$: $b \tan^2 y + a = d \sec^2 y = d(1 + \tan^2 y)$
$b \tan^2 y + a = d + d \tan^2 y \Rightarrow (b - d) \tan^2 y = d - a \Rightarrow \tan^2 y = \frac{d - a}{b - d} = \frac{-(a - d)}{-(d - b)} = \frac{a - d}{d - b}$
Given: $a \tan x = b \tan y \Rightarrow \frac{\tan x}{\tan y} = \frac{b}{a} \Rightarrow \frac{\tan^2 x}{\tan^2 y} = \frac{b^2}{a^2}$
Substituting the values: $\frac{b^2}{a^2} = \frac{(b - c)/(c - a)}{(a - d)/(d - b)} = \frac{(b - c)(d - b)}{(c - a)(a - d)}$
Therefore,$\frac{a^2}{b^2} = \frac{(c - a)(a - d)}{(b - c)(d - b)}$.

(C) Using the sum-to-product formulas:
$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
Substituting these into the expression:
First term: $\left( \frac{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}} \right)^n = \cot^n \left( \frac{A-B}{2} \right)$
Second term: $\left( \frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}} \right)^n = (-\cot \left( \frac{A-B}{2} \right))^n$
Total expression = $\cot^n \left( \frac{A-B}{2} \right) + (-1)^n \cot^n \left( \frac{A-B}{2} \right)$
If $n$ is even,$(-1)^n = 1$,so the result is $2 \cot^n \left( \frac{A-B}{2} \right)$.
If $n$ is odd,$(-1)^n = -1$,so the result is $0$.

(A) Given $\sin \alpha = 1/\sqrt{5}$,then $\cos \alpha = \sqrt{1 - (1/\sqrt{5})^2} = 2/\sqrt{5}$.
Given $\sin \beta = 3/5$,then $\cos \beta = \sqrt{1 - (3/5)^2} = 4/5$.
Using the formula $\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha$:
$\sin(\beta - \alpha) = (3/5)(2/\sqrt{5}) - (4/5)(1/\sqrt{5}) = 6/(5\sqrt{5}) - 4/(5\sqrt{5}) = 2/(5\sqrt{5})$.
Since $5\sqrt{5} \approx 11.18$,$\sin(\beta - \alpha) \approx 2/11.18 \approx 0.1789$.
We know $\sin(0) = 0$ and $\sin(\pi/4) = 1/\sqrt{2} \approx 0.7071$.
Since $0 < 0.1789 < 0.7071$,it follows that $0 < \beta - \alpha < \pi/4$.
Thus,$\beta - \alpha$ lies in the interval $[0, \pi/4]$.

(A) The given equation is $2 \sec 2\alpha = \tan \beta + \cot \beta$.
We can rewrite the right side as:
$\tan \beta + \cot \beta = \frac{\sin \beta}{\cos \beta} + \frac{\cos \beta}{\sin \beta} = \frac{\sin^2 \beta + \cos^2 \beta}{\sin \beta \cos \beta} = \frac{1}{\sin \beta \cos \beta}$.
Multiplying numerator and denominator by $2$,we get $\frac{2}{2 \sin \beta \cos \beta} = \frac{2}{\sin 2\beta}$.
So,the equation becomes $2 \sec 2\alpha = \frac{2}{\sin 2\beta}$,which simplifies to $\frac{2}{\cos 2\alpha} = \frac{2}{\sin 2\beta}$.
This implies $\cos 2\alpha = \sin 2\beta$.
Using the identity $\sin \theta = \cos(\frac{\pi}{2} - \theta)$,we have $\cos 2\alpha = \cos(\frac{\pi}{2} - 2\beta)$.
Therefore,$2\alpha = \frac{\pi}{2} - 2\beta$,which gives $2\alpha + 2\beta = \frac{\pi}{2}$.
Dividing by $2$,we get $\alpha + \beta = \frac{\pi}{4}$.

(B) Let $\frac{x}{\cos \theta} = \frac{y}{\cos \left( \theta - \frac{2\pi}{3} \right)} = \frac{z}{\cos \left( \theta + \frac{2\pi}{3} \right)} = k.$
Then,$x = k \cos \theta, y = k \cos \left( \theta - \frac{2\pi}{3} \right),$ and $z = k \cos \left( \theta + \frac{2\pi}{3} \right).$
Now,$x + y + z = k \left[ \cos \theta + \cos \left( \theta - \frac{2\pi}{3} \right) + \cos \left( \theta + \frac{2\pi}{3} \right) \right].$
Using the formula $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B,$
$x + y + z = k \left[ \cos \theta + 2 \cos \theta \cos \left( \frac{2\pi}{3} \right) \right].$
Since $\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2},$
$x + y + z = k \left[ \cos \theta + 2 \cos \theta \left( -\frac{1}{2} \right) \right] = k [\cos \theta - \cos \theta] = k(0) = 0.$
Therefore,$x + y + z = 0.$

(D) We know that $\sin 6\theta = 2 \sin 3\theta \cos 3\theta$.
Using the triple angle identities $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$ and $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$:
$\sin 6\theta = 2(3\sin \theta - 4\sin^3 \theta)(4\cos^3 \theta - 3\cos \theta)$
$= 2(12\sin \theta \cos^3 \theta - 9\sin \theta \cos \theta - 16\sin^3 \theta \cos^3 \theta + 12\sin^3 \theta \cos \theta)$
$= 24\sin \theta \cos^3 \theta - 18\sin \theta \cos \theta - 32\sin^3 \theta \cos^3 \theta + 24\sin^3 \theta \cos \theta$
Using $\sin^2 \theta = 1 - \cos^2 \theta$ and $\cos^2 \theta = 1 - \sin^2 \theta$,we simplify the expression to match the form $32\cos^5 \theta \sin \theta - 32\cos^3 \theta \sin \theta + 3x$.
After expansion and rearrangement,we get $\sin 6\theta = 32\cos^5 \theta \sin \theta - 32\cos^3 \theta \sin \theta + 3\sin 2\theta$.
Comparing this with the given equation,we find $x = \sin 2\theta$.