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

(B) Let the two points be $P = (\cos \alpha, \sin \alpha)$ and $Q = (\cos (\pi+\alpha), \sin (\pi+\alpha))$.
Using the trigonometric identities $\cos (\pi+\alpha) = -\cos \alpha$ and $\sin (\pi+\alpha) = -\sin \alpha$,we get $Q = (-\cos \alpha, -\sin \alpha)$.
This can be written as $Q = -(\cos \alpha, \sin \alpha) = -P$.
Since the coordinates of $Q$ are the negative of the coordinates of $P$,the vector from the origin to $Q$ is in the opposite direction to the vector from the origin to $P$.
Thus,the points have opposite directions.

(C) Given $\cot (A+B) = 0$.
Since $\cot \theta = 0$ when $\theta = (2n+1) \frac{\pi}{2}$,we have $A+B = \frac{\pi}{2}$ (taking $n=0$ for simplicity).
Thus,$B = \frac{\pi}{2} - A$.
Now,substitute $B$ into the expression $\sin (A+2B)$:
$\sin (A + 2(\frac{\pi}{2} - A)) = \sin (A + \pi - 2A) = \sin (\pi - A)$.
Using the identity $\sin (\pi - \theta) = \sin \theta$,we get $\sin (\pi - A) = \sin A$.

(B) Given $\sec x + \tan x = 3$ $(1)$
We know that $\sec^2 x - \tan^2 x = 1$
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(\sec x - \tan x)(\sec x + \tan x) = 1$
Substituting $(1)$,we have $\sec x - \tan x = \frac{1}{3}$ $(2)$
Adding $(1)$ and $(2)$,we get $2 \sec x = 3 + \frac{1}{3} = \frac{10}{3}$
Thus,$\sec x = \frac{5}{3}$,which implies $\cos x = \frac{3}{5}$
Since $x \in (0, \frac{\pi}{2})$,$\sin x$ is positive.
$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

(B) The range of the trigonometric functions are as follows:
$1$. For $\sin \theta$ and $\cos \theta$,the value must lie in the interval $[-1, 1]$.
$2$. For $\sec \theta$ and $\csc \theta$,the value must lie in $(-\infty, -1] \cup [1, \infty)$.
$3$. For $\tan \theta$ and $\cot \theta$,the value can be any real number $(R)$.
Evaluating the options:
- Option $A$: $\sec \theta = 23$ is possible since $23 \in (-\infty, -1] \cup [1, \infty)$.
- Option $B$: $\cos \theta = \sqrt{2} \approx 1.414$. Since $1.414 > 1$,this value lies outside the range $[-1, 1]$. Thus,$\cos \theta = \sqrt{2}$ has no solution.
- Option $C$: $\tan \theta = 2019$ is possible since the range of $\tan \theta$ is $R$.
- Option $D$: $\sin \theta = -\frac{1}{5} = -0.2$ is possible since $-0.2 \in [-1, 1]$.
Therefore,the correct option is $B$.

(B) Given the equation: $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$: $2 \sin x \cos x = \frac{2}{4} = \frac{1}{2}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = \frac{1}{2}$.
Since $x \in \left(0, \frac{\pi}{2}\right)$,then $2x \in (0, \pi)$.
The solutions for $2x$ in $(0, \pi)$ are $2x = \frac{\pi}{6}$ and $2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Therefore,$x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$.

(A) Given expression: $\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-\theta)}{\sin \theta}$
Since $\sin(-\alpha) = -\sin(\alpha)$,we have $\sin^2(-160^{\circ}) = (-\sin 160^{\circ})^2 = \sin^2 160^{\circ}$.
Also,$\sin(180^{\circ}-\theta) = \sin \theta$.
So,the expression becomes $\frac{\sin^2 160^{\circ}}{\sin^2 70^{\circ}} + \frac{\sin \theta}{\sin \theta} = \frac{\sin^2(180^{\circ}-20^{\circ})}{\sin^2 70^{\circ}} + 1$.
Since $\sin(180^{\circ}-\alpha) = \sin \alpha$,$\sin 160^{\circ} = \sin 20^{\circ}$.
Thus,$\frac{\sin^2 20^{\circ}}{\sin^2 70^{\circ}} + 1 = \frac{\sin^2 20^{\circ}}{\cos^2 20^{\circ}} + 1 = \tan^2 20^{\circ} + 1 = \sec^2 20^{\circ}$.

(B) Let $A = 18^{\circ}$. Then $5A = 90^{\circ}$,which implies $2A = 90^{\circ} - 3A$.
Taking sine on both sides,$\sin 2A = \sin(90^{\circ} - 3A) = \cos 3A$.
Using the double angle and triple angle formulas:
$2 \sin A \cos A = 4 \cos^3 A - 3 \cos A$.
Since $\cos 18^{\circ} \neq 0$,we can divide by $\cos A$:
$2 \sin A = 4 \cos^2 A - 3$.
Substituting $\cos^2 A = 1 - \sin^2 A$:
$2 \sin A = 4(1 - \sin^2 A) - 3$.
$2 \sin A = 4 - 4 \sin^2 A - 3$.
$4 \sin^2 A + 2 \sin A - 1 = 0$.
Using the quadratic formula for $\sin A$:
$\sin A = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.
Since $18^{\circ}$ is in the first quadrant,$\sin 18^{\circ} > 0$.
Therefore,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.

(D) Given $\theta = \frac{17 \pi}{3} = 5 \pi + \frac{2 \pi}{3}$.
Since $\tan(n \pi + x) = \tan x$ and $\cot(n \pi + x) = \cot x$,we have $\tan \theta = \tan \frac{2 \pi}{3} = -\sqrt{3}$ and $\cot \theta = \cot \frac{2 \pi}{3} = -\frac{1}{\sqrt{3}}$.
Now,calculate $(\tan \theta - \cot \theta) = -\sqrt{3} - (-\frac{1}{\sqrt{3}})$.
$= -\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{-3 + 1}{\sqrt{3}} = \frac{-2}{\sqrt{3}}$.

(A) The given expression is $\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \dots \cos 90^{\circ} \dots \cos 179^{\circ}$.
Since $\cos 90^{\circ} = 0$,the entire product becomes $0$ because any number multiplied by $0$ is $0$.
Therefore,$\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \dots \cos 179^{\circ} = 0$.

(B) We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
The given expression is $P = \tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \cdots \tan 44^{\circ} \tan 45^{\circ} \tan 46^{\circ} \cdots \tan 88^{\circ} \tan 89^{\circ}$.
We can write $\tan 89^{\circ} = \tan(90^{\circ} - 1^{\circ}) = \cot 1^{\circ}$,$\tan 88^{\circ} = \cot 2^{\circ}$,and so on.
Thus,$P = (\tan 1^{\circ} \cot 1^{\circ}) \times (\tan 2^{\circ} \cot 2^{\circ}) \times \cdots \times (\tan 44^{\circ} \cot 44^{\circ}) \times \tan 45^{\circ}$.
Since $\tan \theta \times \cot \theta = 1$ and $\tan 45^{\circ} = 1$,we have $P = 1 \times 1 \times \cdots \times 1 \times 1 = 1$.

(A) Given $\sec \theta = \frac{13}{12}$,therefore $\cos \theta = \frac{12}{13}$.
Since $\theta$ lies in the $4^{\text{th}}$ quadrant,$\sin \theta$ is negative.
Using $\sin^2 \theta + \cos^2 \theta = 1$,we get $\sin \theta = -\sqrt{1 - (\frac{12}{13})^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-5/13}{12/13} = -\frac{5}{12}$ and $\operatorname{cosec} \theta = \frac{1}{\sin \theta} = -\frac{13}{5}$.
Substituting these values into the expression: $\tan \theta \times \operatorname{cosec} \theta \times \sin \theta \times \cos \theta = (-\frac{5}{12}) \times (-\frac{13}{5}) \times (-\frac{5}{13}) \times (\frac{12}{13})$.
$= (\frac{5}{12} \times \frac{13}{5}) \times (-\frac{5}{13} \times \frac{12}{13}) = (\frac{13}{12}) \times (-\frac{60}{169}) = -\frac{5}{13}$.

(A) We know that $\sec^{2} \theta = 1 + \tan^{2} \theta$.
Substituting the value of $\tan \theta = 2$,we get $\sec^{2} \theta = 1 + (2)^{2} = 1 + 4 = 5$.
Since $\theta$ lies in the third quadrant,$\sec \theta$ must be negative.
Therefore,$\sec \theta = -\sqrt{5}$.

(D) $(a) \sin 120^{\circ} = \sin(180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$(b) \cos 930^{\circ} = \cos(2 \times 360^{\circ} + 210^{\circ}) = \cos 210^{\circ} = \cos(180^{\circ} + 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$
$(c) \tan 840^{\circ} = \tan(2 \times 360^{\circ} + 120^{\circ}) = \tan 120^{\circ} = \tan(180^{\circ} - 60^{\circ}) = -\tan 60^{\circ} = -\sqrt{3}$
$(d) \cot(-1110^{\circ}) = -\cot(1110^{\circ}) = -\cot(3 \times 360^{\circ} + 30^{\circ}) = -\cot 30^{\circ} = -\sqrt{3}$
Comparing the values,$(c)$ and $(d)$ both equal $-\sqrt{3}$.

(A) $\sin 690^{\circ} \times \sec 240^{\circ}$
$= \sin (2 \times 360^{\circ} - 30^{\circ}) \times \sec (180^{\circ} + 60^{\circ})$
$= \sin (-30^{\circ}) \times (-\sec 60^{\circ})$
$= (-\sin 30^{\circ}) \times (-2)$
$= (-\frac{1}{2}) \times (-2) = 1$

(C) Given $a = \sin 175^{\circ} + \cos 175^{\circ}$.
Using the allied angle formulas:
$\sin 175^{\circ} = \sin(180^{\circ} - 5^{\circ}) = \sin 5^{\circ}$
$\cos 175^{\circ} = \cos(180^{\circ} - 5^{\circ}) = -\cos 5^{\circ}$
Therefore,$a = \sin 5^{\circ} - \cos 5^{\circ}$.
In the interval $0^{\circ} < \theta < 45^{\circ}$,we know that $\sin \theta < \cos \theta$.
Since $5^{\circ}$ lies in this interval,$\sin 5^{\circ} < \cos 5^{\circ}$.
Thus,$\sin 5^{\circ} - \cos 5^{\circ} < 0$,which implies $a < 0$.

(A) Given that $A$ and $B$ are supplementary angles,we have $A + B = 180^{\circ}$.
This implies $B = 180^{\circ} - A$,so $\frac{B}{2} = 90^{\circ} - \frac{A}{2}$.
Substituting this into the expression:
$\sin^{2} \frac{A}{2} + \sin^{2} \frac{B}{2} = \sin^{2} \frac{A}{2} + \sin^{2} (90^{\circ} - \frac{A}{2})$.
Using the identity $\sin(90^{\circ} - \theta) = \cos \theta$,we get:
$\sin^{2} \frac{A}{2} + \cos^{2} \frac{A}{2}$.
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,the value is $1$.

(A) We know that $\tan \frac{\pi}{3} = \sqrt{3}$.
$\tan \frac{2 \pi}{3} = \tan(\pi - \frac{\pi}{3}) = -\tan \frac{\pi}{3} = -\sqrt{3}$.
$\tan \frac{4 \pi}{3} = \tan(\pi + \frac{\pi}{3}) = \tan \frac{\pi}{3} = \sqrt{3}$.
$\tan \frac{8 \pi}{3} = \tan(2 \pi + \frac{2 \pi}{3}) = \tan \frac{2 \pi}{3} = -\sqrt{3}$.
Substituting these values into the expression:
$\sqrt{3} + 2(-\sqrt{3}) + 4(\sqrt{3}) + 8(-\sqrt{3})$
$= \sqrt{3} - 2 \sqrt{3} + 4 \sqrt{3} - 8 \sqrt{3}$
$= (1 - 2 + 4 - 8) \sqrt{3}$
$= -5 \sqrt{3}$.

(A) $\sec 2 \theta - \tan 2 \theta = \frac{1}{\cos 2 \theta} - \frac{\sin 2 \theta}{\cos 2 \theta} = \frac{1 - \sin 2 \theta}{\cos 2 \theta}$
Using the identities $1 - \sin 2 \theta = (\cos \theta - \sin \theta)^2$ and $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = (\cos \theta - \sin \theta)(\cos \theta + \sin \theta)$:
$\frac{(\cos \theta - \sin \theta)^2}{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$
Dividing numerator and denominator by $\cos \theta$:
$\frac{1 - \tan \theta}{1 + \tan \theta} = \frac{\tan \frac{\pi}{4} - \tan \theta}{1 + \tan \frac{\pi}{4} \tan \theta}$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$= \tan \left(\frac{\pi}{4} - \theta\right)$

(C) Given $\sin A - \sin B = 0$.
This implies $\sin A = \sin B$.
We know that for $A, B \in (0, \pi)$,$\sin A = \sin B$ implies either $A = B$ or $A = \pi - B$.
The condition $A = \pi - B$ is equivalent to $A + B = \pi$,which means the angles are supplementary.
Since the problem states that the angles are not supplementary,we must have $A = B$.

(B) We have,$\frac{1-2(\cos 60^{\circ}-\cos 80^{\circ})}{2 \sin 10^{\circ}}$
$= \frac{1-2(\frac{1}{2}-\cos 80^{\circ})}{2 \sin 10^{\circ}}$
$= \frac{1-1+2 \cos 80^{\circ}}{2 \sin 10^{\circ}}$
$= \frac{2 \cos 80^{\circ}}{2 \sin 10^{\circ}}$
$= \frac{\cos(90^{\circ}-10^{\circ})}{\sin 10^{\circ}}$
$= \frac{\sin 10^{\circ}}{\sin 10^{\circ}} = 1$

(A) We know that $\sin(90^{\circ} - \theta) = \cos \theta$.
Given expression: $\sin^{2} 51^{\circ} + \sin^{2} 39^{\circ}$.
We can write $39^{\circ}$ as $(90^{\circ} - 51^{\circ})$.
So,$\sin^{2} 39^{\circ} = \sin^{2}(90^{\circ} - 51^{\circ}) = \cos^{2} 51^{\circ}$.
Substituting this into the expression: $\sin^{2} 51^{\circ} + \cos^{2} 51^{\circ}$.
Using the identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we get $1$.

(C) To convert radians to degrees,we use the formula: $\text{Degree} = \text{Radian} \times \frac{180^{\circ}}{\pi}$.
Given the angle is $\frac{\pi}{32}$ radians.
$\text{Degree measure} = \frac{\pi}{32} \times \frac{180^{\circ}}{\pi} = \frac{180}{32}^{\circ} = \frac{45}{8}^{\circ} = 5 \frac{5}{8}^{\circ}$.
Now,convert the fractional part to minutes: $5^{\circ} + (\frac{5}{8} \times 60)^{\prime} = 5^{\circ} + (\frac{300}{8})^{\prime} = 5^{\circ} + 37.5^{\prime} = 5^{\circ} 37^{\prime} + 0.5^{\prime}$.
Convert the remaining fractional part to seconds: $0.5^{\prime} = (0.5 \times 60)^{\prime \prime} = 30^{\prime \prime}$.
Thus,the final measure is $5^{\circ} 37^{\prime} 30^{\prime \prime}$.

(A) Given expression: $\cos 1200^{\circ} + \tan 1485^{\circ}$
Using the property $\cos(n \times 360^{\circ} + \theta) = \cos \theta$ and $\tan(n \times 360^{\circ} + \theta) = \tan \theta$:
$\cos 1200^{\circ} = \cos(3 \times 360^{\circ} + 120^{\circ}) = \cos 120^{\circ}$
$\tan 1485^{\circ} = \tan(4 \times 360^{\circ} + 45^{\circ}) = \tan 45^{\circ}$
Now,calculate the values:
$\cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$
$\tan 45^{\circ} = 1$
Therefore,$\cos 1200^{\circ} + \tan 1485^{\circ} = -\frac{1}{2} + 1 = \frac{1}{2}$

(A) We evaluate each option:
$A$: $\cos 5 \pi = -1$ and $\cos 4 \pi = 1$. Since $-1 \neq 1$,this statement is incorrect.
$B$: $\sin 2 \pi = 0$ and $\sin (-2 \pi) = 0$. This is correct.
$C$: $\sin 4 \pi = 0$ and $\sin 6 \pi = 0$. This is correct.
$D$: $\tan 45^{\circ} = 1$ and $\tan (-315^{\circ}) = \tan (-315^{\circ} + 360^{\circ}) = \tan 45^{\circ} = 1$. This is correct.
Therefore,the incorrect statement is $A$.

(D) Given,function $y = \tan x$.
In the $II$ quadrant,the angle $x$ ranges from $\frac{\pi}{2}$ to $\pi$.
As $x$ approaches $\frac{\pi}{2}^+$,$\tan x$ approaches $-\infty$.
As $x$ approaches $\pi^-$,$\tan x$ approaches $0$.
Since the function values move from $-\infty$ to $0$ as $x$ increases,the function $y = \tan x$ increases from $-\infty$ to $0$ in the $II$ quadrant.

(B) We have the expression $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$.
Using the identity $\cos \theta = \sin(90^{\circ}-\theta)$ and $\sin \theta = \cos(90^{\circ}-\theta)$:
$\frac{\sin 70^{\circ}+\sin 50^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}}$
Using sum-to-product formulas $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$ and $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$= \frac{2 \sin(\frac{70^{\circ}+50^{\circ}}{2}) \cos(\frac{70^{\circ}-50^{\circ}}{2})}{2 \cos(\frac{70^{\circ}+50^{\circ}}{2}) \cos(\frac{70^{\circ}-50^{\circ}}{2})}$
$= \frac{2 \sin 60^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}$
$= \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \tan 60^{\circ} = \sqrt{3}$.

(C) Given expression: $(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)$
$= (\sin \theta + \cos \theta) \left( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \right)$
$= (\sin \theta + \cos \theta) \left( \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \right)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$= (\sin \theta + \cos \theta) \left( \frac{1}{\sin \theta \cos \theta} \right)$
$= \frac{\sin \theta}{\sin \theta \cos \theta} + \frac{\cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\cos \theta} + \frac{1}{\sin \theta}$
$= \sec \theta + \operatorname{cosec} \theta$

(C) The range of $\sin \theta$ and $\cos \theta$ is $[-1, 1]$. The range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$. The range of $\tan \theta$ is $(-\infty, \infty)$.
$(a)$ For $\sin \theta = \frac{a^{2}+b^{2}}{a^{2}-b^{2}}$,if we take $a=2, b=1$,then $\sin \theta = \frac{5}{3} > 1$,which is impossible.
$(b)$ For $\sec \theta = \frac{4}{5} = 0.8$,which is impossible since $|sec \theta| \geq 1$.
$(c)$ For $\tan \theta = 45$,this is possible as the range of $\tan \theta$ is all real numbers.
$(d)$ For $\cos \theta = \frac{7}{3} > 1$,which is impossible.
Therefore,option $(c)$ is correct.

(C) Let $x = 67 \frac{1}{2}^{\circ} = 67.5^{\circ}$. We need to evaluate $\tan 67.5^{\circ} + \cot 67.5^{\circ}$.
Using the identity $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$.
Here,$\theta = 67.5^{\circ}$,so $2\theta = 135^{\circ}$.
Thus,$\tan 67.5^{\circ} + \cot 67.5^{\circ} = \frac{2}{\sin 135^{\circ}}$.
Since $\sin 135^{\circ} = \sin(180^{\circ} - 45^{\circ}) = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Therefore,the value is $\frac{2}{1/\sqrt{2}} = 2 \sqrt{2}$.

(B) We have,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$
Since $\sin(90^{\circ} - \theta) = \cos \theta$,we can write $\sin(72.5^{\circ}) = \sin(90^{\circ} - 17.5^{\circ}) = \cos(17.5^{\circ})$.
Therefore,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ} = \sin ^{2} 17.5^{\circ} + \cos ^{2} 17.5^{\circ}$.
Using the identity $\sin ^{2} \theta + \cos ^{2} \theta = 1$,we get $1$.
Since $\tan 45^{\circ} = 1$,then $\tan ^{2} 45^{\circ} = 1^{2} = 1$.
Thus,the expression is equal to $\tan ^{2} 45^{\circ}$.

(B) We are given the expression: $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$.
Using the property $\tan(90^{\circ} - \theta) = \cot \theta$,we can rewrite the terms from $46^{\circ}$ to $89^{\circ}$ as:
$\tan 89^{\circ} = \tan(90^{\circ} - 1^{\circ}) = \cot 1^{\circ}$
$\tan 88^{\circ} = \tan(90^{\circ} - 2^{\circ}) = \cot 2^{\circ}$
$\dots$
$\tan 46^{\circ} = \tan(90^{\circ} - 44^{\circ}) = \cot 44^{\circ}$.
Substituting these into the expression:
$= (\tan 1^{\circ} \cot 1^{\circ}) \times (\tan 2^{\circ} \cot 2^{\circ}) \times \dots \times (\tan 44^{\circ} \cot 44^{\circ}) \times \tan 45^{\circ}$.
Since $\tan \theta \cot \theta = 1$ and $\tan 45^{\circ} = 1$,the expression becomes:
$= 1 \times 1 \times \dots \times 1 \times 1 = 1$.

(B) Given expression: $\tan \left(1^{\circ}\right)+\tan \left(89^{\circ}\right)$
Since $\tan \left(90^{\circ}-\theta\right)=\cot \theta$,we have $\tan \left(89^{\circ}\right)=\tan \left(90^{\circ}-1^{\circ}\right)=\cot \left(1^{\circ}\right)$
So,the expression becomes: $\tan \left(1^{\circ}\right)+\cot \left(1^{\circ}\right)$
$= \frac{\sin \left(1^{\circ}\right)}{\cos \left(1^{\circ}\right)}+\frac{\cos \left(1^{\circ}\right)}{\sin \left(1^{\circ}\right)}$
$= \frac{\sin^2 \left(1^{\circ}\right)+\cos^2 \left(1^{\circ}\right)}{\sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}$
$= \frac{1}{\sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}$
Multiply numerator and denominator by $2$:
$= \frac{2}{2 \sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}$
Using the identity $\sin \left(2\theta\right)=2 \sin \theta \cos \theta$:
$= \frac{2}{\sin \left(2^{\circ}\right)}$

(D) We know that $\operatorname{cosech} x = \frac{1}{\sinh x}$.
Given $\operatorname{cosech} x = \frac{4}{5}$,therefore $\sinh x = \frac{5}{4}$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we have $\cosh^2 x = 1 + \sinh^2 x$.
Substituting the value of $\sinh x$: $\cosh^2 x = 1 + (\frac{5}{4})^2 = 1 + \frac{25}{16} = \frac{16 + 25}{16} = \frac{41}{16}$.
Since $\cosh x > 0$ for all real $x$,we take the positive square root: $\cosh x = \sqrt{\frac{41}{16}}$.

(A) We know that $\sin(180^{\circ} + \theta) = -\sin \theta$ and $\sin(540^{\circ} + \theta) = \sin(360^{\circ} + 180^{\circ} + \theta) = \sin(180^{\circ} + \theta) = -\sin \theta$.
Step $1$: Simplify $\sin 210^{\circ} = \sin(180^{\circ} + 30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2}$.
Step $2$: Simplify $\sin 585^{\circ} = \sin(540^{\circ} + 45^{\circ}) = -\sin 45^{\circ} = -\frac{1}{\sqrt{2}}$.
Step $3$: Multiply the values: $(-\frac{1}{2}) \times (-\frac{1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}}$.
Thus,the correct option is $A$.

(C) $\operatorname{cosec} 15^{\circ} + \sec 15^{\circ} = \frac{1}{\sin 15^{\circ}} + \frac{1}{\cos 15^{\circ}}$
$= \frac{\cos 15^{\circ} + \sin 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2(\cos 15^{\circ} + \sin 15^{\circ})}{2 \sin 15^{\circ} \cos 15^{\circ}} = \frac{2(\cos 15^{\circ} + \sin 15^{\circ})}{\sin 30^{\circ}}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have:
$= 4(\cos 15^{\circ} + \sin 15^{\circ})$
Using $\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$:
$= 4 \left( \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{6} - \sqrt{2}}{4} \right)$
$= 4 \left( \frac{2\sqrt{6}}{4} \right) = 2\sqrt{6}$

(A) Given $x = \tan 15^{\circ} = \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \approx 0.268$.
$y = \operatorname{cosec} 75^{\circ} = \frac{1}{\sin(45^{\circ} + 30^{\circ})} = \frac{1}{\sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}} = \frac{1}{\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}} = \frac{2\sqrt{2}}{\sqrt{3} + 1} = \sqrt{6} - \sqrt{2} \approx 2.449 - 1.414 = 1.035$.
$z = 4 \sin 18^{\circ} = 4 \left( \frac{\sqrt{5} - 1}{4} \right) = \sqrt{5} - 1 \approx 2.236 - 1 = 1.236$.
Comparing the values: $0.268 < 1.035 < 1.236$,which implies $x < y < z$.

(D) Given expression: $\frac{\sin \theta}{1-\cot \theta} + \frac{\cos \theta}{1-\tan \theta}$
Substitute $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
$= \frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\sin \theta - \cos \theta} + \frac{\cos^2 \theta}{\cos \theta - \sin \theta}$
$= \frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\cos^2 \theta}{\sin \theta - \cos \theta}$
$= \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta - \cos \theta}$
Using the identity $a^2 - b^2 = (a-b)(a+b)$:
$= \frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta - \cos \theta}$
$= \sin \theta + \cos \theta$

(B) We know that the hyperbolic functions are related by the identity:
$\sinh x = \frac{1}{\operatorname{cosech} x}$
Given that $\operatorname{cosech} x = \frac{4}{5}$,we substitute this value into the identity:
$\sinh x = \frac{1}{4/5} = \frac{5}{4}$

(C) Substituting the definition of $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$,we get:
$\frac{1+\tanh x}{1-\tanh x} = \frac{1 + \frac{e^x - e^{-x}}{e^x + e^{-x}}}{1 - \frac{e^x - e^{-x}}{e^x + e^{-x}}}$
$= \frac{\frac{e^x + e^{-x} + e^x - e^{-x}}{e^x + e^{-x}}}{\frac{e^x + e^{-x} - (e^x - e^{-x})}{e^x + e^{-x}}}$
$= \frac{2e^x}{2e^{-x}}$
$= e^{x - (-x)} = e^{2x}$

(D) Given expression: $f(x) = 1 + \sec ^2 x \sin ^2 x$
Since $\sec x = \frac{1}{\cos x}$,we have $\sec ^2 x = \frac{1}{\cos ^2 x}$.
Substituting this into the expression: $f(x) = 1 + \frac{\sin ^2 x}{\cos ^2 x}$
Since $\tan x = \frac{\sin x}{\cos x}$,we have $\tan ^2 x = \frac{\sin ^2 x}{\cos ^2 x}$.
Therefore,$f(x) = 1 + \tan ^2 x$
Using the identity $1 + \tan ^2 x = \sec ^2 x$,we get $f(x) = \sec ^2 x$.

(A) We evaluate each expression based on the quadrant in which the angle lies:
$I. \sin(-292^{\circ}) = \sin(68^{\circ})$. Since $68^{\circ}$ is in the first quadrant,$\sin(68^{\circ}) > 0$.
$II. \tan(-190^{\circ}) = -\tan(190^{\circ}) = -\tan(180^{\circ} + 10^{\circ}) = -\tan(10^{\circ})$. Since $\tan(10^{\circ}) > 0$,the value is negative.
$III. \cos(-207^{\circ}) = \cos(207^{\circ}) = \cos(180^{\circ} + 27^{\circ}) = -\cos(27^{\circ})$. Since $\cos(27^{\circ}) > 0$,the value is negative.
$IV. \cot(-222^{\circ}) = -\cot(222^{\circ}) = -\cot(180^{\circ} + 42^{\circ}) = -\cot(42^{\circ})$. Since $\cot(42^{\circ}) > 0$,the value is negative.
Thus,$II, III$,and $IV$ are negative.

(B) Given expression: $\sin ^2\left(\frac{2 \pi}{3}\right)+\cos ^2\left(\frac{5 \pi}{6}\right)-\tan ^2\left(\frac{3 \pi}{4}\right)$
Using allied angles: $\sin(\pi-x) = \sin x$,$\cos(\pi-x) = -\cos x$,$\tan(\pi-x) = -\tan x$
$= \sin ^2\left(\pi-\frac{\pi}{3}\right)+\cos ^2\left(\pi-\frac{\pi}{6}\right)-\tan ^2\left(\pi-\frac{\pi}{4}\right)$
$= \sin ^2\left(\frac{\pi}{3}\right)+\left(-\cos \frac{\pi}{6}\right)^2-\left(-\tan \frac{\pi}{4}\right)^2$
$= \left(\frac{\sqrt{3}}{2}\right)^2+\left(-\frac{\sqrt{3}}{2}\right)^2-(-1)^2$
$= \frac{3}{4}+\frac{3}{4}-1$
$= \frac{6}{4}-1 = \frac{3}{2}-1 = \frac{1}{2}$

(D) $\frac{1}{1+\sin \theta} + \frac{1}{1-\sin \theta} = \frac{(1-\sin \theta) + (1+\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}$
$= \frac{2}{1-\sin^2 \theta}$
$= \frac{2}{\cos^2 \theta}$
$= 2 \sec^2 \theta$

(D) We start with the expression: $\frac{\cos x}{1+\sin x} + \tan x$
Multiply the numerator and denominator of the first term by $(1-\sin x)$:
$\frac{\cos x(1-\sin x)}{(1+\sin x)(1-\sin x)} + \tan x$
Using the identity $(1+\sin x)(1-\sin x) = 1-\sin^2 x = \cos^2 x$:
$\frac{\cos x(1-\sin x)}{\cos^2 x} + \tan x$
Simplify the fraction:
$\frac{1-\sin x}{\cos x} + \tan x$
Split the fraction:
$\frac{1}{\cos x} - \frac{\sin x}{\cos x} + \tan x$
Since $\frac{1}{\cos x} = \sec x$ and $\frac{\sin x}{\cos x} = \tan x$:
$\sec x - \tan x + \tan x = \sec x$

(B) We use the property $\cos^2 \theta = \cos^2(180^{\circ} \pm \theta) = \cos^2(360^{\circ} \pm \theta)$.
$\cos 135^{\circ} = \cos(180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} \implies \cos^2 135^{\circ} = \cos^2 45^{\circ}$.
$\cos 225^{\circ} = \cos(180^{\circ} + 45^{\circ}) = -\cos 45^{\circ} \implies \cos^2 225^{\circ} = \cos^2 45^{\circ}$.
$\cos 315^{\circ} = \cos(360^{\circ} - 45^{\circ}) = \cos 45^{\circ} \implies \cos^2 315^{\circ} = \cos^2 45^{\circ}$.
Thus,the expression becomes:
$\cos^2 45^{\circ} + \cos^2 45^{\circ} + \cos^2 45^{\circ} + \cos^2 45^{\circ} = 4 \cos^2 45^{\circ}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\cos^2 45^{\circ} = \frac{1}{2}$.
Therefore,$4 \times \frac{1}{2} = 2$.

(B) We know that $\cot 30^{\circ} = \sqrt{3}$ and $\sec 45^{\circ} = \sqrt{2}$.
Substituting these values into the expression:
$1 + (\sqrt{3})^2 - (\sqrt{2})^2$
$= 1 + 3 - 2$
$= 4 - 2$
$= 2$

(B) Given expression: $\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x}$
Taking the common denominator: $\frac{\sin^2 x + (1+\cos x)^2}{\sin x(1+\cos x)}$
$= \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x(1+\cos x)}$
Since $\sin^2 x + \cos^2 x = 1$,we get: $\frac{1 + 1 + 2\cos x}{\sin x(1+\cos x)}$
$= \frac{2 + 2\cos x}{\sin x(1+\cos x)} = \frac{2(1+\cos x)}{\sin x(1+\cos x)}$
$= \frac{2}{\sin x} = 2 \operatorname{cosec} x$