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

(A) Given expression: $\cos \theta(\operatorname{cosec} \theta - \sec \theta) - \cot \theta$
$= \cos \theta \left( \frac{1}{\sin \theta} - \frac{1}{\cos \theta} \right) - \cot \theta$
$= \frac{\cos \theta}{\sin \theta} - \frac{\cos \theta}{\cos \theta} - \cot \theta$
$= \cot \theta - 1 - \cot \theta$
$= -1$

(C) The point $(-5, 12)$ has a negative $x$-coordinate and a positive $y$-coordinate,which means it lies in the second quadrant.
In the second quadrant,$\sin \theta$ and $\operatorname{cosec} \theta$ are positive,while $\cos \theta, \sec \theta, \tan \theta,$ and $\cot \theta$ are negative.
By the definition of the modulus function,for any value $x$,$|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.
Since $\sin \theta > 0$,$|\sin \theta| = \sin \theta$.
Since $\cos \theta < 0$,$|\cos \theta| = -\cos \theta$.
Since $\tan \theta < 0$,$|\tan \theta| = -\tan \theta$.
Since $\operatorname{cosec} \theta > 0$,$|\operatorname{cosec} \theta| = \operatorname{cosec} \theta$.
Comparing these with the given options,option $C$ is correct.

(C) Using the allied angle formulas:
$\sin(\pi+x) = -\sin x$
$\cos(\frac{\pi}{2}+x) = -\sin x$
$\tan(\frac{3\pi}{2}-x) = \cot x$
$\cot(2\pi-x) = -\cot x$
$\sin(2\pi-x) = -\sin x$
$\cos(2\pi+x) = \cos x$
$\operatorname{cosec}(-x) = -\operatorname{cosec} x$
$\sin(\frac{3\pi}{2}+x) = -\cos x$
Substituting these values into the expression:
$= \frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x)}$
$= \frac{-\sin^2 x \cot^2 x}{\sin x \cos x \operatorname{cosec} x \cos x}$
$= \frac{-\sin^2 x \cot^2 x}{\sin x \cos^2 x \cdot \frac{1}{\sin x}}$
$= \frac{-\sin^2 x \cot^2 x}{\cos^2 x} = -\tan^2 x \cdot \cot^2 x = -1$
Wait,let us re-evaluate the signs carefully:
Numerator: $(-\sin x)(-\sin x)(\cot x)(-\cot x) = -\sin^2 x \cot^2 x$
Denominator: $(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x) = -\sin x \cos^2 x \operatorname{cosec} x = -\cos^2 x$
Result: $\frac{-\sin^2 x \cot^2 x}{-\cos^2 x} = \frac{\sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x}}{\cos^2 x} = \frac{\cos^2 x}{\cos^2 x} = 1$
Thus,the correct option is $C$.

(D) Given that $\theta$ lies in the third quadrant,where $\tan \theta$ is positive.
We know that $\sin^2 \theta + \cos^2 \theta = 1$.
Substituting $\cos \theta = -\frac{3}{5}$,we get $\sin^2 \theta + (-\frac{3}{5})^2 = 1$.
$\sin^2 \theta + \frac{9}{25} = 1 \implies \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$.
Since $\theta$ is in the third quadrant,$\sin \theta$ must be negative,so $\sin \theta = -\frac{4}{5}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-4/5}{-3/5} = \frac{4}{3}$.
Hence,option $D$ is correct.

(B) We know that $\tanh \left(\frac{x}{2}\right) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}$.
Substituting this into the expression:
$\frac{1 + \tanh \left(\frac{x}{2}\right)}{1 - \tanh \left(\frac{x}{2}\right)} = \frac{1 + \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}{1 - \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}$
$= \frac{\frac{e^{x/2} + e^{-x/2} + e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}{\frac{e^{x/2} + e^{-x/2} - (e^{x/2} - e^{-x/2})}{e^{x/2} + e^{-x/2}}}$
$= \frac{2e^{x/2}}{2e^{-x/2}}$
$= e^{x/2 - (-x/2)} = e^x$.

(B) Given $\operatorname{cosec} \theta + \cot \theta = \frac{1}{3}$.
Since $\operatorname{cosec} \theta - \cot \theta = \frac{1}{\operatorname{cosec} \theta + \cot \theta}$,we have $\operatorname{cosec} \theta - \cot \theta = 3$.
Adding the two equations: $(\operatorname{cosec} \theta + \cot \theta) + (\operatorname{cosec} \theta - \cot \theta) = \frac{1}{3} + 3 = \frac{10}{3}$.
$2 \operatorname{cosec} \theta = \frac{10}{3}$ $\Rightarrow \operatorname{cosec} \theta = \frac{5}{3}$ $\Rightarrow \sin \theta = \frac{3}{5}$.
Subtracting the equations: $(\operatorname{cosec} \theta + \cot \theta) - (\operatorname{cosec} \theta - \cot \theta) = \frac{1}{3} - 3 = -\frac{8}{3}$.
$2 \cot \theta = -\frac{8}{3}$ $\Rightarrow \cot \theta = -\frac{4}{3}$ $\Rightarrow \cos \theta = -\frac{4}{5}$.
Since $\sin \theta > 0$ and $\cos \theta < 0$,$\theta$ lies in the $2^{\text{nd}}$ quadrant.

(C) We know that $\cos(\pi - \theta) = -\cos \theta$,so $\cos^2(\pi - \theta) = \cos^2 \theta$.
Also,$\cos(\frac{\pi}{2} - \theta) = \sin \theta$,so $\cos^2(\frac{\pi}{2} - \theta) = \sin^2 \theta$.
Given expression: $E = \cos^2(\frac{7\pi}{8}) + \cos^2(\frac{5\pi}{8}) + \cos^2(\frac{3\pi}{8}) + \cos^2(\frac{\pi}{8})$.
Note that $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$,so $\cos^2(\frac{7\pi}{8}) = \cos^2(\frac{\pi}{8})$.
Note that $\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$,so $\cos^2(\frac{5\pi}{8}) = \cos^2(\frac{3\pi}{8})$.
Thus,$E = 2\cos^2(\frac{\pi}{8}) + 2\cos^2(\frac{3\pi}{8})$.
Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,we have $\cos^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8})$.
Substituting this into $E$: $E = 2\cos^2(\frac{\pi}{8}) + 2\sin^2(\frac{\pi}{8})$.
$E = 2(\cos^2(\frac{\pi}{8}) + \sin^2(\frac{\pi}{8}))$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get $E = 2(1) = 2$.

(D) Given the equation: $\sin(270^{\circ}-x^{\circ})=\cos 292^{\circ}$.
Using the allied angle formula $\sin(270^{\circ}-\theta) = -\cos \theta$,we have:
$-\cos x^{\circ} = \cos 292^{\circ}$.
Since $\cos(180^{\circ}-\theta) = -\cos \theta$,we can write:
$-\cos x^{\circ} = \cos(180^{\circ}-112^{\circ}) = \cos 68^{\circ}$.
Alternatively,using $-\cos \theta = \cos(180^{\circ}+\theta)$:
$-\cos x^{\circ} = \cos(180^{\circ}+x^{\circ})$.
Comparing $\cos(180^{\circ}+x^{\circ}) = \cos 292^{\circ}$,we get:
$180+x = 292 \implies x = 292-180 = 112$.
Thus,$x = 112$.

(A) Given,$\theta$ lies in the first quadrant and $5 \tan \theta = 4$.
Dividing the numerator and denominator of the expression by $\cos \theta$,we get:
$\frac{5 \tan \theta - 3}{\tan \theta + 2}$
Substituting $\tan \theta = \frac{4}{5}$:
$\frac{5(\frac{4}{5}) - 3}{\frac{4}{5} + 2} = \frac{4 - 3}{\frac{4 + 10}{5}} = \frac{1}{\frac{14}{5}} = \frac{5}{14}$.

(B) We use the values of trigonometric functions for allied angles:
$\sin 120^{\circ} = \sin(180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$\cos 150^{\circ} = \cos(180^{\circ} - 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$
$\cos 240^{\circ} = \cos(180^{\circ} + 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$
$\sin 330^{\circ} = \sin(360^{\circ} - 30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2}$
Substituting these values into the expression:
$\left(\frac{\sqrt{3}}{2}\right) \left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right)$
$= -\frac{3}{4} - \frac{1}{4}$
$= -\frac{4}{4} = -1$

(C) Let $x \cos \theta = y \cos \left(\theta + \frac{2 \pi}{3}\right) = z \cos \left(\theta + \frac{4 \pi}{3}\right) = \lambda$ (where $\lambda \neq 0$).
Then,$\frac{1}{x} = \frac{\cos \theta}{\lambda}$,$\frac{1}{y} = \frac{\cos \left(\theta + \frac{2 \pi}{3}\right)}{\lambda}$,and $\frac{1}{z} = \frac{\cos \left(\theta + \frac{4 \pi}{3}\right)}{\lambda}$.
Summing these,we get $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{\lambda} \left[ \cos \theta + \cos \left(\theta + \frac{2 \pi}{3}\right) + \cos \left(\theta + \frac{4 \pi}{3}\right) \right]$.
Using the identity $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ for the first and third terms:
$\cos \theta + \cos \left(\theta + \frac{4 \pi}{3}\right) = 2 \cos \left(\theta + \frac{2 \pi}{3}\right) \cos \left(-\frac{2 \pi}{3}\right) = 2 \cos \left(\theta + \frac{2 \pi}{3}\right) \left(-\frac{1}{2}\right) = -\cos \left(\theta + \frac{2 \pi}{3}\right)$.
Substituting this back into the sum:
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{\lambda} \left[ -\cos \left(\theta + \frac{2 \pi}{3}\right) + \cos \left(\theta + \frac{2 \pi}{3}\right) \right] = \frac{1}{\lambda} (0) = 0$.

(B) Using the sum-to-product formulas: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$ and $\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$.
Applying these to the expression:
$\frac{\cos 10^{\circ} + \cos 80^{\circ}}{\sin 80^{\circ} - \sin 10^{\circ}} = \frac{2 \cos \left( \frac{80^{\circ} + 10^{\circ}}{2} \right) \cos \left( \frac{80^{\circ} - 10^{\circ}}{2} \right)}{2 \cos \left( \frac{80^{\circ} + 10^{\circ}}{2} \right) \sin \left( \frac{80^{\circ} - 10^{\circ}}{2} \right)}$
$= \frac{\cos 45^{\circ} \cos 35^{\circ}}{\cos 45^{\circ} \sin 35^{\circ}}$
$= \cot 35^{\circ} = \tan(90^{\circ} - 35^{\circ}) = \tan 55^{\circ}$.

(B) Given expression: $E = \frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt{3} \sin 250^{\circ}}$
Using allied angles: $\cos 290^{\circ} = \cos(270^{\circ} + 20^{\circ}) = \sin 20^{\circ}$ and $\sin 250^{\circ} = \sin(270^{\circ} - 20^{\circ}) = -\cos 20^{\circ}$.
So,$E = \frac{1}{\sin 20^{\circ}} - \frac{1}{\sqrt{3} \cos 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sqrt{3} \sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$E = \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\frac{\sqrt{3}}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$
$E = \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{\sqrt{3}}{2} \sin 40^{\circ}}$
$E = \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} = \frac{4}{\sqrt{3}}$

(B) Given the expression: $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$.
Using trigonometric identities: $\cos(\frac{\pi}{2}+\alpha) = -\sin \alpha$,$\sin(\frac{3\pi}{2}+\alpha) = -\cos \alpha$,$\sin(\frac{3\pi}{2}-\alpha) = -\cos \alpha$,and $\cos(\frac{3\pi}{2}+\alpha) = \sin \alpha$.
The expression becomes: $(1+\sin \alpha-\cos \alpha)^2+(1+\cos \alpha-\sin \alpha)^2 = a+b \sin ^2(\frac{\pi}{4}+\alpha)$.
Expanding both squares: $(1+\sin^2 \alpha+\cos^2 \alpha+2\sin \alpha-2\sin \alpha \cos \alpha-2\cos \alpha) + (1+\cos^2 \alpha+\sin^2 \alpha+2\cos \alpha-2\sin \alpha \cos \alpha-2\sin \alpha) = a+b \sin^2(\frac{\pi}{4}+\alpha)$.
Simplifying: $4-4\sin \alpha \cos \alpha = a+b(\sin \frac{\pi}{4} \cos \alpha+\cos \frac{\pi}{4} \sin \alpha)^2$.
$4-4\sin \alpha \cos \alpha = a+\frac{b}{2}(\cos \alpha+\sin \alpha)^2 = a+\frac{b}{2}(1+2\sin \alpha \cos \alpha) = (a+\frac{b}{2}) + b\sin \alpha \cos \alpha$.
Comparing coefficients: $b = -4$ and $a+\frac{b}{2} = 4$ $\Rightarrow a-2 = 4$ $\Rightarrow a = 6$.
Thus,$a^2+b^2 = 6^2+(-4)^2 = 36+16 = 52$.

(A) We have the expression $\operatorname{cosec} 750^{\circ} - 2 \cot 765^{\circ}$.
Using the periodicity of trigonometric functions,$\operatorname{cosec}(n \times 360^{\circ} + \theta) = \operatorname{cosec} \theta$ and $\cot(n \times 360^{\circ} + \theta) = \cot \theta$.
$\operatorname{cosec} 750^{\circ} = \operatorname{cosec}(2 \times 360^{\circ} + 30^{\circ}) = \operatorname{cosec} 30^{\circ} = 2$.
$\cot 765^{\circ} = \cot(2 \times 360^{\circ} + 45^{\circ}) = \cot 45^{\circ} = 1$.
Substituting these values into the expression:
$2 - 2(1) = 2 - 2 = 0$.
Thus,the correct option is $A$.

(B) We need to evaluate $S = \sin \frac{\pi}{5} + \sin \frac{2\pi}{5} + \sin \frac{3\pi}{5} + \sin \frac{4\pi}{5}$.
Note that $\sin \frac{4\pi}{5} = \sin(\pi - \frac{\pi}{5}) = \sin \frac{\pi}{5}$ and $\sin \frac{3\pi}{5} = \sin(\pi - \frac{2\pi}{5}) = \sin \frac{2\pi}{5}$.
So,$S = 2(\sin \frac{\pi}{5} + \sin \frac{2\pi}{5})$.
Using the values $\sin \frac{\pi}{5} = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ and $\sin \frac{2\pi}{5} = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$,
$S = 2 \left( \frac{\sqrt{10 - 2\sqrt{5}} + \sqrt{10 + 2\sqrt{5}}}{4} \right) = \frac{\sqrt{10 - 2\sqrt{5}} + \sqrt{10 + 2\sqrt{5}}}{2}$.
Squaring $S$: $S^2 = \frac{1}{4} (10 - 2\sqrt{5} + 10 + 2\sqrt{5} + 2\sqrt{(10 - 2\sqrt{5})(10 + 2\sqrt{5})}) = \frac{1}{4} (20 + 2\sqrt{100 - 20}) = \frac{1}{4} (20 + 2\sqrt{80}) = \frac{1}{4} (20 + 8\sqrt{5}) = 5 + 2\sqrt{5}$.
Thus,$S = \sqrt{5 + 2\sqrt{5}}$.

(C) We are given the expression: $\frac{\sqrt{3} \sin \theta + \cos \theta}{\sin \left(\theta + \frac{\pi}{6}\right)}$.
Using the expansion formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we expand the denominator:
$\sin \left(\theta + \frac{\pi}{6}\right) = \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6}$.
Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{6} = \frac{1}{2}$,the denominator becomes:
$\frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = \frac{1}{2} (\sqrt{3} \sin \theta + \cos \theta)$.
Substituting this back into the original expression:
$\frac{\sqrt{3} \sin \theta + \cos \theta}{\frac{1}{2} (\sqrt{3} \sin \theta + \cos \theta)} = \frac{1}{1/2} = 2$.
Thus,the correct option is $C$.

(D) We have the expression: $\tan \left(-\frac{23 \pi}{3}\right)-\cot \left(\theta-\frac{13 \pi}{3}\right)$
Since $\tan(-x) = -\tan(x)$,we have $\tan \left(-\frac{23 \pi}{3}\right) = -\tan \left(\frac{23 \pi}{3}\right) = -\tan \left(8 \pi - \frac{\pi}{3}\right) = -(-\tan \frac{\pi}{3}) = \sqrt{3}$.
For the second term,$\cot \left(\theta-\frac{13 \pi}{3}\right) = \cot \left(\theta - (4 \pi + \frac{\pi}{3})\right) = \cot \left(\theta - \frac{\pi}{3}\right) = -\cot \left(\frac{\pi}{3} - \theta\right)$.
Substituting these back into the expression: $\sqrt{3} - (-\cot \left(\frac{\pi}{3} - \theta\right)) = \sqrt{3} + \cot \left(\frac{\pi}{3} - \theta\right)$.
Thus,the correct option is $D$.

(A) Let $x = \cot \theta$. The expression becomes $2x^2 - x - 3$.
To factorize $2x^2 - x - 3$,we look for two numbers whose product is $2 \times (-3) = -6$ and whose sum is $-1$.
These numbers are $-3$ and $2$.
$2x^2 - 3x + 2x - 3 = x(2x - 3) + 1(2x - 3) = (2x - 3)(x + 1)$.
Substituting $x = \cot \theta$ back,we get $(2 \cot \theta - 3)(\cot \theta + 1)$.

(C) Given expression: $\tan x + \frac{\cos x}{1 + \sin x}$
$= \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}$
$= \frac{\sin x(1 + \sin x) + \cos^2 x}{\cos x(1 + \sin x)}$
$= \frac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1 + \sin x)}$
Since $\sin^2 x + \cos^2 x = 1$,we get:
$= \frac{\sin x + 1}{\cos x(1 + \sin x)}$
$= \frac{1}{\cos x} = \sec x$

(A) We have $\sin \left(\frac{5 \pi}{3}\right) + \sec \left(\frac{13 \pi}{3}\right)$.
First,simplify $\sin \left(\frac{5 \pi}{3}\right) = \sin \left(2 \pi - \frac{\pi}{3}\right) = -\sin \left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.
Next,simplify $\sec \left(\frac{13 \pi}{3}\right) = \sec \left(4 \pi + \frac{\pi}{3}\right) = \sec \left(\frac{\pi}{3}\right) = 2$.
Adding these values,we get $-\frac{\sqrt{3}}{2} + 2 = 2 - \frac{\sqrt{3}}{2}$.
Thus,the correct option is $A$.

(B) Given the inequality $\sqrt{3} \cos \theta + \sin \theta > 0$.
Divide the entire inequality by $2$:
$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta > 0$.
We know that $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{6} = \frac{1}{2}$.
So,$\cos \frac{\pi}{6} \cos \theta + \sin \frac{\pi}{6} \sin \theta > 0$.
Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we get:
$\cos(\theta - \frac{\pi}{6}) > 0$.
For $\cos x > 0$,$x$ must lie in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ within one period.
So,$-\frac{\pi}{2} < \theta - \frac{\pi}{6} < \frac{\pi}{2}$.
Adding $\frac{\pi}{6}$ to all parts:
$-\frac{\pi}{2} + \frac{\pi}{6} < \theta < \frac{\pi}{2} + \frac{\pi}{6}$.
$-\frac{3\pi}{6} + \frac{\pi}{6} < \theta < \frac{3\pi}{6} + \frac{\pi}{6}$.
$-\frac{2\pi}{6} < \theta < \frac{4\pi}{6}$.
$-\frac{\pi}{3} < \theta < \frac{2\pi}{3}$.
Thus,the correct option is $B$.

(C) We know that $\tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$.
Substituting $y = \log x$,we get:
$\tanh(\log x) = \frac{e^{\log x} - e^{-\log x}}{e^{\log x} + e^{-\log x}}$
Since $e^{\log x} = x$ and $e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}$,the expression becomes:
$\tanh(\log x) = \frac{x - \frac{1}{x}}{x + \frac{1}{x}}$
Multiplying the numerator and denominator by $x$,we get:
$\tanh(\log x) = \frac{x^2 - 1}{x^2 + 1}$

(D) Let $x = \operatorname{Tanh}^{-1}(\sin \theta)$.
Then $\tanh x = \sin \theta$.
We know that $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we have $\operatorname{sech}^2 x = 1 - \tanh^2 x = 1 - \sin^2 \theta = \cos^2 \theta$.
Thus,$\operatorname{sech} x = \cos \theta$,which implies $\cosh x = \sec \theta$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we get $\sinh^2 x = \cosh^2 x - 1 = \sec^2 \theta - 1 = \tan^2 \theta$.
Therefore,$\sinh x = \tan \theta$.
This implies $x = \operatorname{Sinh}^{-1}(\tan \theta)$.
However,checking the options,we use the identity $\operatorname{Tanh}^{-1}(y) = \operatorname{Sinh}^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right)$.
Substituting $y = \sin \theta$,we get $\operatorname{Sinh}^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}\right) = \operatorname{Sinh}^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \operatorname{Sinh}^{-1}(\tan \theta)$.
Wait,checking the options again,let us re-evaluate $\operatorname{Cosh}^{-1}(\sec \theta)$.
Since $\cosh x = \sec \theta$,then $x = \operatorname{Cosh}^{-1}(\sec \theta)$.
Thus,$\operatorname{Tanh}^{-1}(\sin \theta) = \operatorname{Cosh}^{-1}(\sec \theta)$.

(B) We know that,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$\cos 18^{\circ} = \sqrt{1 - \sin^2 18^{\circ}}$
$= \sqrt{1 - \left(\frac{\sqrt{5}-1}{4}\right)^2}$
$= \sqrt{1 - \frac{5+1-2\sqrt{5}}{16}}$
$= \sqrt{\frac{16 - 6 + 2\sqrt{5}}{16}}$
$= \sqrt{\frac{10 + 2\sqrt{5}}{16}}$
$= \frac{\sqrt{10 + 2\sqrt{5}}}{4}$
$= \frac{\sqrt{2(5 + \sqrt{5})}}{4} = \frac{\sqrt{2} \cdot \sqrt{5+\sqrt{5}}}{2 \cdot 2} = \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}$.

(B) Given $\tan A = \frac{-60}{11}$. Since $\tan A$ is negative,$A$ lies in the $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant. Given $A$ does not lie in the $4^{\text{th}}$ quadrant,$A$ must lie in the $2^{\text{nd}}$ quadrant.
In the $2^{\text{nd}}$ quadrant,$\operatorname{cosec} A$ is positive. Using the triangle with sides $60, 11, 61$,we get $\operatorname{cosec} A = \frac{61}{60}$.
Given $\sec B = \frac{41}{9}$. Since $\sec B$ is positive,$B$ lies in the $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. Given $B$ does not lie in the $1^{\text{st}}$ quadrant,$B$ must lie in the $4^{\text{th}}$ quadrant.
In the $4^{\text{th}}$ quadrant,$\cot B$ is negative. Using the triangle with sides $40, 9, 41$,we get $\cot B = -\frac{9}{40}$.
Now,$K = \operatorname{cosec} A + \cot B = \frac{61}{60} - \frac{9}{40}$.
Finding a common denominator $(120)$: $K = \frac{122 - 27}{120} = \frac{95}{120} = \frac{19}{24}$.
Therefore,$24K = 24 \times \frac{19}{24} = 19$.

(B) Given $\cot \theta = -\frac{2}{3}$.
Since $\cot \theta < 0$ and $\theta$ is not in the $4^{\text{th}}$ quadrant,$\theta$ must lie in the $2^{\text{nd}}$ quadrant.
In the $2^{\text{nd}}$ quadrant,$\sin \theta > 0$ and $\cos \theta < 0$.
Using $\cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{2}{3}$,let $\cos \theta = -2k$ and $\sin \theta = 3k$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $(3k)^2 + (-2k)^2 = 1 \implies 13k^2 = 1 \implies k = \frac{1}{\sqrt{13}}$.
Thus,$\sin \theta = \frac{3}{\sqrt{13}}$,$\cos \theta = -\frac{2}{\sqrt{13}}$,and $\tan \theta = -\frac{3}{2}$.
Substituting these values into the expression:
$\frac{(5 \sin \theta + \cos \theta)^2}{\tan \theta + \cot \theta} = \frac{(\frac{15}{\sqrt{13}} - \frac{2}{\sqrt{13}})^2}{-\frac{3}{2} - \frac{2}{3}} = \frac{(\frac{13}{\sqrt{13}})^2}{-\frac{9+4}{6}} = \frac{13}{-\frac{13}{6}} = -6$.

(B) Given expression: $\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}$
$= \frac{1}{\sin(270^{\circ}-20^{\circ})} + \frac{\sqrt{3}}{\cos(270^{\circ}+20^{\circ})}$
$= -\frac{1}{\cos 20^{\circ}} + \frac{\sqrt{3}}{\sin 20^{\circ}}$
$= \frac{-\sin 20^{\circ} + \sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ} \sin 20^{\circ}}$
$= \frac{2 \left( -\frac{1}{2} \sin 20^{\circ} + \frac{\sqrt{3}}{2} \cos 20^{\circ} \right)}{\frac{1}{2} \sin 40^{\circ}}$
$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$
$= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$
$= \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$
Therefore,option $B$ is correct.

(B) Given $\sin A = \frac{-7}{25}$. Since $\sin A < 0$ and $A$ is not in the $3^{\text{rd}}$ quadrant,$A$ must be in the $4^{\text{th}}$ quadrant.
In the $4^{\text{th}}$ quadrant,$\cos A > 0$. Thus,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{-7}{25})^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}$.
Therefore,$\tan A = \frac{\sin A}{\cos A} = \frac{-7/25}{24/25} = \frac{-7}{24}$.
Given $\cos B = \frac{8}{17}$. Since $\cos B > 0$ and $B$ is not in the $1^{\text{st}}$ quadrant,$B$ must be in the $4^{\text{th}}$ quadrant.
In the $4^{\text{th}}$ quadrant,$\sin B < 0$. Thus,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{8}{17})^2} = -\sqrt{1 - \frac{64}{289}} = -\sqrt{\frac{225}{289}} = \frac{-15}{17}$.
Therefore,$\cot B = \frac{\cos B}{\sin B} = \frac{8/17}{-15/17} = \frac{-8}{15}$.
Now,$8 \tan A - 5 \cot B = 8(\frac{-7}{24}) - 5(\frac{-8}{15}) = \frac{-7}{3} + \frac{8}{3} = \frac{1}{3}$.

(B) $\triangle ACB$ is a right-angled triangle at $C$.
By the Pythagoras theorem,$AB^2 = AC^2 + BC^2$.
$AC^2 = AB^2 - BC^2 = 29^2 - 21^2 = 841 - 441 = 400$.
Thus,$AC = \sqrt{400} = 20$ units.
$\cos \theta = \frac{BC}{AB} = \frac{21}{29}$ and $\sin \theta = \frac{AC}{AB} = \frac{20}{29}$.
$\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2 = \frac{441 - 400}{841} = \frac{41}{841}$.

(D) We are given that $\operatorname{cosec} \theta - \cot \theta = 2017$ $(i)$.
Using the identity $\operatorname{cosec}^2 \theta - \cot^2 \theta = 1$,we can write $(\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) = 1$.
Therefore,$\operatorname{cosec} \theta + \cot \theta = \frac{1}{2017}$ (ii).
Adding $(i)$ and (ii):
$2 \operatorname{cosec} \theta = 2017 + \frac{1}{2017} > 0$,which implies $\operatorname{cosec} \theta > 0$.
Since $\operatorname{cosec} \theta > 0$,$\theta$ must lie in the $I$ or $II$ quadrant.
Subtracting $(i)$ from (ii):
$2 \cot \theta = \frac{1}{2017} - 2017 < 0$,which implies $\cot \theta < 0$.
Since $\cot \theta < 0$,$\theta$ must lie in the $II$ or $IV$ quadrant.
Since $\theta$ must satisfy both conditions ($\operatorname{cosec} \theta > 0$ and $\cot \theta < 0$),$\theta$ lies in the $II$ quadrant.

(D) We know that $\tan(90^\circ - \theta) = \cot \theta$.
Given expression is $E = \tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 44^\circ \tan 45^\circ \tan 46^\circ \dots \tan 88^\circ \tan 89^\circ$.
We can pair the terms as:
$(\tan 1^\circ \tan 89^\circ) \times (\tan 2^\circ \tan 88^\circ) \times \dots \times (\tan 44^\circ \tan 46^\circ) \times \tan 45^\circ$.
Since $\tan 89^\circ = \tan(90^\circ - 1^\circ) = \cot 1^\circ$,we have $\tan 1^\circ \tan 89^\circ = \tan 1^\circ \cot 1^\circ = 1$.
Similarly,$\tan k^\circ \tan(90^\circ - k^\circ) = 1$ for all $k$ from $1$ to $44$.
Thus,$E = 1 \times 1 \times \dots \times 1 \times \tan 45^\circ = 1 \times 1 = 1$.

(B) Given expression: $\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^2+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)^2-2 \cos 30^{\circ}$
Since $\cos 55^{\circ} = \cos(90^{\circ}-35^{\circ}) = \sin 35^{\circ}$,we substitute this into the expression:
$= \left(\frac{\sin 35^{\circ}}{\sin 35^{\circ}}\right)^2 + \left(\frac{\sin 35^{\circ}}{\sin 35^{\circ}}\right)^2 - 2 \cos 30^{\circ}$
$= (1)^2 + (1)^2 - 2 \left(\frac{\sqrt{3}}{2}\right)$
$= 1 + 1 - \sqrt{3}$
$= 2 - \sqrt{3}$

(B) Using the formula $\cos C - \cos D = -2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$,we have:
$\cos 36^{\circ} - \cos 72^{\circ} = -2 \sin \left(\frac{36^{\circ}+72^{\circ}}{2}\right) \sin \left(\frac{36^{\circ}-72^{\circ}}{2}\right)$
$= -2 \sin 54^{\circ} \sin (-18^{\circ})$
$= 2 \sin 54^{\circ} \sin 18^{\circ}$
Substituting the values $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:
$= 2 \times \left(\frac{\sqrt{5}+1}{4}\right) \times \left(\frac{\sqrt{5}-1}{4}\right)$
$= 2 \times \frac{(\sqrt{5})^2 - 1^2}{16} = 2 \times \frac{5-1}{16} = 2 \times \frac{4}{16} = 2 \times \frac{1}{4} = \frac{1}{2}$

(D) Given expression: $E = \sin ^4 \frac{\pi}{8} + \cos ^4 \frac{3 \pi}{8} - \sin ^4 \frac{3 \pi}{8} + \sin ^4 \frac{5 \pi}{8} + \cos ^4 \frac{7 \pi}{8} - \sin ^4 \frac{7 \pi}{8}$
Using $\sin(\pi - \theta) = \sin \theta$ and $\cos(\pi - \theta) = -\cos \theta$:
$\sin^4 \frac{5 \pi}{8} = \sin^4(\pi - \frac{3 \pi}{8}) = \sin^4 \frac{3 \pi}{8}$
$\cos^4 \frac{7 \pi}{8} = \cos^4(\pi - \frac{\pi}{8}) = (-\cos \frac{\pi}{8})^4 = \cos^4 \frac{\pi}{8}$
$\sin^4 \frac{7 \pi}{8} = \sin^4(\pi - \frac{\pi}{8}) = \sin^4 \frac{\pi}{8}$
Substituting these into the expression:
$E = \sin ^4 \frac{\pi}{8} + \cos ^4 \frac{3 \pi}{8} - \sin ^4 \frac{3 \pi}{8} + \sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{\pi}{8} - \sin ^4 \frac{\pi}{8}$
$E = \cos ^4 \frac{3 \pi}{8} + \cos ^4 \frac{\pi}{8}$
Using $\cos^2 \theta = \frac{1 + \cos 2 \theta}{2}$:
$E = \left(\frac{1 + \cos \frac{3 \pi}{4}}{2}\right)^2 + \left(\frac{1 + \cos \frac{\pi}{4}}{2}\right)^2$
$E = \frac{1}{4} \left[ (1 - \frac{1}{\sqrt{2}})^2 + (1 + \frac{1}{\sqrt{2}})^2 \right]$
$E = \frac{1}{4} \left[ (1 + \frac{1}{2} - \sqrt{2}) + (1 + \frac{1}{2} + \sqrt{2}) \right]$
$E = \frac{1}{4} [3] = \frac{3}{4}$

(C) Given $\sin A = -\frac{60}{61}$. Since $A$ is not in the $4^{\text{th}}$ quadrant and $\sin A < 0$,$A$ must be in the $3^{\text{rd}}$ quadrant. In the $3^{\text{rd}}$ quadrant,$\cot A > 0$. $\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (-\frac{60}{61})^2} = -\sqrt{\frac{3721 - 3600}{3721}} = -\frac{11}{61}$. Thus,$\cot A = \frac{\cos A}{\sin A} = \frac{-11/61}{-60/61} = \frac{11}{60}$.
Given $\cot B = -\frac{40}{9}$. Since $B$ is not in the $4^{\text{th}}$ quadrant and $\cot B < 0$,$B$ must be in the $2^{\text{nd}}$ quadrant. In the $2^{\text{nd}}$ quadrant,$\sec B < 0$. $\sec B = -\sqrt{1 + \tan^2 B} = -\sqrt{1 + (-\frac{9}{40})^2} = -\sqrt{1 + \frac{81}{1600}} = -\sqrt{\frac{1681}{1600}} = -\frac{41}{40}$.
Now,$6 \cot A + 4 \sec B = 6(\frac{11}{60}) + 4(-\frac{41}{40}) = \frac{11}{10} - \frac{41}{10} = -\frac{30}{10} = -3$.

(D) Given $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$.
$e^y = \tan \left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}$.
We know that $\tanh \left(\frac{y}{2}\right) = \frac{e^y - 1}{e^y + 1}$.
Substituting the value of $e^y$:
$\tanh \left(\frac{y}{2}\right) = \frac{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} - 1}{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} + 1} = \frac{(1 + \tan \frac{x}{2}) - (1 - \tan \frac{x}{2})}{(1 + \tan \frac{x}{2}) + (1 - \tan \frac{x}{2})} = \frac{2 \tan \frac{x}{2}}{2} = \tan \frac{x}{2}$.
Thus,option $D$ is correct.

(B) Given $\cosh x = \frac{\sqrt{14}}{3}$.
Using the identity $\sinh^2 x = \cosh^2 x - 1$,we get:
$\sinh^2 x = \left(\frac{\sqrt{14}}{3}\right)^2 - 1 = \frac{14}{9} - 1 = \frac{5}{9}$.
Since $\sinh x = \cos \theta$,we have $\cos \theta = \frac{\sqrt{5}}{3}$ (taking the positive root as $\sinh x$ is positive for $x > 0$ and $\cos \theta$ is positive in the given range).
Now,$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{5}{9} = \frac{4}{9}$.
Thus,$\sin \theta = \pm \frac{2}{3}$.
Given $-\pi < \theta < -\frac{\pi}{2}$,$\theta$ lies in the second quadrant (or equivalent to the range $(-\pi, -\pi/2)$ which corresponds to the second quadrant in the standard circle).
In the second quadrant,$\sin \theta$ is positive.
However,checking the range $-\pi < \theta < -\frac{\pi}{2}$,this corresponds to the interval where $\sin \theta$ is positive.
Wait,re-evaluating the quadrant: $-\pi < \theta < -\frac{\pi}{2}$ is the second quadrant where $\sin \theta > 0$.
Therefore,$\sin \theta = \frac{2}{3}$.

(B) Given,length of the arc $l = 44 \text{ cm}$.
Radius of the circle $r = 12 \text{ cm}$.
The angle $\theta$ subtended by the arc at the centre in radians is given by $\theta = \frac{l}{r}$.
$\theta = \frac{44}{12} = \frac{11}{3} \text{ radians}$.
To convert the angle from radians to degrees,we multiply by $\frac{180}{\pi}$.
$\theta_{\text{deg}} = \frac{11}{3} \times \frac{180}{\pi} = \frac{11 \times 60}{\pi} = \left(\frac{660}{\pi}\right)^{\circ}$.

(B) We know that $\tanh \theta = \frac{\sinh \theta}{\cosh \theta}$.
Substituting $\theta = \frac{x}{2}$,we have $\tanh \frac{x}{2} = \frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}$.
Substituting this into the expression:
$\frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}} = \frac{1+\frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}}{1-\frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}}$
$= \frac{\cosh \frac{x}{2} + \sinh \frac{x}{2}}{\cosh \frac{x}{2} - \sinh \frac{x}{2}}$
Using the definitions $\cosh \theta = \frac{e^{\theta} + e^{-\theta}}{2}$ and $\sinh \theta = \frac{e^{\theta} - e^{-\theta}}{2}$,we get:
$\cosh \frac{x}{2} + \sinh \frac{x}{2} = e^{x/2}$
$\cosh \frac{x}{2} - \sinh \frac{x}{2} = e^{-x/2}$
Therefore,the expression becomes $\frac{e^{x/2}}{e^{-x/2}} = e^{x/2 - (-x/2)} = e^x$.

(A) Given expression: $(1+\cos \frac{\pi}{6})(1+\cos \frac{\pi}{3})(1+\cos \frac{2\pi}{3})(1+\cos \frac{7\pi}{6})$
Using the values of trigonometric functions:
$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,$\cos \frac{\pi}{3} = \frac{1}{2}$,$\cos \frac{2\pi}{3} = -\frac{1}{2}$,$\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$
Substituting these values:
$= (1 + \frac{\sqrt{3}}{2})(1 + \frac{1}{2})(1 - \frac{1}{2})(1 - \frac{\sqrt{3}}{2})$
Group the terms using the identity $(a+b)(a-b) = a^2 - b^2$:
$= [(1 + \frac{\sqrt{3}}{2})(1 - \frac{\sqrt{3}}{2})] \times [(1 + \frac{1}{2})(1 - \frac{1}{2})]$
$= (1^2 - (\frac{\sqrt{3}}{2})^2) \times (1^2 - (\frac{1}{2})^2)$
$= (1 - \frac{3}{4}) \times (1 - \frac{1}{4})$
$= \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$

(C) Given $\frac{\cos A}{3} = \frac{1}{5} \implies \cos A = \frac{3}{5}$.
Since $-\frac{\pi}{2} < A < 0$,$A$ lies in the fourth quadrant,so $\sin A$ is negative.
$\sin A = -\sqrt{1 - \cos^2 A} = -\sqrt{1 - (\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.
Given $\frac{\cos B}{4} = \frac{1}{5} \implies \cos B = \frac{4}{5}$.
Since $-\frac{\pi}{2} < B < 0$,$B$ lies in the fourth quadrant,so $\sin B$ is negative.
$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{4}{5})^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.
Now,calculate $2 \sin A + 4 \sin B = 2(-\frac{4}{5}) + 4(-\frac{3}{5}) = -\frac{8}{5} - \frac{12}{5} = -\frac{20}{5} = -4$.

(B) We know that $\cot \theta = \tan(90^{\circ} - \theta)$.
Using this identity:
$\cot 54^{\circ} = \tan(90^{\circ} - 54^{\circ}) = \tan 36^{\circ}$.
$\cot 70^{\circ} = \tan(90^{\circ} - 70^{\circ}) = \tan 20^{\circ}$.
Substituting these values into the expression:
$\frac{\tan 36^{\circ}}{\tan 36^{\circ}} + \frac{\tan 20^{\circ}}{\tan 20^{\circ}} = 1 + 1 = 2$.

(D) We know that $\cos 55^{\circ} = \sin(90^{\circ} - 55^{\circ}) = \sin 35^{\circ}$.
Substituting this into the expression,we get $\frac{\sin 55^{\circ} - \sin 35^{\circ}}{\sin 10^{\circ}}$.
Using the formula $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$,we have:
$\sin 55^{\circ} - \sin 35^{\circ} = 2 \cos\left(\frac{55^{\circ} + 35^{\circ}}{2}\right) \sin\left(\frac{55^{\circ} - 35^{\circ}}{2}\right) = 2 \cos 45^{\circ} \sin 10^{\circ}$.
Now,the expression becomes $\frac{2 \cos 45^{\circ} \sin 10^{\circ}}{\sin 10^{\circ}} = 2 \cos 45^{\circ}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the value is $2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.