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

(B) First,simplify each trigonometric term using reduction formulas:
$ \sin(3\pi - \alpha) = \sin(\pi - \alpha) = \sin \alpha $
$ \cos(3\pi + \alpha) = -\cos \alpha $
$ \sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha $
$ \cos(\frac{5\pi}{2} - \alpha) = \cos(\frac{\pi}{2} - \alpha) = \sin \alpha $
Substituting these into the expression:
$ [1 - \sin \alpha - \cos \alpha] [1 - (-\cos \alpha) + \sin \alpha ] $
$ = [1 - (\sin \alpha + \cos \alpha)] [1 + (\sin \alpha + \cos \alpha)] $
Using the identity $(a - b)(a + b) = a^2 - b^2$:
$ = 1^2 - (\sin \alpha + \cos \alpha)^2 $
$ = 1 - (\sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha) $
$ = 1 - (1 + \sin 2\alpha) $
$ = - \sin 2\alpha $

(A) We know that the function $f(x) = sin\ x$ is increasing in the interval $[0^{\circ}, 90^{\circ}]$.
For angles greater than $90^{\circ}$,we use the identity $sin\ \theta = sin(180^{\circ} - \theta)$.
Thus,$sin\ 95^{\circ} = sin(180^{\circ} - 95^{\circ}) = sin\ 85^{\circ}$.
Now we compare $sin\ 85^{\circ}$,$sin\ 63^{\circ}$,and $sin\ 1^{\circ}$.
Since $85^{\circ} > 63^{\circ} > 1^{\circ}$ and the sine function is increasing in the first quadrant,we have $sin\ 85^{\circ} > sin\ 63^{\circ} > sin\ 1^{\circ}$.
Therefore,$sin\ 95^{\circ} > sin\ 63^{\circ} > sin\ 1^{\circ}$.

(D) Given $\sin \left( x + \frac{4\pi}{9} \right) = a$.
We need to find $\cos \left( x + \frac{7\pi}{9} \right)$.
Note that $\frac{7\pi}{9} = \frac{4\pi}{9} + \frac{3\pi}{9} = \frac{4\pi}{9} + \frac{\pi}{3}$.
So,$\cos \left( x + \frac{7\pi}{9} \right) = \cos \left( \left( x + \frac{4\pi}{9} \right) + \frac{\pi}{3} \right)$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$\cos \left( \left( x + \frac{4\pi}{9} \right) + \frac{\pi}{3} \right) = \cos \left( x + \frac{4\pi}{9} \right) \cos \frac{\pi}{3} - \sin \left( x + \frac{4\pi}{9} \right) \sin \frac{\pi}{3}$.
Given $\frac{\pi}{9} < x < \frac{\pi}{3}$,then $\frac{5\pi}{9} < x + \frac{4\pi}{9} < \frac{7\pi}{9}$.
In this interval,the cosine function is negative.
Since $\sin \left( x + \frac{4\pi}{9} \right) = a$,then $\cos \left( x + \frac{4\pi}{9} \right) = -\sqrt{1 - a^2}$.
Substituting the values:
$-\sqrt{1 - a^2} \cdot \frac{1}{2} - a \cdot \frac{\sqrt{3}}{2} = \frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$.

(D) Given expression $E = \frac{\sin 81^o + \cos 81^o}{\sin 81^o - \cos 81^o}$.
Divide numerator and denominator by $\cos 81^o$:
$E = \frac{\tan 81^o + 1}{\tan 81^o - 1}$.
Multiply numerator and denominator by $-1$:
$E = \frac{1 + \tan 81^o}{1 - \tan 81^o}$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = 45^o$ and $B = 81^o$:
$E = \tan(45^o + 81^o) = \tan 126^o$.
Alternatively,note that $\sin 81^o = \cos 9^o$ and $\cos 81^o = \sin 9^o$:
$E = \frac{\cos 9^o + \sin 9^o}{\cos 9^o - \sin 9^o} = \frac{1 + \tan 9^o}{1 - \tan 9^o} = \tan(45^o + 9^o) = \tan 54^o$.

(A) We know that $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + 45^{\circ})$.
For $A$,$\theta = 45^{\circ}$,so $A = \sqrt{2} \sin(45^{\circ} + 45^{\circ}) = \sqrt{2} \sin 90^{\circ} = \sqrt{2}(1) = \sqrt{2}$.
For $B$,$\theta = 44^{\circ}$,so $B = \sqrt{2} \sin(44^{\circ} + 45^{\circ}) = \sqrt{2} \sin 89^{\circ}$.
Since $\sin 89^{\circ} < \sin 90^{\circ}$,it follows that $\sqrt{2} \sin 89^{\circ} < \sqrt{2} \sin 90^{\circ}$.
Therefore,$B < A$ or $A > B$.

(A) Let the expression be $E = \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} - \sqrt{\frac{1 + \sin \alpha}{1 - \sin \alpha}}$.
Rationalizing the denominators:
$E = \frac{\sqrt{(1 - \sin \alpha)^2}}{\sqrt{1 - \sin^2 \alpha}} - \frac{\sqrt{(1 + \sin \alpha)^2}}{\sqrt{1 - \sin^2 \alpha}}$
$E = \frac{|1 - \sin \alpha| - |1 + \sin \alpha|}{\sqrt{\cos^2 \alpha}}$
$E = \frac{(1 - \sin \alpha) - (1 + \sin \alpha)}{|\cos \alpha|}$ (Since $\sin \alpha \in (0, 1)$ in the second quadrant,$1 - \sin \alpha > 0$ and $1 + \sin \alpha > 0$)
$E = \frac{-2 \sin \alpha}{|\cos \alpha|}$.
Since $\alpha$ is in the second quadrant,$\cos \alpha < 0$,so $|\cos \alpha| = -\cos \alpha$.
$E = \frac{-2 \sin \alpha}{-\cos \alpha} = 2 \tan \alpha$.

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.
Using trigonometric identities:
$\cos(180^{\circ} + A) = -\cos A$
$\cos(180^{\circ} - B) = -\cos B$
$\cos(180^{\circ} - C) = -\cos C$
$\sin(90^{\circ} - D) = \cos D$
Substituting these into the expression:
$-\cos A - \cos B - \cos C - \cos D = -(\cos A + \cos C) - (\cos B + \cos D)$
Since $C = 180^{\circ} - A$,$\cos C = \cos(180^{\circ} - A) = -\cos A$,so $\cos A + \cos C = 0$.
Similarly,$D = 180^{\circ} - B$,$\cos D = \cos(180^{\circ} - B) = -\cos B$,so $\cos B + \cos D = 0$.
Therefore,the expression equals $-(0) - (0) = 0$.

(B) We know that $\sin(\pi - \theta) = \sin \theta$.
$\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$
$\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$
Substituting these into the expression:
$\text{Expression} = \sin ^2 \frac{\pi}{8} + \sin ^2 \frac{3\pi}{8} + \sin ^2 \frac{3\pi}{8} + \sin ^2 \frac{\pi}{8}$
$= 2(\sin ^2 \frac{\pi}{8} + \sin ^2 \frac{3\pi}{8})$
Since $\sin \frac{3\pi}{8} = \sin(\frac{\pi}{2} - \frac{\pi}{8}) = \cos \frac{\pi}{8}$,we have:
$= 2(\sin ^2 \frac{\pi}{8} + \cos ^2 \frac{\pi}{8})$
Using the identity $\sin ^2 \theta + \cos ^2 \theta = 1$:
$= 2(1) = 2$

(D) We know that $1 \text{ radian} \approx 57.295^\circ$.
$\sin 1 \text{ rad} = \sin 57.295^\circ \approx 0.841$
$\sin 2 \text{ rad} = \sin 114.591^\circ = \sin(180^\circ - 114.591^\circ) = \sin 65.409^\circ \approx 0.909$
$\sin 3 \text{ rad} = \sin 171.887^\circ = \sin(180^\circ - 171.887^\circ) = \sin 8.113^\circ \approx 0.141$
Comparing the values: $0.141 < 0.841 < 0.909$.
Therefore,$\sin 3 < \sin 1 < \sin 2$.

(C) Consider $\cos 255^o + \sin 195^o$.
Using the reduction formulas:
$\cos 255^o = \cos(270^o - 15^o) = -\sin 15^o$.
$\sin 195^o = \sin(180^o + 15^o) = -\sin 15^o$.
Therefore,the expression becomes:
$-\sin 15^o - \sin 15^o = -2 \sin 15^o$.
Using the value $\sin 15^o = \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$:
$-2 \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) = -\frac{\sqrt{3} - 1}{\sqrt{2}}$.

(A) We know that $180^{\circ} = \pi \text{ radians}$.
First,convert $40^{\circ} 20^{\prime}$ into degrees:
$20^{\prime} = (\frac{20}{60})^{\circ} = (\frac{1}{3})^{\circ}$.
So,$40^{\circ} 20^{\prime} = (40 + \frac{1}{3})^{\circ} = (\frac{120+1}{3})^{\circ} = (\frac{121}{3})^{\circ}$.
Now,convert degrees to radians by multiplying by $\frac{\pi}{180}$:
$\text{Radian measure} = (\frac{121}{3}) \times \frac{\pi}{180} = \frac{121 \pi}{540} \text{ radians}$.
Therefore,$40^{\circ} 20^{\prime} = \frac{121 \pi}{540} \text{ radians}$.

(A) We know that $\pi \text{ radians} = 180^{\circ}$.
Therefore,$6 \text{ radians} = \frac{180}{\pi} \times 6 \text{ degrees} = \frac{1080 \times 7}{22} \text{ degrees} \approx 343.6363^{\circ}$.
Converting the decimal part to minutes: $0.6363^{\circ} = 0.6363 \times 60^{\prime} = 38.1818^{\prime} = 38^{\prime} + 0.1818^{\prime}$.
Converting the remaining decimal part to seconds: $0.1818^{\prime} = 0.1818 \times 60^{\prime \prime} \approx 10.9^{\prime \prime} \approx 11^{\prime \prime}$.
Thus,$6 \text{ radians} \approx 343^{\circ} 38^{\prime} 11^{\prime \prime}$.

(B) Let $r_{1}$ and $r_{2}$ be the radii of the two circles and $l$ be the length of the arc in both cases.
We know that the arc length formula is $l = r \theta$,where $\theta$ is in radians.
Given $\theta_{1} = 65^{\circ}$ and $\theta_{2} = 110^{\circ}$.
Since the arc lengths are equal,$l = r_{1} \theta_{1} = r_{2} \theta_{2}$.
Therefore,$\frac{r_{1}}{r_{2}} = \frac{\theta_{2}}{\theta_{1}}$.
Substituting the values,$\frac{r_{1}}{r_{2}} = \frac{110^{\circ}}{65^{\circ}} = \frac{110}{65} = \frac{22}{13}$.
Thus,the ratio of their radii is $22: 13$.

(A) We know that $180^{\circ} = \pi \text{ radians}$.
Therefore,$1^{\circ} = \frac{\pi}{180} \text{ radians}$.
To convert $25^{\circ}$ into radians,we multiply by $\frac{\pi}{180}$:
$25^{\circ} = 25 \times \frac{\pi}{180} \text{ radians}$.
Simplifying the fraction by dividing both numerator and denominator by $5$:
$25^{\circ} = \frac{5 \pi}{36} \text{ radians}$.

(A) We know that $1^{\circ} = 60^{\prime}$,so $30^{\prime} = (\frac{30}{60})^{\circ} = 0.5^{\circ}$.
Thus,$-47^{\circ} 30^{\prime} = -(47 + 0.5)^{\circ} = -47.5^{\circ} = -\frac{95}{2}^{\circ}$.
Since $180^{\circ} = \pi \text{ radians}$,$1^{\circ} = \frac{\pi}{180} \text{ radians}$.
Therefore,$-\frac{95}{2}^{\circ} = (-\frac{95}{2}) \times \frac{\pi}{180} \text{ radians}$.
Simplifying the expression: $(-\frac{95}{360}) \pi = -\frac{19}{72} \pi \text{ radians}$.
Hence,$-47^{\circ} 30^{\prime} = -\frac{19}{72} \pi \text{ radians}$.

(A) We know that $180^{\circ} = \pi \text{ radians}$.
To convert degrees to radians,we use the formula: $\text{Radian measure} = \text{Degree measure} \times \frac{\pi}{180^{\circ}}$.
Therefore,$240^{\circ} = 240 \times \frac{\pi}{180} \text{ radians}$.
Simplifying the fraction: $\frac{240}{180} \pi = \frac{4}{3} \pi \text{ radians}$.

(A) We know that $180^{\circ} = \pi \text{ radian}$.
To convert degrees to radians,we multiply the degree measure by $\frac{\pi}{180}$.
$\therefore 520^{\circ} = 520 \times \frac{\pi}{180} \text{ radian}$.
$= \frac{520}{180} \pi \text{ radian}$.
$= \frac{52}{18} \pi \text{ radian} = \frac{26 \pi}{9} \text{ radian}$.

(A) We know that $\pi \text{ radian} = 180^{\circ}$.
$\therefore \frac{11}{16} \text{ radian} = \frac{180}{\pi} \times \frac{11}{16} \text{ degree}$.
Using $\pi = \frac{22}{7}$,we get: $\frac{180 \times 7}{22} \times \frac{11}{16} = \frac{180 \times 7}{2 \times 16} = \frac{90 \times 7}{16} = \frac{45 \times 7}{8} = \frac{315}{8} \text{ degree}$.
$= 39 \frac{3}{8} \text{ degree} = 39^{\circ} + \frac{3}{8} \times 60^{\prime} \quad [\because 1^{\circ} = 60^{\prime}]$.
$= 39^{\circ} + \frac{180}{8}^{\prime} = 39^{\circ} + 22.5^{\prime} = 39^{\circ} + 22^{\prime} + 0.5^{\prime}$.
$= 39^{\circ} + 22^{\prime} + 0.5 \times 60^{\prime \prime} \quad [\because 1^{\prime} = 60^{\prime \prime}]$.
$= 39^{\circ} 22^{\prime} 30^{\prime \prime}$.

(A) We know that $\pi \text{ radians} = 180^{\circ}$.
$-4 \text{ radians} = \frac{180}{\pi} \times (-4) \text{ degrees}$.
Using $\pi = \frac{22}{7}$,we get:
$= \frac{180 \times 7 \times (-4)}{22} \text{ degrees} = \frac{-2520}{11} \text{ degrees} = -229 \frac{1}{11} \text{ degrees}$.
$= -229^{\circ} + \left( \frac{1}{11} \times 60 \right) \text{ minutes} \quad [\because 1^{\circ} = 60^{\prime}]$.
$= -229^{\circ} + 5 \frac{5}{11} \text{ minutes} = -229^{\circ} 5^{\prime} + \left( \frac{5}{11} \times 60 \right) \text{ seconds} \quad [\because 1^{\prime} = 60^{\prime \prime}]$.
$= -229^{\circ} 5^{\prime} + 27.27^{\prime \prime} \approx -229^{\circ} 5^{\prime} 27^{\prime \prime}$.

(A) We know that $1 \text{ radian} = \left( \frac{180}{\pi} \right)^{\circ}$.
Therefore,to convert $\frac{5 \pi}{3}$ radians to degrees,we multiply by $\frac{180}{\pi}$:
$\text{Degree measure} = \frac{5 \pi}{3} \times \left( \frac{180}{\pi} \right)^{\circ}$
$= \frac{5}{3} \times 180^{\circ}$
$= 5 \times 60^{\circ}$
$= 300^{\circ}$

(A) We know that $\pi \text{ radian} = 180^{\circ}$.
To convert radians to degrees,we multiply by $\frac{180}{\pi}$.
$\frac{7 \pi}{6} \text{ radian} = \frac{7 \pi}{6} \times \frac{180}{\pi}^{\circ}$.
$= \frac{7}{6} \times 180^{\circ}$.
$= 7 \times 30^{\circ} = 210^{\circ}$.

(B) Number of revolutions made by the wheel in $1$ minute $= 360$.
Since $1$ minute $= 60$ seconds,the number of revolutions made by the wheel in $1$ second $= \frac{360}{60} = 6$.
In one complete revolution,the wheel turns an angle of $2 \pi$ radians.
Hence,in $6$ complete revolutions,it will turn an angle of $6 \times 2 \pi = 12 \pi$ radians.
Thus,in one second,the wheel turns an angle of $12 \pi$ radians.

(A) We know that in a circle of radius $r$,if an arc of length $l$ subtends an angle $\theta$ in radians at the centre,then $\theta = \frac{l}{r}$.
Given $r = 100 \, cm$ and $l = 22 \, cm$,we have:
$\theta = \frac{22}{100} \, \text{radians}$.
To convert radians to degrees,we use the relation $1 \, \text{radian} = \frac{180^{\circ}}{\pi}$.
$\theta = \frac{22}{100} \times \frac{180}{\pi} \, \text{degrees}$.
Substituting $\pi = \frac{22}{7}$:
$\theta = \frac{22}{100} \times 180 \times \frac{7}{22} = \frac{180 \times 7}{100} = \frac{1260}{100} = 12.6^{\circ}$.
Converting $0.6^{\circ}$ to minutes:
$0.6^{\circ} = 0.6 \times 60^{\prime} = 36^{\prime}$.
Thus,the required angle is $12^{\circ} 36^{\prime}$.

(A) We know that in a circle of radius $r$ units,if an arc of length $l$ units subtends an angle $\theta$ radians at the center,then $\theta = \frac{l}{r}$.
Given that the length of the pendulum $r = 75 \, cm$ and the arc length $l = 10 \, cm$.
Substituting these values into the formula:
$\theta = \frac{10}{75} \, \text{radians}$.
Simplifying the fraction,we get $\theta = \frac{2}{15} \, \text{radians}$.

(A) We know that in a circle of radius $r$ unit,if an arc of length $l$ unit subtends an angle $\theta$ radian at the centre,then $\theta = \frac{l}{r}$.
Given that the length of the pendulum $r = 75\, cm$ and the arc length $l = 15\, cm$.
Substituting the values in the formula:
$\theta = \frac{15}{75} \text{ radian}$.
$\theta = \frac{1}{5} \text{ radian}$.

(A) We know that in a circle of radius $r$,if an arc of length $l$ subtends an angle $\theta$ in radians at the center,then $\theta = \frac{l}{r}$.
Given that the length of the pendulum $r = 75 \, cm$ and the arc length $l = 21 \, cm$.
Substituting these values into the formula:
$\theta = \frac{21}{75} \, \text{radians}$.
Simplifying the fraction by dividing both numerator and denominator by $3$:
$\theta = \frac{7}{25} \, \text{radians}$.

Given $\cos x = -\frac{3}{5}$.
Since $\sec x = \frac{1}{\cos x}$,we have $\sec x = -\frac{5}{3}$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we get $\sin^2 x = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Thus,$\sin x = \pm \frac{4}{5}$.
Since $x$ lies in the third quadrant,$\sin x$ must be negative,so $\sin x = -\frac{4}{5}$.
Consequently,$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$.
Now,$\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \frac{4}{3}$.
Finally,$\cot x = \frac{1}{\tan x} = \frac{3}{4}$.

(A) Given $\cot x = -\frac{5}{12}$. Since $x$ is in the second quadrant,$\tan x = \frac{1}{\cot x} = -\frac{12}{5}$.
Using the identity $\sec^2 x = 1 + \tan^2 x = 1 + (-\frac{12}{5})^2 = 1 + \frac{144}{25} = \frac{169}{25}$.
Since $x$ is in the second quadrant,$\sec x$ must be negative,so $\sec x = -\frac{13}{5}$.
Consequently,$\cos x = \frac{1}{\sec x} = -\frac{5}{13}$.
Now,$\sin x = \tan x \cdot \cos x = (-\frac{12}{5}) \cdot (-\frac{5}{13}) = \frac{12}{13}$.
Finally,$\csc x = \frac{1}{\sin x} = \frac{13}{12}$.

(A) We know that the value of $\sin x$ repeats after an interval of $2 \pi$.
Therefore,we can write:
$\sin \frac{31 \pi}{3} = \sin \left( 10 \pi + \frac{\pi}{3} \right)$
Since $\sin(2n \pi + \theta) = \sin \theta$ for any integer $n$,we have:
$\sin \left( 10 \pi + \frac{\pi}{3} \right) = \sin \frac{\pi}{3}$
Finally,the value is:
$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

(A) We know that the value of $\cos(x)$ repeats after an interval of $2\pi$ or $360^{\circ}$.
Therefore,$\cos \left(-1710^{\circ}\right) = \cos \left(-1710^{\circ} + n \times 360^{\circ}\right)$.
To find the smallest positive angle coterminal with $-1710^{\circ}$,we add multiples of $360^{\circ}$:
$-1710^{\circ} + 5 \times 360^{\circ} = -1710^{\circ} + 1800^{\circ} = 90^{\circ}$.
Thus,$\cos \left(-1710^{\circ}\right) = \cos(90^{\circ})$.
Since $\cos(90^{\circ}) = 0$,the value is $0$.

Given $\cos x = -\frac{1}{2}$.
$\sec x = \frac{1}{\cos x} = \frac{1}{(-1/2)} = -2$.
Using the identity $\sin^2 x + \cos^2 x = 1$:
$\sin^2 x = 1 - \cos^2 x = 1 - (-1/2)^2 = 1 - 1/4 = 3/4$.
Since $x$ lies in the third quadrant,$\sin x$ must be negative.
$\sin x = -\frac{\sqrt{3}}{2}$.
$\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.
$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$.

Given $\sin x = \frac{3}{5}$.
$\csc x = \frac{1}{\sin x} = \frac{1}{\left(\frac{3}{5}\right)} = \frac{5}{3}$.
Using the identity $\sin^{2} x + \cos^{2} x = 1$:
$\cos^{2} x = 1 - \sin^{2} x = 1 - \left(\frac{3}{5}\right)^{2} = 1 - \frac{9}{25} = \frac{16}{25}$.
Since $x$ lies in the $2^{\text{nd}}$ quadrant,$\cos x$ is negative:
$\cos x = -\frac{4}{5}$.
$\sec x = \frac{1}{\cos x} = \frac{1}{(-\frac{4}{5})} = -\frac{5}{4}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{(\frac{3}{5})}{(-\frac{4}{5})} = -\frac{3}{4}$.
$\cot x = \frac{1}{\tan x} = -\frac{4}{3}$.

Given $\cot x = \frac{3}{4}$.
$\tan x = \frac{1}{\cot x} = \frac{1}{3/4} = \frac{4}{3}$.
Using the identity $1 + \tan^2 x = \sec^2 x$:
$1 + (4/3)^2 = \sec^2 x$
$1 + \frac{16}{9} = \sec^2 x$
$\frac{25}{9} = \sec^2 x$
$\sec x = \pm \frac{5}{3}$.
Since $x$ lies in the third quadrant,$\sec x$ is negative:
$\sec x = -\frac{5}{3}$.
$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$.
Since $\tan x = \frac{\sin x}{\cos x}$,we have $\sin x = \tan x \cdot \cos x$:
$\sin x = \left(\frac{4}{3}\right) \cdot \left(-\frac{3}{5}\right) = -\frac{4}{5}$.
$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$.

(N/A) Given $\sec x = \frac{13}{5}$.
Since $\cos x = \frac{1}{\sec x}$,we have $\cos x = \frac{5}{13}$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we get $\sin^2 x = 1 - \cos^2 x = 1 - (\frac{5}{13})^2 = 1 - \frac{25}{169} = \frac{144}{169}$.
Thus,$\sin x = \pm \frac{12}{13}$.
Since $x$ lies in the $4^{\text{th}}$ quadrant,$\sin x$ is negative,so $\sin x = -\frac{12}{13}$.
Now,$\csc x = \frac{1}{\sin x} = -\frac{13}{12}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$.
$\cot x = \frac{1}{\tan x} = -\frac{5}{12}$.

Given $\tan x = -\frac{5}{12}$.
$\cot x = \frac{1}{\tan x} = \frac{1}{(-\frac{5}{12})} = -\frac{12}{5}$.
Using the identity $1 + \tan^2 x = \sec^2 x$:
$1 + (-\frac{5}{12})^2 = \sec^2 x$
$1 + \frac{25}{144} = \sec^2 x$
$\frac{169}{144} = \sec^2 x$
$\sec x = \pm \frac{13}{12}$.
Since $x$ lies in the $2^{\text{nd}}$ quadrant,$\sec x$ is negative.
$\therefore \sec x = -\frac{13}{12}$.
$\cos x = \frac{1}{\sec x} = \frac{1}{(-\frac{13}{12})} = -\frac{12}{13}$.
Since $\tan x = \frac{\sin x}{\cos x}$,we have:
$\sin x = \tan x \times \cos x = (-\frac{5}{12}) \times (-\frac{12}{13}) = \frac{5}{13}$.
$\csc x = \frac{1}{\sin x} = \frac{1}{(\frac{5}{13})} = \frac{13}{5}$.

(C) It is known that the values of $\sin x$ repeat after an interval of $360^{\circ}$ (or $2\pi$ radians).
$\sin 765^{\circ} = \sin (2 \times 360^{\circ} + 45^{\circ})$
Since $\sin (n \times 360^{\circ} + \theta) = \sin \theta$,we have:
$\sin 765^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.

(B) We know that the trigonometric function $\csc(x)$ has a period of $360^{\circ}$.
Therefore,$\csc(-1410^{\circ}) = \csc(-1410^{\circ} + n \times 360^{\circ})$.
For $n = 4$,we have $-1410^{\circ} + 4 \times 360^{\circ} = -1410^{\circ} + 1440^{\circ} = 30^{\circ}$.
Thus,$\csc(-1410^{\circ}) = \csc(30^{\circ})$.
Since $\csc(30^{\circ}) = \frac{1}{\sin(30^{\circ})} = \frac{1}{1/2} = 2$.

(A) The period of the tangent function $\tan x$ is $\pi$.
We can write $\frac{19 \pi}{3}$ as $6 \pi + \frac{\pi}{3}$.
Using the property $\tan(n \pi + \theta) = \tan \theta$,where $n \in \mathbb{Z}$:
$\tan \frac{19 \pi}{3} = \tan \left(6 \pi + \frac{\pi}{3}\right) = \tan \frac{\pi}{3}$.
Since $\tan \frac{\pi}{3} = \tan 60^{\circ} = \sqrt{3}$,the value is $\sqrt{3}$.

(A) We know that the sine function is periodic with a period of $2\pi$. Therefore,$\sin(x + 2n\pi) = \sin(x)$ for any integer $n$.
$\sin \left(-\frac{11 \pi}{3}\right) = \sin \left(-\frac{11 \pi}{3} + 4\pi\right)$
$= \sin \left(\frac{-11\pi + 12\pi}{3}\right)$
$= \sin \left(\frac{\pi}{3}\right)$
$= \frac{\sqrt{3}}{2}$

(A) We know that the trigonometric function $\cot(x)$ has a period of $\pi$.
$\therefore \cot \left(-\frac{15 \pi}{4}\right) = \cot \left(-\frac{15 \pi}{4} + 4 \pi\right)$
$= \cot \left(\frac{-15 \pi + 16 \pi}{4}\right)$
$= \cot \left(\frac{\pi}{4}\right)$
$= 1$

(N/A) We have the $L.H.S. = 3 \sin \frac{\pi}{6} \sec \frac{\pi}{3} - 4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}$.
Using the values $\sin \frac{\pi}{6} = \frac{1}{2}$,$\sec \frac{\pi}{3} = 2$,$\sin \frac{5 \pi}{6} = \sin(\pi - \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2}$,and $\cot \frac{\pi}{4} = 1$:
$L.H.S. = 3 \times \frac{1}{2} \times 2 - 4 \times \frac{1}{2} \times 1$
$L.H.S. = 3 - 2 = 1$
Since $L.H.S. = R.H.S. = 1$,the identity is proved.

(N/A) We start with the $L.H.S$: $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
Substituting the standard trigonometric values: $\sin \frac{\pi}{6} = \frac{1}{2}$,$\cos \frac{\pi}{3} = \frac{1}{2}$,and $\tan \frac{\pi}{4} = 1$
$= (\frac{1}{2})^{2} + (\frac{1}{2})^{2} - (1)^{2}$
$= \frac{1}{4} + \frac{1}{4} - 1$
$= \frac{2}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$
$= R.H.S$
Hence,the identity is proved.

$L.H.S. = 2 \sin ^{2} \frac{\pi}{6} + \csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}$
$= 2 \left( \frac{1}{2} \right)^{2} + \csc ^{2} \left( \pi + \frac{\pi}{6} \right) \left( \frac{1}{2} \right)^{2}$
$= 2 \times \frac{1}{4} + \left( -\csc \frac{\pi}{6} \right)^{2} \left( \frac{1}{4} \right)$
$= \frac{1}{2} + (-2)^{2} \left( \frac{1}{4} \right)$
$= \frac{1}{2} + \frac{4}{4} = \frac{1}{2} + 1 = \frac{3}{2}$
$= R.H.S.$

(N/A) $L.H.S. = \cot ^{2} \frac{\pi}{6} + \csc \frac{5 \pi}{6} + 3 \tan ^{2} \frac{\pi}{6}$
$= (\sqrt{3})^{2} + \csc \left(\pi - \frac{\pi}{6}\right) + 3 \left(\frac{1}{\sqrt{3}}\right)^{2}$
$= 3 + \csc \frac{\pi}{6} + 3 \times \frac{1}{3}$
$= 3 + 2 + 1 = 6$
$= R.H.S.$

(N/A) $L.H.S. = 2 \sin ^{2} \frac{3 \pi}{4} + 2 \cos ^{2} \frac{\pi}{4} + 2 \sec ^{2} \frac{\pi}{3}$
$= 2 \left\{ \sin \left( \pi - \frac{\pi}{4} \right) \right\}^{2} + 2 \left( \frac{1}{\sqrt{2}} \right)^{2} + 2 (2)^{2}$
$= 2 \left( \sin \frac{\pi}{4} \right)^{2} + 2 \times \frac{1}{2} + 2(4)$
$= 2 \left( \frac{1}{\sqrt{2}} \right)^{2} + 1 + 8$
$= 2 \left( \frac{1}{2} \right) + 1 + 8$
$= 1 + 1 + 8 = 10$
$= R.H.S.$

(N/A) We start with the $L.H.S.$: $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos (\frac{\pi}{2}+x)}$
Using the allied angle formulas:
$\cos (\pi+x) = -\cos x$
$\cos (-x) = \cos x$
$\sin (\pi-x) = \sin x$
$\cos (\frac{\pi}{2}+x) = -\sin x$
Substituting these values into the expression:
$= \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}$
$= \frac{-\cos^{2} x}{-\sin^{2} x}$
$= \frac{\cos^{2} x}{\sin^{2} x}$
$= \cot^{2} x = R.H.S.$

(N/A) $L.H.S. = \cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x) \left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
Using allied angle formulas:
$\cos \left(\frac{3 \pi}{2}+x\right) = \sin x$
$\cos (2 \pi+x) = \cos x$
$\cot \left(\frac{3 \pi}{2}-x\right) = \tan x$
$\cot (2 \pi+x) = \cot x$
Substituting these values:
$= \sin x \cos x [\tan x + \cot x]$
$= \sin x \cos x \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right)$
$= \sin x \cos x \left[\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\right]$
$= \sin^2 x + \cos^2 x = 1 = R.H.S.$

(B) Let $\angle A = \theta$.
The area of $\triangle ABC$ is given by $\text{ar}(\triangle ABC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin \theta$.
The area of $\triangle ADE$ is given by $\text{ar}(\triangle ADE) = \frac{1}{2} \cdot AD \cdot AE \cdot \sin \theta$.
Therefore,the ratio of the areas is:
$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle ADE)} = \frac{\frac{1}{2} \cdot AB \cdot AC \cdot \sin \theta}{\frac{1}{2} \cdot AD \cdot AE \cdot \sin \theta} = \frac{AB}{AD} \times \frac{AC}{AE}$.
Given $AD : AB = 3 : 5$,we have $\frac{AB}{AD} = \frac{5}{3}$.
Given $AE : AC = 2 : 3$,we have $\frac{AC}{AE} = \frac{3}{2}$.
Thus,the ratio is $\frac{5}{3} \times \frac{3}{2} = \frac{5}{2} = 2.5$.
The value $2.5$ lies in the interval $\left(2, \frac{5}{2}\right]$.
Hence,the correct option is $B$.

(A) Given $\sin x = -\frac{3}{5}$ and $\pi < x < \frac{3\pi}{2}$ (which is the third quadrant).
In the third quadrant,$\tan x$ is positive and $\cos x$ is negative.
Using $\cos^2 x = 1 - \sin^2 x$,we get $\cos^2 x = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Since $x$ is in the third quadrant,$\cos x = -\frac{4}{5}$.
Now,$\tan x = \frac{\sin x}{\cos x} = \frac{-3/5}{-4/5} = \frac{3}{4}$.
Therefore,$\tan^2 x = (\frac{3}{4})^2 = \frac{9}{16}$.
Substituting these values into the expression $80(\tan^2 x - \cos x)$:
$80(\frac{9}{16} - (-\frac{4}{5})) = 80(\frac{9}{16} + \frac{4}{5})$.
$= 80(\frac{45 + 64}{80}) = 45 + 64 = 109$.

(C) We know that $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.
Substituting these values into the expression:
$\frac{3(\frac{\sqrt{5}+1}{4}) + 5(\frac{\sqrt{5}-1}{4})}{5(\frac{\sqrt{5}+1}{4}) - 3(\frac{\sqrt{5}-1}{4})} = \frac{3\sqrt{5}+3+5\sqrt{5}-5}{5\sqrt{5}+5-3\sqrt{5}+3} = \frac{8\sqrt{5}-2}{2\sqrt{5}+8} = \frac{4\sqrt{5}-1}{\sqrt{5}+4}$.
Rationalizing the denominator:
$\frac{4\sqrt{5}-1}{\sqrt{5}+4} \times \frac{4-\sqrt{5}}{4-\sqrt{5}} = \frac{16\sqrt{5}-20-4+\sqrt{5}}{16-5} = \frac{17\sqrt{5}-24}{11}$.
Comparing this with $\frac{a\sqrt{5}-b}{c}$,we get $a=17, b=24, c=11$.
Since $\gcd(17, 11)=1$,the values are valid.
Thus,$a+b+c = 17+24+11 = 52$.