Trigonometry Questions in English

(D) Given that $\sin x + \sin^2 x = 1.$
This implies $\sin x = 1 - \sin^2 x = \cos^2 x.$
Now,we need to evaluate $\cos^8 x + 2\cos^6 x + \cos^4 x.$
Substituting $\cos^2 x = \sin x$ into the expression:
$= (\cos^2 x)^4 + 2(\cos^2 x)^3 + (\cos^2 x)^2$
$= (\sin x)^4 + 2(\sin x)^3 + (\sin x)^2$
$= \sin^4 x + 2\sin^3 x + \sin^2 x$
$= (\sin^2 x + \sin x)^2.$
Since $\sin x + \sin^2 x = 1,$ the expression becomes $(1)^2 = 1.$

(C) Given equations are:
$x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ --- $(i)$
$x \sin \alpha - y \cos \alpha = 0$ --- $(ii)$
From equation $(ii)$,we have $x \sin \alpha = y \cos \alpha$.
Substitute $x \sin \alpha = y \cos \alpha$ into equation $(i)$:
$(x \sin \alpha) \sin^2 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$
$(y \cos \alpha) \sin^2 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$
$y \cos \alpha (\sin^2 \alpha + \cos^2 \alpha) = \sin \alpha \cos \alpha$
Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we get:
$y \cos \alpha = \sin \alpha \cos \alpha$
Assuming $\cos \alpha \neq 0$,we get $y = \sin \alpha$.
Substituting $y = \sin \alpha$ into $x \sin \alpha = y \cos \alpha$,we get $x \sin \alpha = \sin \alpha \cos \alpha$,which implies $x = \cos \alpha$.
Therefore,$x^2 + y^2 = \cos^2 \alpha + \sin^2 \alpha = 1$.

(B) Let $x = (1 + \sin A)(1 + \sin B)(1 + \sin C)$ and $y = (1 - \sin A)(1 - \sin B)(1 - \sin C)$.
Given $x = y$.
Multiply both sides by $y$:
$xy = y^2$
$(1 + \sin A)(1 - \sin A)(1 + \sin B)(1 - \sin B)(1 + \sin C)(1 - \sin C) = y^2$
$(1 - \sin^2 A)(1 - \sin^2 B)(1 - \sin^2 C) = y^2$
$\cos^2 A \cos^2 B \cos^2 C = y^2$
Taking the square root of both sides:
$y = \pm \cos A \cos B \cos C$
Since $x = y$,each side is equal to $\pm \cos A \cos B \cos C$.

(A) Given: $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma ) = \tan \alpha \tan \beta \tan \gamma$ ... $(i)$
Let $x = (\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma )$ ... $(ii)$
Multiplying equations $(i)$ and $(ii)$,we get:
$(\sec^2 \alpha - \tan^2 \alpha )(\sec^2 \beta - \tan^2 \beta )(\sec^2 \gamma - \tan^2 \gamma ) = x \cdot (\tan \alpha \tan \beta \tan \gamma )$
Since $\sec^2 \theta - \tan^2 \theta = 1$ for any angle $\theta$,the left side becomes $(1)(1)(1) = 1$.
Therefore,$1 = x \cdot (\tan \alpha \tan \beta \tan \gamma )$.
$x = \frac{1}{\tan \alpha \tan \beta \tan \gamma } = \cot \alpha \cot \beta \cot \gamma $.

(D) Given that $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$.
Since the maximum value of the sine function is $1$,this equation can only hold true if each term is equal to $1$.
Therefore,$\sin \theta_1 = 1$,$\sin \theta_2 = 1$,and $\sin \theta_3 = 1$.
This implies that $\theta_1 = \theta_2 = \theta_3 = \frac{\pi}{2}$.
Now,substituting these values into the expression $\cos \theta_1 + \cos \theta_2 + \cos \theta_3$:
$\cos(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 0 + 0 + 0 = 0$.

(D) We know that the range of $\sin^2 \theta$ is $0 \le \sin^2 \theta \le 1$.
Given the equation $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,it must satisfy the condition $\frac{x^2 + y^2 + 1}{2x} \le 1$.
Assuming $x > 0$,we have $x^2 + y^2 + 1 \le 2x$.
Rearranging the terms,we get $x^2 - 2x + 1 + y^2 \le 0$.
This simplifies to $(x - 1)^2 + y^2 \le 0$.
Since the sum of squares of real numbers can only be non-negative,the only way for this inequality to hold is if $(x - 1)^2 = 0$ and $y^2 = 0$.
This implies $x = 1$ and $y = 0$.
Since the value of $x$ depends on the value of $y$ (i.e.,$x$ is not fixed as $1$ unless $y=0$),and the options provided do not explicitly state this dependency,the correct choice is "None of these".

(D) Given that $\tan \theta - \cot \theta = a$ $(i)$ and $\sin \theta + \cos \theta = b$ $(ii)$.
Now,consider the expression $({b^2} - 1)^2({a^2} + 4)$.
Substitute $b = \sin \theta + \cos \theta$:
$b^2 - 1 = (\sin \theta + \cos \theta)^2 - 1 = 1 + 2 \sin \theta \cos \theta - 1 = 2 \sin \theta \cos \theta = \sin 2\theta$.
So,$(b^2 - 1)^2 = \sin^2 2\theta$.
Substitute $a = \tan \theta - \cot \theta$:
$a^2 + 4 = (\tan \theta - \cot \theta)^2 + 4 = \tan^2 \theta + \cot^2 \theta - 2 + 4 = \tan^2 \theta + \cot^2 \theta + 2 = (\tan \theta + \cot \theta)^2$.
Since $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$.
Therefore,$(a^2 + 4) = (\frac{2}{\sin 2\theta})^2 = \frac{4}{\sin^2 2\theta}$.
Multiplying the two parts:
$(b^2 - 1)^2(a^2 + 4) = \sin^2 2\theta \times \frac{4}{\sin^2 2\theta} = 4$.
Alternatively,since the expression is independent of $\theta$,let $\theta = 45^\circ$. Then $a = \tan 45^\circ - \cot 45^circ = 1 - 1 = 0$ and $b = \sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
Substituting these values: $((\sqrt{2})^2 - 1)^2(0^2 + 4) = (2 - 1)^2(4) = 1^2 \times 4 = 4$.

(C) Let $x = \tan^2 \alpha$,$y = \tan^2 \beta$,and $z = \tan^2 \gamma$.
The given equation is $xy + yz + zx + 2xyz = 1$.
We want to find the value of $S = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Since $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$,we have $S = \frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z}$.
$S = \frac{x(1+y)(1+z) + y(1+x)(1+z) + z(1+x)(1+y)}{(1+x)(1+y)(1+z)}$.
Expanding the numerator: $x(1+y+z+yz) + y(1+x+z+xz) + z(1+x+y+xy) = x+xy+xz+xyz + y+yx+yz+xyz + z+zx+zy+xyz = (x+y+z) + 2(xy+yz+zx) + 3xyz$.
Using the identity $(1+x)(1+y)(1+z) = 1 + (x+y+z) + (xy+yz+zx) + xyz$,and substituting $xy+yz+zx = 1 - 2xyz$,we get:
$S = \frac{(x+y+z) + 2(1-2xyz) + 3xyz}{1 + (x+y+z) + (1-2xyz) + xyz} = \frac{x+y+z + 2 - 4xyz + 3xyz}{2 + x+y+z - xyz} = \frac{x+y+z + 2 - xyz}{2 + x+y+z - xyz} = 1$.

(A) The given expression is a product of cosine values from $1^\circ$ to $179^\circ$.
We know that the sequence of angles includes $90^\circ$.
Since $\cos 90^\circ = 0$,the entire product becomes zero because any number multiplied by zero is zero.
Therefore,$\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \dots \cos 90^\circ \dots \cos 179^\circ = 0$.

(A) We use the trigonometric identities: $\tan(90^\circ - \theta) = \cot \theta$ and $\cot(90^\circ - \theta) = \tan \theta$.
Given expression: $\frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ}$.
Step $1$: Convert the denominator of the first term using $\tan 36^\circ = \tan(90^\circ - 54^\circ) = \cot 54^\circ$.
Step $2$: Convert the denominator of the second term using $\cot 70^\circ = \cot(90^\circ - 20^\circ) = \tan 20^\circ$.
Step $3$: Substitute these values into the expression:
$\frac{\cot 54^\circ}{\cot 54^\circ} + \frac{\tan 20^\circ}{\tan 20^\circ} = 1 + 1 = 2$.

(B) The given expression is a sum of sines of angles in an arithmetic progression: $S = \sum_{k=1}^{36} \sin(k \cdot 10^\circ)$.
Using the formula for the sum of sines of angles in arithmetic progression: $\sum_{k=1}^{n} \sin(a + (k-1)d) = \frac{\sin(n d / 2)}{\sin(d / 2)} \cdot \sin\left(a + \frac{(n-1)d}{2}\right)$.
Here,$a = 10^\circ$,$d = 10^\circ$,and $n = 36$.
$S = \frac{\sin(36 \cdot 10^\circ / 2)}{\sin(10^\circ / 2)} \cdot \sin\left(10^\circ + \frac{35 \cdot 10^\circ}{2}\right)$.
$S = \frac{\sin(180^\circ)}{\sin(5^\circ)} \cdot \sin(10^\circ + 175^\circ)$.
Since $\sin(180^\circ) = 0$,the entire expression evaluates to $0$.

(C) The given expression is $S = \cos 1^\circ + \cos 2^\circ + \dots + \cos 179^\circ + \cos 180^\circ$.
We know that $\cos(180^\circ - \theta) = -\cos \theta$.
Therefore,$\cos 179^\circ = -\cos 1^\circ$,$\cos 178^\circ = -\cos 2^\circ$,and so on.
Pairing the terms from $1^\circ$ to $179^\circ$ as $(\cos 1^\circ + \cos 179^\circ) + (\cos 2^\circ + \cos 178^\circ) + \dots + (\cos 89^\circ + \cos 91^\circ) + \cos 90^\circ$.
Since $\cos(180^\circ - \theta) + \cos \theta = 0$,each pair sums to $0$.
Also,$\cos 90^\circ = 0$.
Thus,the sum of the first $179$ terms is $0$.
The remaining term is $\cos 180^\circ = -1$.
Therefore,the total sum is $0 + (-1) = -1$.

(A) Given $\alpha = 22^\circ 30' = 22.5^\circ$.
We need to evaluate $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.
Note that $3\alpha = 67.5^\circ$,$5\alpha = 112.5^\circ = 180^\circ - 67.5^\circ$,and $7\alpha = 157.5^\circ = 180^\circ - 22.5^\circ$.
Using $\cos(180^\circ - \theta) = -\cos \theta$,we have:
$1 + \cos 5\alpha = 1 - \cos 67.5^\circ = 1 - \cos 3\alpha$
$1 + \cos 7\alpha = 1 - \cos 22.5^\circ = 1 - \cos \alpha$
Substituting these into the expression:
$P = (1 + \cos \alpha)(1 - \cos \alpha)(1 + \cos 3\alpha)(1 - \cos 3\alpha)$
$P = (1 - \cos^2 \alpha)(1 - \cos^2 3\alpha) = \sin^2 \alpha \cdot \sin^2 3\alpha$
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$\sin^2 22.5^\circ = \frac{1 - \cos 45^\circ}{2} = \frac{1 - 1/\sqrt{2}}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$
$\sin^2 67.5^\circ = \frac{1 - \cos 135^\circ}{2} = \frac{1 - (-1/\sqrt{2})}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}}$
$P = \left( \frac{\sqrt{2} - 1}{2\sqrt{2}} \right) \left( \frac{\sqrt{2} + 1}{2\sqrt{2}} \right) = \frac{(\sqrt{2})^2 - 1^2}{8} = \frac{2 - 1}{8} = \frac{1}{8}$.

(C) Given expression: $6(\sin^6 \theta + \cos^6 \theta) - 9(\sin^4 \theta + \cos^4 \theta) + 4$
Using the identities $\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)^3 - 3\sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1 - 3\sin^2 \theta \cos^2 \theta$ and $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$.
Substituting these into the expression:
$= 6(1 - 3\sin^2 \theta \cos^2 \theta) - 9(1 - 2\sin^2 \theta \cos^2 \theta) + 4$
$= 6 - 18\sin^2 \theta \cos^2 \theta - 9 + 18\sin^2 \theta \cos^2 \theta + 4$
$= 6 - 9 + 4$
$= 1$.

(A) Given expression: $\sin 15^\circ + \cos 105^\circ$
Using the trigonometric identity $\cos(90^\circ + \theta) = -\sin \theta$,we can rewrite $\cos 105^\circ$ as:
$\cos 105^\circ = \cos(90^\circ + 15^\circ) = -\sin 15^\circ$
Substituting this back into the original expression:
$\sin 15^\circ + (-\sin 15^\circ) = \sin 15^\circ - \sin 15^\circ = 0$
Therefore,the correct option is $A$.

(D) We are given the expression $\cos 105^\circ + \sin 105^\circ$.
Using the trigonometric identities $\cos(90^\circ + \theta) = -\sin \theta$ and $\sin(90^\circ + \theta) = \cos \theta$:
$\cos 105^\circ = \cos(90^\circ + 15^\circ) = -\sin 15^\circ$
$\sin 105^\circ = \sin(90^\circ + 15^\circ) = \cos 15^\circ$
Substituting these values into the expression:
$\cos 105^\circ + \sin 105^\circ = \cos 15^\circ - \sin 15^\circ$
Using the values $\cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$ and $\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$:
$= \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}}$
$= \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2\sqrt{2}}$
$= \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(D) Given expression: $E = \cos y \cos \left( \frac{\pi}{2} - x \right) - \cos \left( \frac{\pi}{2} - y \right) \cos x + \sin y \cos \left( \frac{\pi}{2} - x \right) + \cos x \sin \left( \frac{\pi}{2} - y \right)$.
Using the identities $\cos \left( \frac{\pi}{2} - \theta \right) = \sin \theta$ and $\sin \left( \frac{\pi}{2} - \theta \right) = \cos \theta$,we get:
$E = \cos y \sin x - \sin y \cos x + \sin y \sin x + \cos x \cos x$.
Wait,let us re-evaluate the expression: $E = \sin x \cos y - \cos x \sin y + \sin y \sin x + \cos x \cos x$.
$E = \sin(x - y) + \cos(x - y)$.
To make $E = 0$,we have $\sin(x - y) + \cos(x - y) = 0$.
$\Rightarrow \sin(x - y) = -\cos(x - y)$.
$\Rightarrow \tan(x - y) = -1$.
$\Rightarrow x - y = n\pi - \frac{\pi}{4}$.
$\Rightarrow x = n\pi - \frac{\pi}{4} + y$,where $n \in I$.

(C) We know that $\frac{\pi}{10} = 18^\circ$ and $\frac{3\pi}{10} = 54^\circ$.
Thus,the expression is $\sin 18^\circ \sin 54^\circ$.
We know that $\sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ$.
Using the standard values: $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.
Multiplying these values: $\left( \frac{\sqrt{5} - 1}{4} \right) \left( \frac{\sqrt{5} + 1}{4} \right) = \frac{(\sqrt{5})^2 - 1^2}{16} = \frac{5 - 1}{16} = \frac{4}{16} = \frac{1}{4}$.

(C) Given the equation: $x \sin 45^\circ \cos^2 60^\circ = \frac{\tan^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$
Substitute the trigonometric values: $\sin 45^\circ = \frac{1}{\sqrt{2}}$,$\cos 60^\circ = \frac{1}{2}$,$\tan 60^\circ = \sqrt{3}$,$\csc 30^\circ = 2$,$\sec 45^\circ = \sqrt{2}$,$\cot 30^\circ = \sqrt{3}$.
$x \cdot \frac{1}{\sqrt{2}} \cdot (\frac{1}{2})^2 = \frac{(\sqrt{3})^2 \cdot 2}{\sqrt{2} \cdot (\sqrt{3})^2}$
$x \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{4} = \frac{3 \cdot 2}{\sqrt{2} \cdot 3}$
$\frac{x}{4\sqrt{2}} = \frac{6}{3\sqrt{2}}$
$\frac{x}{4\sqrt{2}} = \frac{2}{\sqrt{2}}$
$x = 2 \cdot 4 = 8$.

(A) Given $x = \sin 130^\circ + \cos 130^\circ$.
Using the identity $\sin A = \cos(90^\circ - A)$,we have $\sin 130^\circ = \cos(90^\circ - 130^\circ) = \cos(-40^\circ) = \cos 40^\circ$.
So,$x = \cos 40^\circ + \cos 130^\circ$.
Using the sum-to-product formula $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$x = 2 \cos(\frac{40^\circ + 130^\circ}{2}) \cos(\frac{40^\circ - 130^\circ}{2})$
$x = 2 \cos(85^\circ) \cos(-45^\circ) = 2 \cos 85^\circ \cos 45^\circ$.
Since $85^\circ$ is in the first quadrant,$\cos 85^\circ > 0$,and $\cos 45^\circ > 0$.
Therefore,$x > 0$.

(B) We use the trigonometric reduction formulas:
$1$. $\sin (270^\circ + A) = -\cos A$
$2$. $\sin (270^\circ - A) = -\cos A$
$3$. $\cos (180^\circ + A) = -\cos A$
Substituting these values into the expression:
$\cos A + (-\cos A) - (-\cos A) + (-\cos A)$
$= \cos A - \cos A + \cos A - \cos A$
$= 0$

(B) Given expression: $E = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}}$
Taking the common denominator:
$E = \frac{(1 - \cos \alpha) + (1 + \cos \alpha)}{\sqrt{(1 + \cos \alpha)(1 - \cos \alpha)}}$
$E = \frac{2}{\sqrt{1 - \cos^2 \alpha}} = \frac{2}{\sqrt{\sin^2 \alpha}} = \frac{2}{|\sin \alpha|}$
Since $\pi < \alpha < \frac{3\pi}{2}$,the angle $\alpha$ lies in the third quadrant.
In the third quadrant,$\sin \alpha$ is negative.
Therefore,$|\sin \alpha| = -\sin \alpha$.
Thus,$E = \frac{2}{-\sin \alpha} = -\frac{2}{\sin \alpha}$.

(A) We use the expansion formulas for $\tan(A+B)$ and $\tan(A-B)$:
$\tan \left( \frac{\pi }{4} + \theta \right) = \frac{\tan(\pi/4) + \tan \theta}{1 - \tan(\pi/4)\tan \theta} = \frac{1 + \tan \theta}{1 - \tan \theta}$
$\tan \left( \frac{\pi }{4} - \theta \right) = \frac{\tan(\pi/4) - \tan \theta}{1 + \tan(\pi/4)\tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}$
Now,subtract the two expressions:
$\frac{1 + \tan \theta}{1 - \tan \theta} - \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{(1 + \tan \theta)^2 - (1 - \tan \theta)^2}{(1 - \tan \theta)(1 + \tan \theta)}$
$= \frac{(1 + 2\tan \theta + \tan^2 \theta) - (1 - 2\tan \theta + \tan^2 \theta)}{1 - \tan^2 \theta}$
$= \frac{4\tan \theta}{1 - \tan^2 \theta}$
$= 2 \left( \frac{2\tan \theta}{1 - \tan^2 \theta} \right)$
Since $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$,the expression becomes $2\tan 2\theta$.

(B) We use the trigonometric identities for supplementary and allied angles:
$\sin (\pi + \theta ) = -\sin \theta $
$\sin (\pi - \theta ) = \sin \theta $
Substituting these into the expression:
$= (-\sin \theta )(\sin \theta ) \csc^2 \theta $
$= -\sin^2 \theta \cdot \frac{1}{\sin^2 \theta }$
$= -1$

(C) We know that $\cot(A) = \tan(90^\circ - A)$.
Applying this to the first term: $\cot(45^\circ + \theta) = \tan(90^\circ - (45^\circ + \theta)) = \tan(45^\circ - \theta)$.
Substituting this back into the expression:
$\cot(45^\circ + \theta) \cot(45^\circ - \theta) = \tan(45^\circ - \theta) \cot(45^\circ - \theta)$.
Since $\tan(x) \cot(x) = 1$ for any angle $x$,we have:
$\tan(45^\circ - \theta) \cot(45^\circ - \theta) = 1$.

(A) We use the trigonometric reduction formulas:
$1$. $\cot (180^\circ + A) = \cot A$ (Since it is in the third quadrant where $\cot$ is positive).
$2$. $\cot (90^\circ + A) = -\tan A$ (Since it is in the second quadrant where $\cot$ is negative).
$3$. $\cot (360^\circ - A) = -\cot A$ (Since it is in the fourth quadrant where $\cot$ is negative).
Substituting these values into the expression:
$\tan A + \cot A - \tan A - \cot A$
$= (\tan A - \tan A) + (\cot A - \cot A)$
$= 0 + 0 = 0$.

(D) Given expression: $\tan \theta \sin \left( \frac{\pi }{2} + \theta \right) \cos \left( \frac{\pi }{2} - \theta \right)$
Using trigonometric identities:
$\sin \left( \frac{\pi }{2} + \theta \right) = \cos \theta$
$\cos \left( \frac{\pi }{2} - \theta \right) = \sin \theta$
Substituting these into the expression:
$= \tan \theta \cdot \cos \theta \cdot \sin \theta$
$= \frac{\sin \theta}{\cos \theta} \cdot \cos \theta \cdot \sin \theta$
$= \sin^2 \theta$
Since $\sin^2 \theta$ is not equal to any of the given constant options,the correct answer is None of these.

(A) Let the two parts be $A$ and $B$ such that $A + B = \theta$ and $A - B = \phi$.
Given that $\tan A = k \tan B$,which implies $\frac{\tan A}{\tan B} = k$.
Using the sine and cosine definitions: $\frac{\sin A \cos B}{\cos A \sin B} = k$.
Applying the componendo and dividendo rule:
$\frac{k + 1}{k - 1} = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}$.
Using the trigonometric identity $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:
$\frac{k + 1}{k - 1} = \frac{\sin(A + B)}{\sin(A - B)} = \frac{\sin \theta}{\sin \phi}$.
Therefore,$\sin \theta = \frac{k + 1}{k - 1} \sin \phi$.

(B) Given the equation $x = y \cos \frac{2\pi}{3} = z \cos \frac{4\pi}{3}$.
We know that $\cos \frac{2\pi}{3} = \cos(120^\circ) = -\frac{1}{2}$ and $\cos \frac{4\pi}{3} = \cos(240^\circ) = -\frac{1}{2}$.
Substituting these values,we get $x = y(-\frac{1}{2}) = z(-\frac{1}{2})$.
Let $x = y(-\frac{1}{2}) = z(-\frac{1}{2}) = k$,where $k$ is a constant.
Then $x = k$,$y = -2k$,and $z = -2k$.
Now,calculate the expression $xy + yz + zx$:
$xy + yz + zx = (k)(-2k) + (-2k)(-2k) + (-2k)(k)$
$= -2k^2 + 4k^2 - 2k^2$
$= 0$.

(A) Given $\pi < \alpha < \frac{3\pi}{2},$ $\alpha$ lies in the third quadrant.
Simplify the expression: $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)$
$= \sqrt{4\sin^4 \alpha + 4\sin^2 \alpha \cos^2 \alpha} + 2 \left[ 2\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right) \right]$
$= \sqrt{4\sin^2 \alpha (\sin^2 \alpha + \cos^2 \alpha)} + 2 \left[ 1 + \cos \left( \frac{\pi}{2} - \alpha \right) \right]$
$= \sqrt{4\sin^2 \alpha} + 2(1 + \sin \alpha) = 2|\sin \alpha| + 2 + 2\sin \alpha$
Since $\pi < \alpha < \frac{3\pi}{2},$ $\sin \alpha$ is negative,so $|\sin \alpha| = -\sin \alpha$.
Expression $= 2(-\sin \alpha) + 2 + 2\sin \alpha = 2$.

(D) Given expression: $S = \sin ^2 \frac{\pi }{8} + \sin ^2 \frac{3\pi }{8} + \sin ^2 \frac{5\pi }{8} + \sin ^2 \frac{7\pi }{8}$
Using the identity $\sin(\pi - \theta) = \sin \theta$,we have:
$\sin \frac{5\pi }{8} = \sin(\pi - \frac{3\pi }{8}) = \sin \frac{3\pi }{8}$
$\sin \frac{7\pi }{8} = \sin(\pi - \frac{\pi }{8}) = \sin \frac{\pi }{8}$
Substituting these into the expression:
$S = \sin ^2 \frac{\pi }{8} + \sin ^2 \frac{3\pi }{8} + \sin ^2 \frac{3\pi }{8} + \sin ^2 \frac{\pi }{8}$
$S = 2(\sin ^2 \frac{\pi }{8} + \sin ^2 \frac{3\pi }{8})$
Using $\sin^2 \theta + \sin^2(\frac{\pi}{2} - \theta) = \sin^2 \theta + \cos^2 \theta = 1$:
Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,then $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$.
$S = 2(\sin ^2 \frac{\pi }{8} + \cos ^2 \frac{\pi }{8}) = 2(1) = 2$.

(A) Given expression: $(\sec A + \tan A - 1)(\sec A - \tan A + 1) - 2\tan A$
Rewrite the first part as: $(\sec A + (\tan A - 1))(\sec A - (\tan A - 1)) - 2\tan A$
Using the identity $(x+y)(x-y) = x^2 - y^2$,where $x = \sec A$ and $y = \tan A - 1$:
$= \sec^2 A - (\tan A - 1)^2 - 2\tan A$
$= \sec^2 A - (\tan^2 A - 2\tan A + 1) - 2\tan A$
$= \sec^2 A - \tan^2 A + 2\tan A - 1 - 2\tan A$
Since $\sec^2 A - \tan^2 A = 1$:
$= 1 - 1 = 0$

(B) Given that $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$.
We know the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting the values: $\tan(A + B) = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \frac{5/6}{1 - 1/6} = \frac{5/6}{5/6} = 1$.
Since $\tan(A + B) = 1$,we have $A + B = 45^\circ$.
Therefore,$2A = 90^\circ - 2B$.
Taking cosine on both sides: $\cos 2A = \cos(90^\circ - 2B)$.
Using the identity $\cos(90^\circ - \theta) = \sin \theta$,we get $\cos 2A = \sin 2B$.

(D) We use the trigonometric reduction formulas:
$1$. $\cos (270^\circ + \theta) = \sin \theta$
$2$. $\cos (90^\circ - \theta) = \sin \theta$
$3$. $\sin (270^\circ - \theta) = -\cos \theta$
Substituting these into the expression:
$\cos (270^\circ + \theta) \cos (90^\circ - \theta) - \sin (270^\circ - \theta) \cos \theta$
$= (\sin \theta)(\sin \theta) - (-\cos \theta)(\cos \theta)$
$= \sin^2 \theta + \cos^2 \theta$
Since $\sin^2 \theta + \cos^2 \theta = 1$,the final value is $1$.

(D) Given $\cos(\alpha - \beta) = 1$. Since $-\pi < \alpha, \beta < \pi$,the range for the difference is $-2\pi < \alpha - \beta < 2\pi$.
For $\cos(\alpha - \beta) = 1$,the possible values for $\alpha - \beta$ are $0$.
Thus,$\alpha - \beta = 0$,which implies $\alpha = \beta$.
Substituting $\alpha = \beta$ into the second equation: $\cos(\alpha + \alpha) = \cos(2\alpha) = \frac{1}{e}$.
Since $-\pi < \alpha < \pi$,we have $-2\pi < 2\alpha < 2\pi$.
In the interval $(-2\pi, 2\pi)$,the equation $\cos(2\alpha) = \frac{1}{e}$ (where $0 < \frac{1}{e} < 1$) has four distinct solutions for $2\alpha$ (two in the first/fourth quadrants and two in the second/third quadrants).
Therefore,there are $4$ ordered pairs $(\alpha, \beta)$.

(D) Given $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$.
Since $A$ and $B$ are acute angles,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Similarly,$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$\sin(A + B) = \left(\frac{1}{\sqrt{10}}\right) \left(\frac{2}{\sqrt{5}}\right) + \left(\frac{3}{\sqrt{10}}\right) \left(\frac{1}{\sqrt{5}}\right)$
$\sin(A + B) = \frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\sin(A + B) = \frac{1}{\sqrt{2}}$,we have $A + B = \frac{\pi}{4}$.

(C) Given that $\tan A = 2\tan B + \cot B.$
We need to find the value of $2\tan (A - B).$
Using the formula $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B},$
$2\tan (A - B) = 2 \left( \frac{\tan A - \tan B}{1 + \tan A \tan B} \right).$
Substitute $\tan A = 2\tan B + \cot B$ into the expression:
$2\tan (A - B) = 2 \left( \frac{(2\tan B + \cot B) - \tan B}{1 + (2\tan B + \cot B)\tan B} \right)$
$= 2 \left( \frac{\tan B + \cot B}{1 + 2\tan^2 B + \cot B \tan B} \right)$
Since $\cot B \tan B = 1,$
$= 2 \left( \frac{\tan B + \cot B}{1 + 2\tan^2 B + 1} \right) = 2 \left( \frac{\tan B + \cot B}{2 + 2\tan^2 B} \right)$
$= 2 \left( \frac{\tan B + \cot B}{2(1 + \tan^2 B)} \right) = \frac{\tan B + \cot B}{\sec^2 B}$
$= \frac{\tan B}{\sec^2 B} + \frac{\cot B}{\sec^2 B} = \frac{\sin B}{\cos B} \cdot \cos^2 B + \frac{\cos B}{\sin B} \cdot \cos^2 B$
$= \sin B \cos B + \frac{\cos^3 B}{\sin B} = \frac{\sin^2 B \cos B + \cos^3 B}{\sin B}$
$= \frac{\cos B (\sin^2 B + \cos^2 B)}{\sin B} = \frac{\cos B (1)}{\sin B} = \cot B.$

(D) Given: $\sin A + \sin B = C$ and $\cos A + \cos B = D$.
Dividing the two equations:
$\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{C}{D}$
Using the sum-to-product formulas:
$\frac{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}}{2 \cos \frac{A + B}{2} \cos \frac{A - B}{2}} = \frac{C}{D}$
$\tan \frac{A + B}{2} = \frac{C}{D}$
Now,using the double angle formula in terms of tangent:
$\sin (A + B) = \frac{2 \tan \frac{A + B}{2}}{1 + \tan^2 \frac{A + B}{2}}$
Substituting $\tan \frac{A + B}{2} = \frac{C}{D}$:
$\sin (A + B) = \frac{2(C/D)}{1 + (C/D)^2} = \frac{2C/D}{(D^2 + C^2)/D^2} = \frac{2CD}{C^2 + D^2}$.

(A) Given that $\sin A = \sin B$ and $\cos A = \cos B$.
Squaring both equations and adding them:
$\sin^2 A + \cos^2 A = \sin^2 B + \cos^2 B$
$1 = 1$ (This is always true).
Alternatively,consider $\cos A = \cos B$ and $\sin A = \sin B$.
Subtracting the equations: $\cos A - \cos B = 0$ and $\sin A - \sin B = 0$.
Using sum-to-product formulas:
$-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} = 0$
$2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} = 0$
For both to be zero,$\sin \frac{A-B}{2}$ must be $0$.

(B) Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$,we have:
$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$
$= 2 \cos \left( \frac{50^\circ + 70^\circ}{2} \right) \sin \left( \frac{50^\circ - 70^\circ}{2} \right) + \sin 10^\circ$
$= 2 \cos(60^\circ) \sin(-10^\circ) + \sin 10^\circ$
Since $\sin(-\theta) = -\sin \theta$ and $\cos 60^\circ = 1/2$:
$= 2 \left( \frac{1}{2} \right) (-\sin 10^\circ) + \sin 10^\circ$
$= -\sin 10^\circ + \sin 10^\circ = 0.$

(B) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cdot \cos(A - B)$.
Substituting $A = 48^\circ$ and $B = 12^\circ$:
$\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ) \cdot \cos(48^\circ - 12^\circ)$
$= \cos(60^\circ) \cdot \cos(36^\circ)$
We know that $\cos(60^\circ) = \frac{1}{2}$ and $\cos(36^\circ) = \frac{\sqrt{5} + 1}{4}$.
Therefore,the expression becomes $\frac{1}{2} \cdot \left( \frac{\sqrt{5} + 1}{4} \right) = \frac{\sqrt{5} + 1}{8}$.

(C) Given: $A - B = \frac{\pi}{4}$.
Taking tangent on both sides: $\tan(A - B) = \tan\left(\frac{\pi}{4}\right) = 1$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$\frac{\tan A - \tan B}{1 + \tan A \tan B} = 1$.
$\tan A - \tan B = 1 + \tan A \tan B$.
$\tan A - \tan B - \tan A \tan B = 1$.
Adding $1$ to both sides:
$1 + \tan A - \tan B - \tan A \tan B = 1 + 1$.
$(1 + \tan A) - \tan B(1 + \tan A) = 2$.
$(1 + \tan A)(1 - \tan B) = 2$.
Since $y = (1 + \tan A)(1 - \tan B)$,we have $y = 2$.
Therefore,$(y + 1)^{y + 1} = (2 + 1)^{2 + 1} = 3^3 = 27$.

(B) To find the value of $\sin 75^\circ$,we can express $75^\circ$ as the sum of two standard angles: $45^\circ + 30^\circ$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$\sin 75^\circ = \sin(45^\circ + 30^\circ)$
$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
$= (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} \times \frac{1}{2})$
$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$
$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$.

(B) We are given $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$:
$\tan(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - (\frac{m}{m + 1})(\frac{1}{2m + 1})}$
$= \frac{\frac{m(2m + 1) + 1(m + 1)}{(m + 1)(2m + 1)}}{\frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)}}$
$= \frac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}$
$= \frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$
Since $\tan(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

(B) We know the trigonometric identity for the tangent of a sum: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Let $A = 20^\circ$ and $B = 40^\circ$.
Then,$\tan(20^\circ + 40^\circ) = \tan 60^\circ = \sqrt{3}$.
Substituting these into the identity:
$\sqrt{3} = \frac{\tan 20^\circ + \tan 40^\circ}{1 - \tan 20^\circ \tan 40^\circ}$.
Cross-multiplying gives:
$\sqrt{3}(1 - \tan 20^\circ \tan 40^\circ) = \tan 20^\circ + \tan 40^\circ$.
$\sqrt{3} - \sqrt{3} \tan 20^\circ \tan 40^\circ = \tan 20^\circ + \tan 40^\circ$.
Rearranging the terms,we get:
$\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = \sqrt{3}$.

(D) Given expression: $\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ]$
Multiply and divide by $2$ inside the bracket:
$= \frac{1}{2} [\frac{\sqrt{3}}{2} \cos 23^\circ - \frac{1}{2} \sin 23^\circ]$
We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$.
Substituting these values:
$= \frac{1}{2} [\cos 30^\circ \cos 23^\circ - \sin 30^\circ \sin 23^\circ]$
Using the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$= \frac{1}{2} \cos(30^\circ + 23^\circ)$
$= \frac{1}{2} \cos 53^\circ$
Since $\frac{1}{2} \cos 53^\circ$ is not equal to any of the given options,the correct answer is $(d)$.

(A) We know that $\tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
Rationalizing the denominator: $\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.
Since $\cot 75^\circ = \frac{1}{\tan 75^\circ} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.
Therefore,$\tan 75^\circ - \cot 75^\circ = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}$.

(B) We are given $\tan A = - \frac{1}{2}$ and $\tan B = - \frac{1}{3}$.
Using the formula for the tangent of a sum: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting the values: $\tan(A + B) = \frac{-\frac{1}{2} - \frac{1}{3}}{1 - (-\frac{1}{2})(-\frac{1}{3})} = \frac{-\frac{5}{6}}{1 - \frac{1}{6}} = \frac{-\frac{5}{6}}{\frac{5}{6}} = -1$.
Since $\tan(A + B) = -1$,and considering the range of values for $A$ and $B$ in the second and fourth quadrants,we have $A + B = \frac{3\pi}{4}$.

(D) Let $X = \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B}$.
Converting to tangent form: $X = \frac{1/\tan A}{1 + 1/\tan A} \cdot \frac{1/\tan B}{1 + 1/\tan B} = \frac{1}{\tan A + 1} \cdot \frac{1}{\tan B + 1} = \frac{1}{1 + \tan A + \tan B + \tan A \tan B}$.
Given $A + B = 225^\circ$,we have $\tan(A + B) = \tan(225^\circ) = 1$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$,we get $\tan A + \tan B = 1 - \tan A \tan B$.
Substituting this into the expression for $X$:
$X = \frac{1}{1 + (1 - \tan A \tan B) + \tan A \tan B} = \frac{1}{1 + 1} = \frac{1}{2}$.

(D) Given: $\sin A = \frac{4}{5}$ ($A$ is in the first quadrant,so $\cos A$ is positive) and $\cos B = -\frac{12}{13}$ ($B$ is in the third quadrant,so $\sin B$ is negative).
Step $1$: Find $\cos A$.
$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Step $2$: Find $\sin B$.
$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (-\frac{12}{13})^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.
Step $3$: Use the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$.
$\cos(A + B) = (\frac{3}{5})(-\frac{12}{13}) - (\frac{4}{5})(-\frac{5}{13})$
$= -\frac{36}{65} + \frac{20}{65}$
$= -\frac{16}{65}$.