Fundamental trigonometrical ratios and functions, Trigonometrical ratio of allied angles Questions in English

(A) We know that $\text{sech}(iz) = \sec(z)$.
Substituting $z = \pi$,we get $\text{sech}(i\pi) = \sec(\pi)$.
Since $\sec(\pi) = \frac{1}{\cos(\pi)} = \frac{1}{-1} = -1$.
Therefore,the value is $-1$.

(C) The range of $\sin \theta$ is $[-1, 1]$,so $\sin \theta = -\frac{1}{5}$ is possible.
The range of $\cos \theta$ is $[-1, 1]$,so $\cos \theta = 1$ is possible.
The range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$. Since $\frac{1}{2} < 1$,the statement $\sec \theta = \frac{1}{2}$ is incorrect.
The range of $\tan \theta$ is $(-\infty, \infty)$,so $\tan \theta = 20$ is possible.
Therefore,the incorrect statement is $C$.

(B) For any real angle $\theta$:
$1$. The range of $\sin \theta$ and $\cos \theta$ is $[-1, 1]$. Thus,$\sin \theta = \frac{5}{3} > 1$ is impossible.
$2$. The range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$. Thus,$\sec \theta = \frac{1}{2}$ is impossible because $\frac{1}{2} < 1$.
$3$. For $\cos \theta = \frac{1 + p^2}{1 - p^2}$,let $p = 2$. Then $\cos \theta = \frac{1+4}{1-4} = \frac{5}{-3} = -1.66$,which is less than $-1$,so it is impossible.
$4$. The range of $\tan \theta$ is $(-\infty, \infty)$. Therefore,$\tan \theta = 1002$ is possible.

(A) The values $1$ and $2$ are in radians.
Since $1 \text{ radian} \approx 57.3^{\circ}$,which lies in the first quadrant,$\tan 1 > 0$.
Since $2 \text{ radians} \approx 114.6^{\circ}$,which lies in the second quadrant,$\tan 2 < 0$.
Therefore,$\tan 1 > 0 > \tan 2$.
Thus,$\tan 1 > \tan 2$ is correct.

(B) The correct relation is $\sin 1 > \sin 1^\circ$.
In the interval $[0, \frac{\pi}{2}]$,the function $f(x) = \sin x$ is an increasing function.
We know that $1 \text{ radian} \approx 57.3^\circ$.
Since $57.3^\circ > 1^\circ$ and the sine function is increasing in the first quadrant,it follows that $\sin(57.3^\circ) > \sin(1^\circ)$,which means $\sin 1 > \sin 1^\circ$.

(A) We know that $\tan(90^\circ - \theta) = \cot \theta$.
Thus,$\tan 89^\circ = \tan(90^\circ - 1^\circ) = \cot 1^\circ$.
Similarly,$\tan 88^\circ = \cot 2^\circ$,and so on.
The expression is $P = \tan 1^\circ \tan 2^\circ \dots \tan 44^\circ \tan 45^\circ \tan 46^\circ \dots \tan 88^\circ \tan 89^\circ$.
Pairing terms from both ends: $(\tan 1^\circ \tan 89^\circ) = (\tan 1^\circ \cot 1^\circ) = 1$.
$(\tan 2^\circ \tan 88^\circ) = (\tan 2^\circ \cot 2^\circ) = 1$.
Continuing this up to $\tan 44^\circ \tan 46^\circ = 1$.
The middle term is $\tan 45^\circ = 1$.
Therefore,the product is $1 \times 1 \times \dots \times 1 = 1$.

(C) Given $\sin \theta = \frac{24}{25}$.
Since $\theta$ lies in the second quadrant,$\cos \theta$ and $\tan \theta$ are negative.
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\cos^2 \theta = 1 - (\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$.
Thus,$\cos \theta = -\frac{7}{25}$.
Then,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{24/25}{-7/25} = -\frac{24}{7}$.
Also,$\sec \theta = \frac{1}{\cos \theta} = -\frac{25}{7}$.
Therefore,$\sec \theta + \tan \theta = -\frac{25}{7} - \frac{24}{7} = -\frac{49}{7} = -7$.

(C) Given $5 \tan \theta = 4$,we have $\tan \theta = \frac{4}{5}$.
Divide the numerator and the denominator of the expression by $\cos \theta$:
$\frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = \frac{5 \frac{\sin \theta}{\cos \theta} - 3}{5 \frac{\sin \theta}{\cos \theta} + 2}$
Substitute $\tan \theta = \frac{4}{5}$ into the expression:
$= \frac{5(\frac{4}{5}) - 3}{5(\frac{4}{5}) + 2}$
$= \frac{4 - 3}{4 + 2} = \frac{1}{6}$.

(C) Given that $\tan \theta = \frac{20}{21}$.
We know that $\sec^2 \theta = 1 + \tan^2 \theta$.
$\sec^2 \theta = 1 + (\frac{20}{21})^2 = 1 + \frac{400}{441} = \frac{441 + 400}{441} = \frac{841}{441}$.
Therefore,$\sec \theta = \pm \sqrt{\frac{841}{441}} = \pm \frac{29}{21}$.
Since $\cos \theta = \frac{1}{\sec \theta}$,we have $\cos \theta = \pm \frac{21}{29}$.

(B) Given $\sin x = -\frac{24}{25}$.
Since $\sin^2 x + \cos^2 x = 1$,we have $\cos^2 x = 1 - \sin^2 x$.
$\cos^2 x = 1 - \left(-\frac{24}{25}\right)^2 = 1 - \frac{576}{625} = \frac{625 - 576}{625} = \frac{49}{625}$.
Thus,$\cos x = \pm \frac{7}{25}$.
Since $\tan x = \frac{\sin x}{\cos x}$,if $\cos x = \frac{7}{25}$,then $\tan x = \frac{-24/25}{7/25} = -\frac{24}{7}$.
If $\cos x = -\frac{7}{25}$,then $\tan x = \frac{-24/25}{-7/25} = \frac{24}{7}$.
Given the options,the correct value is $-\frac{24}{7}$.

(B) Given $\tan \theta = -\frac{4}{3}.$
We know that $\sec^2 \theta = 1 + \tan^2 \theta = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}.$
Thus,$\cos^2 \theta = \frac{1}{\sec^2 \theta} = \frac{9}{25}.$
Using $\sin^2 \theta = 1 - \cos^2 \theta,$ we get $\sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}.$
Therefore,$\sin \theta = \pm \frac{4}{5}.$
Since $\tan \theta$ is negative,$\theta$ lies in the $2^{nd}$ or $4^{th}$ quadrant. In the $2^{nd}$ quadrant,$\sin \theta$ is positive $(4/5)$,and in the $4^{th}$ quadrant,$\sin \theta$ is negative $(-4/5)$.
Thus,both values are possible.

(C) Given that $\sin \theta = -\frac{1}{\sqrt{2}}$ and $\tan \theta = 1$.
In the Cartesian coordinate system,the signs of trigonometric functions in different quadrants are as follows:
- $I$ quadrant: All positive
- $II$ quadrant: $\sin$ is positive
- $III$ quadrant: $\tan$ is positive
- $IV$ quadrant: $\cos$ is positive
Since $\sin \theta$ is negative and $\tan \theta$ is positive,$\theta$ must lie in the $III$ quadrant,where both $\sin$ and $\cos$ are negative,making $\tan \theta = \frac{\sin \theta}{\cos \theta}$ positive.

(A) Given $\sin (\alpha - \beta ) = \frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$,we have $\alpha - \beta = 30^\circ$ $(i)$.
Given $\cos (\alpha + \beta ) = \frac{1}{2}$. Since $\cos 60^\circ = \frac{1}{2}$,we have $\alpha + \beta = 60^\circ$ $(ii)$.
Adding equations $(i)$ and $(ii)$ gives $2\alpha = 90^\circ$,so $\alpha = 45^\circ$.
Substituting $\alpha = 45^\circ$ into $(ii)$,we get $45^\circ + \beta = 60^\circ$,so $\beta = 15^\circ$.
Thus,$\alpha = 45^\circ$ and $\beta = 15^\circ$.

(C) Given $\tan \theta = -\frac{1}{\sqrt{10}}$.
Since $\theta$ lies in the fourth quadrant,$\cos \theta$ must be positive.
We use the identity $1 + \tan^2 \theta = \sec^2 \theta$.
Substituting the value of $\tan \theta$:
$1 + \left(-\frac{1}{\sqrt{10}}\right)^2 = \sec^2 \theta$
$1 + \frac{1}{10} = \sec^2 \theta$
$\sec^2 \theta = \frac{11}{10}$
Since $\cos \theta = \frac{1}{\sec \theta}$,we have $\cos^2 \theta = \frac{1}{\sec^2 \theta} = \frac{10}{11}$.
Since $\theta$ is in the fourth quadrant,$\cos \theta$ is positive,so $\cos \theta = \sqrt{\frac{10}{11}}$.

(D) Given $3\tan A + 4 = 0,$ so $\tan A = -\frac{4}{3}$.
Since $A$ lies in the second quadrant,$\tan A$ is negative,$\sin A$ is positive,and $\cos A$ is negative.
Using the identity $\sec^2 A = 1 + \tan^2 A = 1 + (-\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$,we get $\sec A = -\frac{5}{3}$ (as $A$ is in the second quadrant).
Thus,$\cos A = \frac{1}{\sec A} = -\frac{3}{5}$.
Also,$\sin A = \tan A \times \cos A = (-\frac{4}{3}) \times (-\frac{3}{5}) = \frac{4}{5}$.
And $\cot A = \frac{1}{\tan A} = -\frac{3}{4}$.
Substituting these values into the expression $2\cot A - 5\cos A + \sin A$:
$= 2(-\frac{3}{4}) - 5(-\frac{3}{5}) + \frac{4}{5}$
$= -\frac{3}{2} + 3 + \frac{4}{5}$
$= \frac{-15 + 30 + 8}{10} = \frac{23}{10}$.

(B) Let $E = \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$.
Simplifying the expression:
$E = \frac{\sqrt{1 - \sin \theta} \cdot \sqrt{1 - \sin \theta} + \sqrt{1 + \sin \theta} \cdot \sqrt{1 + \sin \theta}}{\sqrt{(1 + \sin \theta)(1 - \sin \theta)}}$
$E = \frac{(1 - \sin \theta) + (1 + \sin \theta)}{\sqrt{1 - \sin^2 \theta}}$
$E = \frac{2}{\sqrt{\cos^2 \theta}} = \frac{2}{|\cos \theta|}$.
Since $\theta$ lies in the second quadrant,$\cos \theta$ is negative.
Therefore,$|\cos \theta| = -\cos \theta$.
Thus,$E = \frac{2}{-\cos \theta} = -2 \sec \theta$.

(B) Given that $x = \sec \theta + \tan \theta$.
We know that $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Therefore,$\frac{1}{x} = \frac{1}{\sec \theta + \tan \theta} = \sec \theta - \tan \theta$.
Now,$x + \frac{1}{x} = (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta)$.
$x + \frac{1}{x} = 2 \sec \theta$.

(A) Given $\tan \theta = \frac{a}{b}.$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{a}{b},$ we have $\sin \theta = \pm \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos \theta = \pm \frac{b}{\sqrt{a^2 + b^2}}.$
Now,consider the expression $E = \frac{\sin \theta}{\cos^8 \theta} + \frac{\cos \theta}{\sin^8 \theta}.$
Substituting the values of $\sin \theta$ and $\cos \theta$:
$E = \frac{\pm \frac{a}{\sqrt{a^2 + b^2}}}{\left( \pm \frac{b}{\sqrt{a^2 + b^2}} \right)^8} + \frac{\pm \frac{b}{\sqrt{a^2 + b^2}}}{\left( \pm \frac{a}{\sqrt{a^2 + b^2}} \right)^8}$
Since $(\pm x)^8 = x^8,$ the signs simplify to $\pm$ for the whole expression:
$E = \pm \left[ \frac{a \cdot (a^2 + b^2)^4}{b^8 \cdot \sqrt{a^2 + b^2}} + \frac{b \cdot (a^2 + b^2)^4}{a^8 \cdot \sqrt{a^2 + b^2}} \right]$
$E = \pm \frac{(a^2 + b^2)^4}{\sqrt{a^2 + b^2}} \left( \frac{a}{b^8} + \frac{b}{a^8} \right).$

(A) The given expression is a product of terms from $\cos 1^\circ$ to $\cos 179^\circ$.
We know that the product contains the term $\cos 90^\circ$ as one of its factors.
Since $\cos 90^\circ = 0$,any product containing this term will result in $0$.
Therefore,$\cos 1^\circ \cdot \cos 2^\circ \cdot \dots \cdot \cos 90^\circ \cdot \dots \cdot \cos 179^\circ = 0$.

(C) The given expression is $S = \sum_{k=1}^{180} \cos k^\circ$.
We know that $\cos(180^\circ - \theta) = -\cos \theta$.
Thus,$\cos 179^\circ = -\cos 1^\circ$,$\cos 178^\circ = -\cos 2^\circ$,and so on.
Pairing the terms: $(\cos 1^\circ + \cos 179^\circ) + (\cos 2^\circ + \cos 178^\circ) + \dots + (\cos 89^\circ + \cos 91^\circ) + \cos 90^\circ + \cos 180^\circ$.
Each pair sums to $0$ because $\cos \theta + \cos(180^\circ - \theta) = 0$.
Since $\cos 90^\circ = 0$ and $\cos 180^\circ = -1$,the total sum is $0 + 0 + \dots + 0 + 0 + (-1) = -1$.

(A) Given expression: $\sin 15^\circ + \cos 105^\circ $
Using the trigonometric identity $\cos(90^\circ + \theta) = -\sin \theta$,we can write:
$\cos 105^\circ = \cos(90^\circ + 15^\circ) = -\sin 15^\circ $
Substituting this into the expression:
$\sin 15^\circ + (-\sin 15^\circ) = 0$

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

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

(B) We use the following trigonometric identities:
$\sin (270^\circ + A) = -\cos A$
$\sin (270^\circ - A) = -\cos A$
$\cos (180^\circ + A) = -\cos A$
Substituting these into the expression:
$\cos A + (-\cos A) - (-\cos A) + (-\cos A)$
$= \cos A - \cos A + \cos A - \cos A$
$= 0$

(B) We know that $\sin (\pi + \theta ) = -\sin \theta$ and $\sin (\pi - \theta ) = \sin \theta$.
Substituting these values into the expression:
$\sin (\pi + \theta )\sin (\pi - \theta )\csc^2 \theta = (-\sin \theta )(\sin \theta )\left(\frac{1}{\sin^2 \theta}\right)$
$= -\sin^2 \theta \times \frac{1}{\sin^2 \theta}$
$= -1$.

(C) We know that $\cot(90^\circ - A) = \tan A$.
Given expression: $\cot (45^\circ + \theta ) \cot (45^\circ - \theta )$.
Using the identity $\cot(45^\circ + \theta) = \tan(90^\circ - (45^\circ + \theta)) = \tan(45^\circ - \theta)$.
Substituting this into the expression:
$\tan (45^\circ - \theta ) \cot (45^\circ - \theta ) = 1$ (since $\tan A \cot A = 1$).
Thus,the correct option is $C$.

(A) Given expression: $\tan A + \cot (180^\circ + A) + \cot (90^\circ + A) + \cot (360^\circ - A)$
Using trigonometric reduction formulas:
$\cot (180^\circ + A) = \cot A$
$\cot (90^\circ + A) = -\tan A$
$\cot (360^\circ - A) = -\cot A$
Substituting these values into the expression:
$= \tan A + \cot A - \tan A - \cot A$
$= 0$

(D) We are given the expression: $\tan \theta \sin \left( \frac{\pi }{2} + \theta \right) \cos \left( \frac{\pi }{2} - \theta \right)$.
Using the trigonometric identities $\sin \left( \frac{\pi }{2} + \theta \right) = \cos \theta$ and $\cos \left( \frac{\pi }{2} - \theta \right) = \sin \theta$,the expression becomes:
$= \tan \theta \cdot \cos \theta \cdot \sin \theta$
$= \left( \frac{\sin \theta}{\cos \theta} \right) \cdot \cos \theta \cdot \sin \theta$
$= \sin \theta \cdot \sin \theta$
$= \sin^2 \theta$.

(B) Given that $x = y \cos \frac{2\pi}{3} = z \cos \frac{4\pi}{3}$.
We know that $\cos \frac{2\pi}{3} = -\frac{1}{2}$ and $\cos \frac{4\pi}{3} = -\frac{1}{2}$.
So,$x = y(-\frac{1}{2}) = z(-\frac{1}{2})$.
This implies $x = -\frac{y}{2} = -\frac{z}{2}$.
Let $x = \lambda$,then $y = -2\lambda$ and $z = -2\lambda$.
Now,calculate $xy + yz + zx$:
$xy + yz + zx = (\lambda)(-2\lambda) + (-2\lambda)(-2\lambda) + (-2\lambda)(\lambda)$
$= -2\lambda^2 + 4\lambda^2 - 2\lambda^2$
$= 0$.

(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) We know that $\tan(-\theta) = -\tan(\theta)$.
So,$\tan(-945^\circ) = -\tan(945^\circ)$.
Since $945^\circ = 2 \times 360^\circ + 225^\circ$,we have $\tan(945^\circ) = \tan(225^\circ)$.
Now,$\tan(225^\circ) = \tan(180^\circ + 45^\circ) = \tan(45^\circ) = 1$.
Therefore,$\tan(-945^\circ) = -1$.

(C) Given $A = \frac{5\pi}{4}$.
We need to find the value of $\cos A - \sin A$.
$\cos\left(\frac{5\pi}{4}\right) - \sin\left(\frac{5\pi}{4}\right) = \cos\left(\pi + \frac{\pi}{4}\right) - \sin\left(\pi + \frac{\pi}{4}\right)$.
Since $\cos(\pi + \theta) = -\cos \theta$ and $\sin(\pi + \theta) = -\sin \theta$,we have:
$= -\cos\left(\frac{\pi}{4}\right) - (-\sin\left(\frac{\pi}{4}\right))$
$= -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 0$.

(D) We use the trigonometric reduction formulas:
$\cos (270^\circ + \theta ) = \sin \theta$
$\cos (90^\circ - \theta ) = \sin \theta$
$\sin (270^\circ - \theta ) = -\cos \theta$
Substituting these into the expression:
$\sin \theta \cdot \sin \theta - (-\cos \theta ) \cdot \cos \theta$
$= \sin^2 \theta + \cos^2 \theta$
$= 1$

(B) Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$
$= 2 \cos \frac{50^\circ + 70^\circ}{2} \sin \frac{50^\circ - 70^\circ}{2} + \sin 10^\circ$
$= 2 \cos 60^\circ \sin(-10^\circ) + \sin 10^\circ$
$= -2 \cos 60^\circ \sin 10^\circ + \sin 10^\circ$
$= -2 \times \frac{1}{2} \times \sin 10^\circ + \sin 10^\circ$
$= -\sin 10^\circ + \sin 10^\circ = 0.$

(C) Given expression: $\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ}$
Using the identity $\cos \theta = \sin(90^\circ - \theta)$:
$\cos 40^\circ = \sin 50^\circ$ and $\cos 70^\circ = \sin 20^\circ$
Substituting these values:
$= \frac{\sin 70^\circ + \sin 50^\circ}{\sin 20^\circ + \sin 40^\circ}$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
Numerator: $\sin 70^\circ + \sin 50^\circ = 2 \sin 60^\circ \cos 10^\circ$
Denominator: $\sin 20^\circ + \sin 40^\circ = 2 \sin 30^\circ \cos(-10^\circ) = 2 \sin 30^\circ \cos 10^\circ$
$= \frac{2 \sin 60^\circ \cos 10^\circ}{2 \sin 30^\circ \cos 10^\circ}$
$= \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(A) We use the reduction formulas:
$\sin 600^\circ = \sin(720^\circ - 120^\circ) = -\sin 120^\circ = -\sin(180^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$
$\cos 330^\circ = \cos(360^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$
$\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$
$\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$
Substituting these values into the expression:
$(- \frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) + (- \frac{1}{2})(\frac{1}{2}) = -\frac{3}{4} - \frac{1}{4} = -\frac{4}{4} = -1$.

(B) We are given the expression: $\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}}$
Substitute $\cot 15^\circ = \frac{{\cos 15^\circ}}{{\sin 15^\circ}}$:
$= \frac{{\frac{{\cos^2 15^\circ}}{{\sin^2 15^\circ}} - 1}}{{\frac{{\cos^2 15^\circ}}{{\sin^2 15^\circ}} + 1}}$
Simplify the numerator and denominator by taking the common denominator $\sin^2 15^\circ$:
$= \frac{{\cos^2 15^\circ - \sin^2 15^\circ}}{{\cos^2 15^\circ + \sin^2 15^\circ}}$
Using the trigonometric identities $\cos^2 \theta - \sin^2 \theta = \cos(2\theta)$ and $\cos^2 \theta + \sin^2 \theta = 1$:
$= \cos(2 \times 15^\circ) = \cos 30^\circ$
Since $\cos 30^\circ = \frac{{\sqrt{3}}}{2}$,the final answer is $\frac{{\sqrt{3}}}{2}$.

(C) Given,the diameter of the circular wire is $10 \, cm$.
Therefore,the length of the wire (arc length $s$) is equal to the circumference of the wire: $s = \pi \times d = 10\pi \, cm$.
The diameter of the circle on which the wire is placed is $1 \, m = 100 \, cm$.
Therefore,the radius $r$ of this circle is $r = \frac{100}{2} = 50 \, cm$.
The angle $\theta$ subtended by an arc of length $s$ at the centre of a circle of radius $r$ is given by $\theta = \frac{s}{r}$.
Substituting the values,$\theta = \frac{10\pi}{50} = \frac{\pi}{5} \, \text{radian}$.

(A) Given $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.
Dividing numerator and denominator by $\cos \alpha$,we get $\tan \theta = \frac{\tan \alpha - 1}{\tan \alpha + 1} = \tan(\alpha - \frac{\pi}{4})$.
Thus,$\theta = \alpha - \frac{\pi}{4}$,which implies $\alpha = \theta + \frac{\pi}{4}$.
Now,$\sin \alpha + \cos \alpha = \sin(\theta + \frac{\pi}{4}) + \cos(\theta + \frac{\pi}{4}) = (\sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4}) + (\cos \theta \cos \frac{\pi}{4} - \sin \theta \sin \frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\sin \theta + \cos \theta + \cos \theta - \sin \theta) = \frac{2 \cos \theta}{\sqrt{2}} = \sqrt{2} \cos \theta$.
Similarly,$\sin \alpha - \cos \alpha = \sin(\theta + \frac{\pi}{4}) - \cos(\theta + \frac{\pi}{4}) = (\sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4}) - (\cos \theta \cos \frac{\pi}{4} - \sin \theta \sin \frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\sin \theta + \cos \theta - \cos \theta + \sin \theta) = \frac{2 \sin \theta}{\sqrt{2}} = \sqrt{2} \sin \theta$.

(B) Given $x = \sin 130^\circ \cos 80^\circ$ and $y = \sin 80^\circ \cos 130^\circ$.
Since $130^\circ$ is in the second quadrant,$\sin 130^\circ = \sin(180^\circ - 50^\circ) = \sin 50^\circ = \cos 40^\circ > 0$ and $\cos 130^\circ = -\sin 40^\circ < 0$.
Also,$\cos 80^\circ > 0$ and $\sin 80^\circ > 0$.
Therefore,$x = (\sin 130^\circ)(\cos 80^\circ) = (\text{positive})(\text{positive}) > 0$.
And $y = (\sin 80^\circ)(\cos 130^\circ) = (\text{positive})(\text{negative}) < 0$.
Since $x > 0$ and $y < 0$,their product $xy < 0$.
Given $z = 1 + xy$,since $xy < 0$,it follows that $z < 1$.
Also,$xy = (\sin 130^\circ \cos 80^\circ)(\sin 80^\circ \cos 130^\circ) = (\sin 130^\circ \cos 130^\circ)(\sin 80^\circ \cos 80^\circ) = \frac{1}{2} \sin 260^\circ \cdot \frac{1}{2} \sin 160^\circ = \frac{1}{4} (-\sin 80^\circ)(\sin 20^\circ) = -\frac{1}{4} \sin 80^\circ \sin 20^\circ$.
Since $\sin 80^\circ \sin 20^\circ$ is a small positive value,$xy$ is a small negative value,so $0 < z < 1$.

(A) Given $\sin (A + B) = 1$ and $\cos (A - B) = \frac{\sqrt{3}}{2}$.
Since $\sin 90^\circ = 1$,we have $A + B = 90^\circ$.
Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$,we have $A - B = 30^\circ$.
Adding the two equations: $(A + B) + (A - B) = 90^\circ + 30^\circ$ $\Rightarrow 2A = 120^\circ$ $\Rightarrow A = 60^\circ$.
Substituting $A = 60^\circ$ in $A + B = 90^\circ$,we get $60^\circ + B = 90^\circ \Rightarrow B = 30^\circ$.
Thus,the smallest positive values are $A = 60^\circ$ and $B = 30^\circ$.

(A) Let $AC = BC = a$. Since $D$ is the midpoint of $AC$,$CD = a/2$.
In the right-angled triangle $\triangle BCD$,$\angle C = 90^{\circ}$.
Let $\angle DBC = \alpha$.
Then,$\tan \alpha = \frac{CD}{BC} = \frac{a/2}{a} = \frac{1}{2}$.
Therefore,$\cot \alpha = \frac{1}{\tan \alpha} = 2$.

(B) The given function is $f(x) = \cos (x/3)$.
We know that the range of the basic cosine function $y = \cos(\theta)$ is the interval $[-1, 1]$ for any real value of $\theta$.
Here,$\theta = x/3$. Since $x$ can be any real number,$x/3$ also takes all real values.
Therefore,the range of $\cos(x/3)$ remains $[-1, 1]$.
Thus,the correct option is $B$.

(C) Given the Cartesian coordinates $(x, y) = (-\sqrt{3}, -1)$.
We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
For the angle $\theta$,we have $\cos \theta = \frac{x}{r} = \frac{-\sqrt{3}}{2}$ and $\sin \theta = \frac{y}{r} = \frac{-1}{2}$.
Since both $\sin \theta$ and $\cos \theta$ are negative,the point lies in the third quadrant.
The reference angle is $\alpha = \tan^{-1}\left|\frac{-1}{-\sqrt{3}}\right| = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.
In the third quadrant,$\theta = -(\pi - \frac{\pi}{6}) = -\frac{5\pi}{6}$.
Thus,the polar coordinates are $(2, -5\pi/6)$.

(D) function $f(x)$ is said to be monotonic on an interval if it is either entirely non-increasing or non-decreasing on that interval.
Trigonometric functions like $\sin(x)$,$\cos(x)$,$\tan(x)$,etc.,are periodic functions.
For example,$\sin(x)$ increases on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ and decreases on the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$.
Since these functions oscillate between values,they are not monotonic over their entire domain $\mathbb{R}$.
Therefore,trigonometric functions are not necessarily monotonic.

(B) Given $\sin \theta = \frac{1}{2}$,since $\theta$ is acute,$\theta = \frac{\pi}{6}$.
Given $\cos \phi = \frac{1}{3}$,since $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\cos \frac{\pi}{2} = 0$,and $\frac{1}{3}$ lies between $0$ and $\frac{1}{2}$,we have $\frac{\pi}{3} < \phi < \frac{\pi}{2}$.
Adding $\theta = \frac{\pi}{6}$ to the inequality $\frac{\pi}{3} < \phi < \frac{\pi}{2}$ gives $\frac{\pi}{6} + \frac{\pi}{3} < \theta + \phi < \frac{\pi}{6} + \frac{\pi}{2}$.
This simplifies to $\frac{\pi}{2} < \theta + \phi < \frac{2\pi}{3}$.

(A) The symbol $ \alpha $ represents the alpha character. Therefore,the correct option is $ A $.

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

(A) Given $\cos(\alpha + \beta) = 0$.
Since $\cos(\theta) = 0$ implies $\theta = \frac{\pi}{2} + n\pi$,we take the principal value $\alpha + \beta = \frac{\pi}{2}$.
Thus,$\beta = \frac{\pi}{2} - \alpha$.
Now,substitute $\beta$ into the expression $\sin(\alpha + 2\beta)$:
$\sin(\alpha + 2(\frac{\pi}{2} - \alpha)) = \sin(\alpha + \pi - 2\alpha) = \sin(\pi - \alpha)$.
Using the identity $\sin(\pi - \theta) = \sin \theta$,we get $\sin(\pi - \alpha) = \sin \alpha$.

(A) Using trigonometric identities:
$\tan \left( \frac{3\pi}{2} - \alpha \right) = \cot \alpha$
$\cos \left( \frac{3\pi}{2} - \alpha \right) = -\sin \alpha$
$\cos (2\pi - \alpha ) = \cos \alpha$
$\cos \left( \alpha - \frac{\pi}{2} \right) = \sin \alpha$
$\sin (\pi - \alpha ) = \sin \alpha$
$\cos (\pi + \alpha ) = -\cos \alpha$
$\sin \left( \alpha - \frac{\pi}{2} \right) = -\cos \alpha$
Substituting these into the expression:
$= \frac{(\cot \alpha )(-\sin \alpha )}{\cos \alpha } + (\sin \alpha )(\sin \alpha ) + (-\cos \alpha )(-\cos \alpha )$
$= \frac{(\frac{\cos \alpha }{\sin \alpha })(-\sin \alpha )}{\cos \alpha } + \sin^2 \alpha + \cos^2 \alpha$
$= -1 + 1 = 0$