Trigonometry Questions in English

(D) Let the expression be $E = \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} + \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$.
Taking the common denominator,we get:
$E = \frac{\sqrt{1+\cos \theta} \cdot \sqrt{1+\cos \theta} + \sqrt{1-\cos \theta} \cdot \sqrt{1-\cos \theta}}{\sqrt{1-\cos \theta} \cdot \sqrt{1+\cos \theta}}$
$E = \frac{(1+\cos \theta) + (1-\cos \theta)}{\sqrt{(1-\cos \theta)(1+\cos \theta)}}$
$E = \frac{2}{\sqrt{1-\cos^2 \theta}}$
Since $1-\cos^2 \theta = \sin^2 \theta$,we have $\sqrt{\sin^2 \theta} = |\sin \theta|$.
Therefore,$E = \frac{2}{|\sin \theta|}$.

(A) Given that $\operatorname{cosec} \theta - \cot \theta = p$ (Equation $1$).
We know the trigonometric identity $\operatorname{cosec}^2 \theta - \cot^2 \theta = 1$.
This can be written as $(\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) = 1$.
Substituting the value of $p$ from Equation $1$,we get $p(\operatorname{cosec} \theta + \cot \theta) = 1$.
Therefore,$\operatorname{cosec} \theta + \cot \theta = \frac{1}{p}$ (Equation $2$).
Adding Equation $1$ and Equation $2$:
$(\operatorname{cosec} \theta - \cot \theta) + (\operatorname{cosec} \theta + \cot \theta) = p + \frac{1}{p}$.
$2 \operatorname{cosec} \theta = p + \frac{1}{p}$.
$\operatorname{cosec} \theta = \frac{1}{2} \left( p + \frac{1}{p} \right)$.

(B) The given expression is $E = \tan 1^{\circ} \tan 2^{\circ} \ldots \tan 44^{\circ} \tan 45^{\circ} \tan 46^{\circ} \ldots \tan 88^{\circ} \tan 89^{\circ}$.
We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
Therefore,$\tan 89^{\circ} = \tan(90^{\circ} - 1^{\circ}) = \cot 1^{\circ}$,$\tan 88^{\circ} = \cot 2^{\circ}$,and so on.
Substituting these values,we get $E = (\tan 1^{\circ} \cot 1^{\circ}) (\tan 2^{\circ} \cot 2^{\circ}) \ldots (\tan 44^{\circ} \cot 44^{\circ}) \tan 45^{\circ}$.
Since $\tan \theta \cdot \cot \theta = 1$ and $\tan 45^{\circ} = 1$,the expression becomes $1 \cdot 1 \cdot \ldots \cdot 1 = 1$.

(C) We know that for any real angle $\theta$,the value of $\operatorname{cosec}^{2} \theta \geq 1$.
Given $\operatorname{cosec}^{2} \theta = \frac{4xy}{(x+y)^{2}}$,therefore $\frac{4xy}{(x+y)^{2}} \geq 1$.
This implies $4xy \geq (x+y)^{2}$ (assuming $(x+y)^2 > 0$).
Rearranging the terms,we get $4xy - (x^{2} + 2xy + y^{2}) \geq 0$.
This simplifies to $-(x^{2} - 2xy + y^{2}) \geq 0$,which is $-(x-y)^{2} \geq 0$.
Multiplying by $-1$,we get $(x-y)^{2} \leq 0$.
Since the square of any real number cannot be negative,the only possibility is $(x-y)^{2} = 0$.
Therefore,$x - y = 0$,which means $x = y$.

(C) We evaluate each trigonometric term using reduction formulas:
$\sin 300^{\circ} = \sin(360^{\circ} - 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$
$\tan 240^{\circ} = \tan(180^{\circ} + 60^{\circ}) = \tan 60^{\circ} = \sqrt{3}$
$\sec(-420^{\circ}) = \sec(420^{\circ}) = \sec(360^{\circ} + 60^{\circ}) = \sec 60^{\circ} = 2$
$\cot(-315^{\circ}) = -\cot(315^{\circ}) = -\cot(360^{\circ} - 45^{\circ}) = -(-\cot 45^{\circ}) = 1$
$\cos 210^{\circ} = \cos(180^{\circ} + 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$
$\operatorname{cosec}(-315^{\circ}) = -\operatorname{cosec}(315^{\circ}) = -\operatorname{cosec}(360^{\circ} - 45^{\circ}) = -(-\operatorname{cosec} 45^{\circ}) = \sqrt{2}$
Substituting these values into the expression:
$= \frac{(-\frac{\sqrt{3}}{2}) \cdot (\sqrt{3}) \cdot (2)}{(1) \cdot (-\frac{\sqrt{3}}{2}) \cdot (\sqrt{2})} = \frac{-\sqrt{3} \cdot \sqrt{3}}{-\frac{\sqrt{3}}{2} \cdot \sqrt{2}} = \frac{-3}{-\frac{\sqrt{6}}{2}} = \frac{6}{\sqrt{6}} = \sqrt{6}$

(C) The formula for the length of an arc $l$ is given by $l = r\theta$,where $r$ is the radius and $\theta$ is the angle in radians.
Given radius $r = 6 \text{ cm}$ and angle $\theta = 18^{\circ}$.
First,convert the angle from degrees to radians: $\theta = 18^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{10} \text{ radians}$.
Now,substitute the values into the formula: $l = 6 \times \frac{\pi}{10} = \frac{6\pi}{10} = \frac{3\pi}{5} \text{ cm}$.
Thus,the correct option is $C$.

(C) Given the equation $x+\frac{1}{x}=2 \cos \theta$.
Multiplying by $x$,we get $x^2 - 2x \cos \theta + 1 = 0$.
Since $x$ is a real number,the discriminant $D$ of this quadratic equation must be greater than or equal to $0$.
$D = b^2 - 4ac = (-2 \cos \theta)^2 - 4(1)(1) = 4 \cos^2 \theta - 4$.
Setting $D \geq 0$,we have $4 \cos^2 \theta - 4 \geq 0$,which implies $\cos^2 \theta \geq 1$.
Since the range of $\cos \theta$ is $[-1, 1]$,the only possible values satisfying $\cos^2 \theta \geq 1$ are $\cos \theta = 1$ or $\cos \theta = -1$.
Thus,$\cos \theta = \pm 1$.

(C) We know that $1^{c} = \left(\frac{180}{\pi}\right)^{\circ} \approx 57.3^{\circ}$.
Therefore,$\sin 1 = \sin(57.3^{\circ})$.
Since the sine function is strictly increasing in the interval $[0, 90^{\circ}]$,and $1^{\circ} < 57.3^{\circ}$,it follows that $\sin 1^{\circ} < \sin 57.3^{\circ}$.
Thus,$\sin 1^{\circ} < \sin 1$.

(D) The value $1$ represents $1$ radian.
We know that $1 \text{ radian} = \frac{180^{\circ}}{\pi} \approx 57.3^{\circ}$.
Since $0^{\circ} < 57.3^{\circ} < 90^{\circ}$,$\tan 1$ is positive (in the first quadrant).
Now,$2 \text{ radians} = 2 \times \frac{180^{\circ}}{\pi} \approx 114.6^{\circ}$.
Since $90^{\circ} < 114.6^{\circ} < 180^{\circ}$,$\tan 2$ is negative (in the second quadrant).
Since any positive value is greater than any negative value,we have $\tan 1 > \tan 2$.

(D) We can write the expression as: $\cos ^{2} \theta+\sec ^{2} \theta = (\cos \theta - \sec \theta)^{2} + 2 \cos \theta \sec \theta$.
Since $\cos \theta \sec \theta = 1$,the expression becomes: $(\cos \theta - \sec \theta)^{2} + 2$.
Because the square of any real number is always non-negative,$(\cos \theta - \sec \theta)^{2} \ge 0$.
Therefore,$(\cos \theta - \sec \theta)^{2} + 2 \ge 2$.
Thus,the value of $\cos ^{2} \theta+\sec ^{2} \theta$ is always greater than or equal to $2$.

(A) Given $\sin \alpha = \frac{2pq}{p^2 + q^2}$.
We know that $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 + q^2)^2 - 4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 - q^2)^2}{(p^2 + q^2)^2}$.
Thus,$\cos \alpha = \frac{p^2 - q^2}{p^2 + q^2}$.
Then,$\sec \alpha = \frac{1}{\cos \alpha} = \frac{p^2 + q^2}{p^2 - q^2}$ and $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{2pq}{p^2 - q^2}$.
Now,$\sec \alpha - \tan \alpha = \frac{p^2 + q^2}{p^2 - q^2} - \frac{2pq}{p^2 - q^2} = \frac{p^2 + q^2 - 2pq}{p^2 - q^2} = \frac{(p-q)^2}{(p-q)(p+q)} = \frac{p-q}{p+q}$.

(B) Given $13 \sin A = 12 \Rightarrow \sin A = \frac{12}{13}$. Since $\frac{\pi}{2} < A < \pi$,$A$ lies in the second quadrant where $\tan A$ is negative.
Using $\cos^2 A = 1 - \sin^2 A = 1 - (\frac{12}{13})^2 = 1 - \frac{144}{169} = \frac{25}{169}$,we get $\cos A = -\frac{5}{13}$.
Thus,$\tan A = \frac{\sin A}{\cos A} = \frac{12/13}{-5/13} = -\frac{12}{5}$.
Given $5 \sec B = 13 \Rightarrow \sec B = \frac{13}{5}$. Since $\frac{3\pi}{2} < B < 2\pi$,$B$ lies in the fourth quadrant where $\tan B$ is negative.
Using $\tan^2 B = \sec^2 B - 1 = (\frac{13}{5})^2 - 1 = \frac{169}{25} - 1 = \frac{144}{25}$,we get $\tan B = -\frac{12}{5}$.
Now,substitute these values into the expression: $5 \tan A + 3 \tan^2 B = 5(-\frac{12}{5}) + 3(\frac{144}{25}) = -12 + \frac{432}{25} = \frac{-300 + 432}{25} = \frac{132}{25}$.
Note: Based on the original provided solution logic,if we assume $\tan B = -4/3$ (from a $3-4-5$ triangle),then $5(-12/5) + 3(-4/3)^2 = -12 + 3(16/9) = -12 + 16/3 = -20/3$.

(A) We know that $\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Also,$\cos 105^{\circ} = \cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$.
Adding these two values:
$\sin 105^{\circ} + \cos 105^{\circ} = \frac{\sqrt{3} + 1}{2\sqrt{2}} + \frac{1 - \sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3} + 1 + 1 - \sqrt{3}}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(B) We use the formula for the tangent of a sum of two angles: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting the given values $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$:
$\tan(A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - (\frac{1}{2})(\frac{1}{3})}$
$= \frac{\frac{3+2}{6}}{1 - \frac{1}{6}}$
$= \frac{\frac{5}{6}}{\frac{5}{6}} = 1$.
Since $\tan(A + B) = 1$,we have $A + B = \tan^{-1}(1)$.
Therefore,$A + B = \frac{\pi}{4}$.

(A) Given $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{7}{24}$.
Substituting $\tan A = \frac{4}{3}$,we get $\frac{\frac{4}{3} - \tan B}{1 + \frac{4}{3} \tan B} = \frac{7}{24}$.
Cross-multiplying: $24(\frac{4}{3} - \tan B) = 7(1 + \frac{4}{3} \tan B)$.
$32 - 24 \tan B = 7 + \frac{28}{3} \tan B$.
$25 = (24 + \frac{28}{3}) \tan B = \frac{72 + 28}{3} \tan B = \frac{100}{3} \tan B$.
$\tan B = 25 \times \frac{3}{100} = \frac{3}{4}$.
Now,we need to find $A+B$. Since $\tan A = \frac{4}{3}$ and $\tan B = \frac{3}{4}$,we have $\cot A = \frac{3}{4}$ and $\cot B = \frac{4}{3}$.
Using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{4}{3} + \frac{3}{4}}{1 - (\frac{4}{3})(\frac{3}{4})} = \frac{\frac{16+9}{12}}{1-1} = \frac{25/12}{0}$,which is undefined.
This implies $A+B = \frac{\pi}{2}$.

(D) The given expression is in the form of the trigonometric identity $\tan(A + B) = \frac{(\tan A + \tan B)}{(1 - \tan A \tan B)}$.
Here,$A = 69^{\circ}$ and $B = 66^{\circ}$.
Therefore,the expression becomes $\tan(69^{\circ} + 66^{\circ})$.
$\tan(69^{\circ} + 66^{\circ}) = \tan(135^{\circ})$.
We can write $\tan(135^{\circ})$ as $\tan(180^{\circ} - 45^{\circ})$.
Using the identity $\tan(180^{\circ} - \theta) = -\tan \theta$,we get $-\tan 45^{\circ}$.
Since $\tan 45^{\circ} = 1$,the final value is $-1$.

(A) We use the trigonometric identity $\sin^{2} A - \sin^{2} B = \sin(A+B) \sin(A-B)$.
Here,$A = 75^{\circ}$ and $B = 15^{\circ}$.
Substituting these values into the identity:
$= \sin(75^{\circ} + 15^{\circ}) \sin(75^{\circ} - 15^{\circ})$
$= \sin(90^{\circ}) \sin(60^{\circ})$
Since $\sin(90^{\circ}) = 1$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,
$= 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

(C) Given $\sin \alpha = \frac{8}{17}$. Since $0 < \alpha < 90^{\circ}$,$\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.
Given $\tan \beta = \frac{5}{12}$. Since $0 < \beta < 90^{\circ}$,we can form a right triangle with opposite side $5$ and adjacent side $12$. The hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Thus,$\sin \beta = \frac{5}{13}$ and $\cos \beta = \frac{12}{13}$.
Using the formula $\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:
$\cos (\alpha - \beta) = (\frac{15}{17})(\frac{12}{13}) + (\frac{8}{17})(\frac{5}{13})$
$= \frac{180}{221} + \frac{40}{221} = \frac{220}{221}$.

(C) Given expression: $E = \sin^{2} \theta + \sin^{2}(\theta + 60^{\circ}) + \sin^{2}(\theta - 60^{\circ})$
Using the identity $\sin^{2} A = \frac{1 - \cos 2A}{2}$:
$E = \frac{1 - \cos 2\theta}{2} + \frac{1 - \cos(2\theta + 120^{\circ})}{2} + \frac{1 - \cos(2\theta - 120^{\circ})}{2}$
$E = \frac{1}{2} [3 - (\cos 2\theta + \cos(2\theta + 120^{\circ}) + \cos(2\theta - 120^{\circ}))]$
Using $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$:
$E = \frac{1}{2} [3 - (\cos 2\theta + 2 \cos 2\theta \cos 120^{\circ})]$
Since $\cos 120^{\circ} = -\frac{1}{2}$:
$E = \frac{1}{2} [3 - (\cos 2\theta + 2 \cos 2\theta (-1/2))]$
$E = \frac{1}{2} [3 - (\cos 2\theta - \cos 2\theta)] = \frac{3}{2}$

(C) We use the formula for the tangent of a sum of two angles: $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.
Substituting the given values:
$\tan(\alpha + \beta) = \frac{\frac{m}{m+1} + \frac{1}{2m+1}}{1 - \left(\frac{m}{m+1}\right) \left(\frac{1}{2m+1}\right)}$
Simplifying the numerator: $\frac{m(2m+1) + 1(m+1)}{(m+1)(2m+1)} = \frac{2m^2 + m + m + 1}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}$.
Simplifying the denominator: $1 - \frac{m}{(m+1)(2m+1)} = \frac{(m+1)(2m+1) - m}{(m+1)(2m+1)} = \frac{2m^2 + 3m + 1 - m}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}$.
Dividing the numerator by the denominator:
$\tan(\alpha + \beta) = \frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$.
Since $\tan(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

(B) Given expression $= \frac{\cos 10^{\circ} - \sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$
Multiply the numerator and denominator by $2$:
$= \frac{2 \left( \frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ} \right)}{\sin 10^{\circ} \cos 10^{\circ}}$
Using $\sin 30^{\circ} = \frac{1}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$:
$= \frac{2 (\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$
Using the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{2 \sin(30^{\circ} - 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}} = \frac{2 \sin 20^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$
Multiply numerator and denominator by $2$ to use the identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \frac{4 \sin 20^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}} = \frac{4 \sin 20^{\circ}}{\sin 20^{\circ}} = 4$

(C) Given expression $= \sqrt{2+\sqrt{2(1+\cos 4A)}}$.
Using the trigonometric identity $1+\cos 2\theta = 2\cos^2 \theta$,we have $1+\cos 4A = 2\cos^2 2A$.
Substituting this into the expression,we get $\sqrt{2+\sqrt{2(2\cos^2 2A)}} = \sqrt{2+\sqrt{4\cos^2 2A}}$.
This simplifies to $\sqrt{2+2\cos 2A}$.
Again,using the identity $1+\cos 2\theta = 2\cos^2 \theta$,we have $2+2\cos 2A = 2(1+\cos 2A) = 2(2\cos^2 A) = 4\cos^2 A$.
Therefore,the expression becomes $\sqrt{4\cos^2 A} = 2\cos A$.

(A) Given $\tan A = \frac{1 - \cos B}{\sin B}$.
Using the trigonometric identities $1 - \cos B = 2 \sin^2 \left( \frac{B}{2} \right)$ and $\sin B = 2 \sin \left( \frac{B}{2} \right) \cos \left( \frac{B}{2} \right)$,we get:
$\tan A = \frac{2 \sin^2 \left( \frac{B}{2} \right)}{2 \sin \left( \frac{B}{2} \right) \cos \left( \frac{B}{2} \right)}$
$\tan A = \tan \left( \frac{B}{2} \right)$.
This implies $A = \frac{B}{2}$,which means $2A = B$.
Therefore,$\tan 2A = \tan B$.

(B) We know that $\cos 2\theta = \sin \left(\frac{\pi}{2} - 2\theta\right)$ and $\sin 2\theta = \cos \left(\frac{\pi}{2} - 2\theta\right)$.
Substituting these into the expression:
$\frac{\cos 2\theta}{1 - \sin 2\theta} = \frac{\sin \left(\frac{\pi}{2} - 2\theta\right)}{1 - \cos \left(\frac{\pi}{2} - 2\theta\right)}$
Using the double angle identities $\sin 2A = 2 \sin A \cos A$ and $1 - \cos 2A = 2 \sin^2 A$,where $A = \frac{\pi}{4} - \theta$:
$= \frac{2 \sin \left(\frac{\pi}{4} - \theta\right) \cos \left(\frac{\pi}{4} - \theta\right)}{2 \sin^2 \left(\frac{\pi}{4} - \theta\right)}$
$= \frac{\cos \left(\frac{\pi}{4} - \theta\right)}{\sin \left(\frac{\pi}{4} - \theta\right)}$
$= \cot \left(\frac{\pi}{4} - \theta\right)$.

(A) Given expression is $\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\cot 70^{\circ} \cot 50^{\circ}}$.
Using the identity $\cot(90^{\circ}-\theta) = \tan \theta$,we can rewrite the denominator:
$\cot 70^{\circ} = \cot(90^{\circ}-20^{\circ}) = \tan 20^{\circ}$
$\cot 50^{\circ} = \cot(90^{\circ}-40^{\circ}) = \tan 40^{\circ}$
Substituting these into the expression:
$= \frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}}$
Using the trigonometric identity $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$,where $A = 40^{\circ}$ and $B = 20^{\circ}$:
$= \tan(40^{\circ}+20^{\circ}) = \tan 60^{\circ}$
Since $\tan 60^{\circ} = \sqrt{3}$,the final value is $\sqrt{3}$.

(B) Given expression is $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$.
$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply the numerator and denominator by $2$:
$= \frac{2 [(\frac{\sqrt{3}}{2}) \cos 20^{\circ} - (\frac{1}{2}) \sin 20^{\circ}]}{\sin 20^{\circ} \cos 20^{\circ}}$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin 40^{\circ}$.
Also,$\frac{\sqrt{3}}{2} = \sin 60^{\circ}$ and $\frac{1}{2} = \cos 60^{\circ}$.
$= \frac{2 [\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}]}{\frac{1}{2} \sin 40^{\circ}}$
Using the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$.

(C) Given expression is $E = \tan 81^{\circ} + \tan 9^{\circ} - (\tan 63^{\circ} + \tan 27^{\circ})$.
Using $\tan 81^{\circ} = \cot 9^{\circ}$ and $\tan 63^{\circ} = \cot 27^{\circ}$,we get:
$E = (\cot 9^{\circ} + \tan 9^{\circ}) - (\cot 27^{\circ} + \tan 27^{\circ})$.
Since $\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$,we have:
$E = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.
Using values $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$,we get:
$E = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = \frac{8(\sqrt{5}+1) - 8(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{8\sqrt{5} + 8 - 8\sqrt{5} + 8}{5-1} = \frac{16}{4} = 4$.

(A) We know that $5x = 3x + 2x$.
Taking tangent on both sides,we get $\tan(5x) = \tan(3x + 2x)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan 5x = \frac{\tan 3x + \tan 2x}{1 - \tan 3x \tan 2x}$.
Cross-multiplying gives:
$\tan 5x(1 - \tan 3x \tan 2x) = \tan 3x + \tan 2x$.
$\tan 5x - \tan 5x \tan 3x \tan 2x = \tan 3x + \tan 2x$.
Rearranging the terms,we get:
$\tan 5x - \tan 3x - \tan 2x = \tan 5x \tan 3x \tan 2x$.

(B) We use the trigonometric identity: $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substitute the given values: $\tan (A+B) = \frac{\frac{n}{n+1} + \frac{1}{2n+1}}{1 - (\frac{n}{n+1} \times \frac{1}{2n+1})}$.
Simplify the numerator: $\frac{n(2n+1) + 1(n+1)}{(n+1)(2n+1)} = \frac{2n^2 + n + n + 1}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)}$.
Simplify the denominator: $1 - \frac{n}{(n+1)(2n+1)} = \frac{(n+1)(2n+1) - n}{(n+1)(2n+1)} = \frac{2n^2 + 3n + 1 - n}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)}$.
Dividing the numerator by the denominator: $\tan (A+B) = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)} \times \frac{(n+1)(2n+1)}{2n^2 + 2n + 1} = 1$.

(B) Given that $A$ and $B$ are positive and acute angles $(0 < A, B < 90^{\circ})$.
Since $\sin A = \frac{1}{\sqrt{10}}$,then $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Since $\sin B = \frac{1}{\sqrt{5}}$,then $\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$\sin(A + B) = \left(\frac{1}{\sqrt{10}}\right) \left(\frac{2}{\sqrt{5}}\right) + \left(\frac{3}{\sqrt{10}}\right) \left(\frac{1}{\sqrt{5}}\right)$
$= \frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\sin(A + B) = \frac{1}{\sqrt{2}}$ and $A, B$ are acute,$A + B = \frac{\pi}{4}$.

(B) We use the identities: $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$,$1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right)$,and $\sin \theta = 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)$.
Substituting these into the expression:
$= \frac{2 \sin^2 \left(\frac{\theta}{2}\right) + 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos^2 \left(\frac{\theta}{2}\right) + 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}$
$= \frac{2 \sin \left(\frac{\theta}{2}\right) [\sin \left(\frac{\theta}{2}\right) + \cos \left(\frac{\theta}{2}\right)]}{2 \cos \left(\frac{\theta}{2}\right) [\cos \left(\frac{\theta}{2}\right) + \sin \left(\frac{\theta}{2}\right)]}$
$= \frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)} = \tan \left(\frac{\theta}{2}\right)$.

(A) We use the half-angle formula: $\tan A = \frac{1 - \cos 2A}{\sin 2A}$.
Setting $A = 7 \frac{1}{2}^{\circ}$,we get $2A = 15^{\circ}$.
Thus,$\tan 7 \frac{1}{2}^{\circ} = \frac{1 - \cos 15^{\circ}}{\sin 15^{\circ}}$.
We know that $\cos 15^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ and $\sin 15^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$.
Substituting these values:
$\tan 7 \frac{1}{2}^{\circ} = \frac{1 - \frac{\sqrt{3} + 1}{2 \sqrt{2}}}{\frac{\sqrt{3} - 1}{2 \sqrt{2}}} = \frac{\frac{2 \sqrt{2} - (\sqrt{3} + 1)}{2 \sqrt{2}}}{\frac{\sqrt{3} - 1}{2 \sqrt{2}}} = \frac{2 \sqrt{2} - (1 + \sqrt{3})}{\sqrt{3} - 1}$.

(A) Given expression: $\frac{\cos 3A + \sin 3A}{\cos A - \sin A} = 1 - K \sin 2A$
Using the identities $\cos 3A = 4 \cos^3 A - 3 \cos A$ and $\sin 3A = 3 \sin A - 4 \sin^3 A$:
Numerator $= (4 \cos^3 A - 3 \cos A) + (3 \sin A - 4 \sin^3 A)$
$= 4(\cos^3 A - \sin^3 A) - 3(\cos A - \sin A)$
$= 4(\cos A - \sin A)(\cos^2 A + \sin^2 A + \sin A \cos A) - 3(\cos A - \sin A)$
$= 4(\cos A - \sin A)(1 + \sin A \cos A) - 3(\cos A - \sin A)$
Dividing by $(\cos A - \sin A)$:
$= 4(1 + \sin A \cos A) - 3$
$= 4 + 4 \sin A \cos A - 3$
$= 1 + 2(2 \sin A \cos A)$
$= 1 + 2 \sin 2A$
Comparing this with $1 - K \sin 2A$,we get $-K = 2$,so $K = -2$.

(A) We know the trigonometric identity for the tangent of a difference: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Let $A = 57^{\circ}$ and $B = 12^{\circ}$.
Then,$\tan(57^{\circ} - 12^{\circ}) = \frac{\tan 57^{\circ} - \tan 12^{\circ}}{1 + \tan 57^{\circ} \tan 12^{\circ}}$.
Since $57^{\circ} - 12^{\circ} = 45^{\circ}$,we have $\tan 45^{\circ} = 1$.
Therefore,$1 = \frac{\tan 57^{\circ} - \tan 12^{\circ}}{1 + \tan 57^{\circ} \tan 12^{\circ}}$.
Multiplying both sides by $(1 + \tan 57^{\circ} \tan 12^{\circ})$,we get:
$1 + \tan 57^{\circ} \tan 12^{\circ} = \tan 57^{\circ} - \tan 12^{\circ}$.
Rearranging the terms to match the given expression:
$\tan 57^{\circ} - \tan 12^{\circ} - \tan 57^{\circ} \tan 12^{\circ} = 1$.

(A) Given expression: $E = \sqrt{4 \sin^{4} \theta + \sin^{2} 2\theta} + 4 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$.
Using $\sin 2\theta = 2 \sin \theta \cos \theta,$ we have $\sin^{2} 2\theta = 4 \sin^{2} \theta \cos^{2} \theta$.
So,$\sqrt{4 \sin^{4} \theta + 4 \sin^{2} \theta \cos^{2} \theta} = \sqrt{4 \sin^{2} \theta (\sin^{2} \theta + \cos^{2} \theta)} = \sqrt{4 \sin^{2} \theta} = 2 |\sin \theta|$.
Using $2 \cos^{2} A = 1 + \cos 2A,$ the second term is $2 [1 + \cos(\frac{\pi}{2} - \theta)] = 2(1 + \sin \theta)$.
Thus,$E = 2 |\sin \theta| + 2 + 2 \sin \theta$.
Since $180^{\circ} < \theta < 270^{\circ},$ $\sin \theta$ is negative,so $|\sin \theta| = -\sin \theta$.
Therefore,$E = 2(-\sin \theta) + 2 + 2 \sin \theta = 2$.

(D) We use the tangent addition formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Let $A = 100^{\circ}$ and $B = 125^{\circ}$.
Then $A + B = 225^{\circ}$.
Substituting these values into the formula:
$\tan(225^{\circ}) = \frac{\tan 100^{\circ} + \tan 125^{\circ}}{1 - \tan 100^{\circ} \tan 125^{\circ}}$.
Since $\tan(225^{\circ}) = \tan(180^{\circ} + 45^{\circ}) = \tan 45^{\circ} = 1$,we have:
$1 = \frac{\tan 100^{\circ} + \tan 125^{\circ}}{1 - \tan 100^{\circ} \tan 125^{\circ}}$.
Multiplying both sides by $(1 - \tan 100^{\circ} \tan 125^{\circ})$:
$1 - \tan 100^{\circ} \tan 125^{\circ} = \tan 100^{\circ} + \tan 125^{\circ}$.
Rearranging the terms:
$\tan 100^{\circ} + \tan 125^{\circ} + \tan 100^{\circ} \tan 125^{\circ} = 1$.

(D) To simplify the expression $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}},$ we multiply the numerator and denominator inside the square root by $(1+\sin \theta).$
$= \sqrt{\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}}$
$= \sqrt{\frac{(1+\sin \theta)^2}{1-\sin^2 \theta}}$
Since $1-\sin^2 \theta = \cos^2 \theta,$
$= \sqrt{\frac{(1+\sin \theta)^2}{\cos^2 \theta}}$
$= \frac{1+\sin \theta}{\cos \theta}$
$= \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}$
$= \sec \theta + \tan \theta$

(B) Divide the numerator and denominator by $\cos 15^{\circ}$:
$\tan \theta = \frac{1 + \tan 15^{\circ}}{1 - \tan 15^{\circ}}$
Since $\tan 45^{\circ} = 1$,we can write this as:
$\tan \theta = \frac{\tan 45^{\circ} + \tan 15^{\circ}}{1 - \tan 45^{\circ} \tan 15^{\circ}}$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\tan \theta = \tan(45^{\circ} + 15^{\circ})$
$\tan \theta = \tan 60^{\circ}$
Since $60^{\circ} = \frac{\pi}{3}$ radians,we have:
$\theta = \frac{\pi}{3}$

(C) We know the formula for the tangent of a difference: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Let $A = 56^{\circ}$ and $B = 11^{\circ}$. Then $A - B = 45^{\circ}$.
Substituting these values into the formula:
$\tan 45^{\circ} = \frac{\tan 56^{\circ} - \tan 11^{\circ}}{1 + \tan 56^{\circ} \tan 11^{\circ}}$.
Since $\tan 45^{\circ} = 1$,we have:
$1 = \frac{\tan 56^{\circ} - \tan 11^{\circ}}{1 + \tan 56^{\circ} \tan 11^{\circ}}$.
Multiplying both sides by $(1 + \tan 56^{\circ} \tan 11^{\circ})$:
$1 + \tan 56^{\circ} \tan 11^{\circ} = \tan 56^{\circ} - \tan 11^{\circ}$.
Rearranging the terms to match the expression in the question:
$\tan 56^{\circ} - \tan 11^{\circ} - \tan 56^{\circ} \tan 11^{\circ} = 1$.

(C) Given that $A+B=45^{\circ}$.
Taking $\cot$ on both sides,we get $\cot(A+B)=\cot 45^{\circ}$.
Using the formula $\cot(A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}$,we have:
$\frac{\cot A \cot B-1}{\cot A+\cot B}=1$.
$\cot A \cot B-1 = \cot A+\cot B$.
Rearranging the terms,we get $\cot A \cot B - \cot A - \cot B = 1$.
Adding $1$ to both sides to factorize:
$\cot A \cot B - \cot A - \cot B + 1 = 1 + 1$.
$(\cot A-1)(\cot B-1) = 2$.
Given that $(\cot A-1)(\cot B-1)=4K$,we equate the two expressions:
$4K = 2$.
$K = \frac{2}{4} = \frac{1}{2}$.

(A) In a triangle $\Delta XYZ$,the sum of all interior angles is $180^{\circ}$.
Given that $\angle Y = 90^{\circ}$ and $\angle X = 60^{\circ}$.
Therefore,$\angle Z = 180^{\circ} - (90^{\circ} + 60^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Now,we need to find the value of $\sec Z + \frac{2}{\sqrt{3}}$.
Since $\angle Z = 30^{\circ}$,$\sec Z = \sec 30^{\circ} = \frac{1}{\cos 30^{\circ}} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$.
Substituting this value into the expression:
$\sec Z + \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

(B) In $\Delta XYZ$,$\angle Y = 90^\circ$.
Given $\cos X = \frac{3}{5}$.
By definition,$\cos X = \frac{\text{Adjacent side to } X}{\text{Hypotenuse}} = \frac{XY}{XZ} = \frac{3}{5}$.
Now,for $\angle Z$,the side $XY$ is the opposite side and $XZ$ is the hypotenuse.
Therefore,$\sin Z = \frac{\text{Opposite side to } Z}{\text{Hypotenuse}} = \frac{XY}{XZ} = \frac{3}{5}$.
We know that $\operatorname{cosec} Z = \frac{1}{\sin Z}$.
Thus,$\operatorname{cosec} Z = \frac{1}{3/5} = \frac{5}{3}$.

(B) Given expression: $\frac{\sin (y-z)+\sin (y+z)+2 \sin y}{\sin (x-z)+\sin (x+z)+2 \sin x}$
Using the trigonometric identity $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$:
Numerator: $\sin (y-z) + \sin (y+z) + 2 \sin y = 2 \sin y \cos z + 2 \sin y = 2 \sin y (\cos z + 1)$
Denominator: $\sin (x-z) + \sin (x+z) + 2 \sin x = 2 \sin x \cos z + 2 \sin x = 2 \sin x (\cos z + 1)$
Dividing the numerator by the denominator:
$= \frac{2 \sin y (\cos z + 1)}{2 \sin x (\cos z + 1)}$
$= \frac{\sin y}{\sin x}$

(D) Given: $\sec \theta(\cos \theta+\sin \theta)=\sqrt{2}$
Since $\sec \theta = \frac{1}{\cos \theta}$,we have:
$\frac{1}{\cos \theta}(\cos \theta+\sin \theta)=\sqrt{2}$
$1+\tan \theta=\sqrt{2}$
$\tan \theta=\sqrt{2}-1$ .....$(1)$
Now,we need to find the value of $\frac{2 \sin \theta}{\cos \theta-\sin \theta}$.
Dividing the numerator and the denominator by $\cos \theta$:
$\frac{2 \tan \theta}{1-\tan \theta}$
Substituting the value of $\tan \theta$ from equation $(1)$:
$= \frac{2(\sqrt{2}-1)}{1-(\sqrt{2}-1)}$
$= \frac{2(\sqrt{2}-1)}{2-\sqrt{2}}$
$= \frac{2(\sqrt{2}-1)}{\sqrt{2}(\sqrt{2}-1)}$
$= \frac{2}{\sqrt{2}} = \sqrt{2}$

(D) Given $\cos \theta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
$\sin \theta = \sqrt{1 - \left(\frac{x^2 - y^2}{x^2 + y^2}\right)^2} = \sqrt{\frac{(x^2 + y^2)^2 - (x^2 - y^2)^2}{(x^2 + y^2)^2}}$.
Using $(a+b)^2 - (a-b)^2 = 4ab$,we get $\sin \theta = \sqrt{\frac{4x^2y^2}{(x^2 + y^2)^2}} = \frac{2xy}{x^2 + y^2}$.
Now,$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{(x^2 - y^2) / (x^2 + y^2)}{(2xy) / (x^2 + y^2)}$.
$\cot \theta = \frac{x^2 - y^2}{2xy}$.

(D) We know the trigonometric identity for the tangent of a difference of two angles:
$\tan (A - B) = \frac{\sin (A - B)}{\cos (A - B)}$
Using the expansion formulas for sine and cosine of a difference:
$\sin (A - B) = \sin A \cos B - \cos A \sin B$
$\cos (A - B) = \cos A \cos B + \sin A \sin B$
Substituting these into the expression for $\tan (A - B)$:
$\tan (A - B) = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B + \sin A \sin B}$
Dividing both the numerator and the denominator by $\cos A \cos B$:
$\tan (A - B) = \frac{\frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} + \frac{\sin A \sin B}{\cos A \cos B}}$
Simplifying the terms:
$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
Thus,$x = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

(B) Given: $x = \operatorname{cosec} \theta - \sin \theta = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^{2} \theta}{\sin \theta} = \frac{\cos^{2} \theta}{\sin \theta}$.
Given: $y = \sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^{2} \theta}{\cos \theta} = \frac{\sin^{2} \theta}{\cos \theta}$.
Now,consider the expression $x^{2} y^{2} (x^{2} + y^{2} + 3)$:
$x^{2} y^{2} = \left( \frac{\cos^{4} \theta}{\sin^{2} \theta} \right) \left( \frac{\sin^{4} \theta}{\cos^{2} \theta} \right) = \cos^{2} \theta \sin^{2} \theta$.
$x^{2} + y^{2} + 3 = \frac{\cos^{4} \theta}{\sin^{2} \theta} + \frac{\sin^{4} \theta}{\cos^{2} \theta} + 3 = \frac{\cos^{6} \theta + \sin^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta}{\sin^{2} \theta \cos^{2} \theta}$.
Using the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$,let $a = \cos^{2} \theta$ and $b = \sin^{2} \theta$:
$\cos^{6} \theta + \sin^{6} \theta = (\cos^{2} \theta + \sin^{2} \theta)(\cos^{4} \theta - \cos^{2} \theta \sin^{2} \theta + \sin^{4} \theta) = 1 \cdot (\cos^{4} \theta - \cos^{2} \theta \sin^{2} \theta + \sin^{4} \theta)$.
Substituting this back: $x^{2} + y^{2} + 3 = \frac{(\cos^{4} \theta - \cos^{2} \theta \sin^{2} \theta + \sin^{4} \theta) + 3 \sin^{2} \theta \cos^{2} \theta}{\sin^{2} \theta \cos^{2} \theta} = \frac{\cos^{4} \theta + 2 \sin^{2} \theta \cos^{2} \theta + \sin^{4} \theta}{\sin^{2} \theta \cos^{2} \theta} = \frac{(\cos^{2} \theta + \sin^{2} \theta)^{2}}{\sin^{2} \theta \cos^{2} \theta} = \frac{1}{\sin^{2} \theta \cos^{2} \theta}$.
Therefore,$x^{2} y^{2} (x^{2} + y^{2} + 3) = (\cos^{2} \theta \sin^{2} \theta) \cdot \left( \frac{1}{\sin^{2} \theta \cos^{2} \theta} \right) = 1$.

(B) Let $E = 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1$.
We know that $\sin^2 \theta + \cos^2 \theta = 1$.
First,express $\sin^4 \theta + \cos^4 \theta$ as:
$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$.
Next,express $\sin^6 \theta + \cos^6 \theta$ as:
$\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta)$.
Substituting $\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$,we get:
$\sin^6 \theta + \cos^6 \theta = 1 \cdot (1 - 2\sin^2 \theta \cos^2 \theta - \sin^2 \theta \cos^2 \theta) = 1 - 3\sin^2 \theta \cos^2 \theta$.
Now,substitute these into the expression $E$:
$E = 2(1 - 3\sin^2 \theta \cos^2 \theta) - 3(1 - 2\sin^2 \theta \cos^2 \theta) + 1$.
$E = 2 - 6\sin^2 \theta \cos^2 \theta - 3 + 6\sin^2 \theta \cos^2 \theta + 1$.
$E = 2 - 3 + 1 = 0$.
Thus,the value of the expression is $0$.

(D) We know the trigonometric identity for $\cot 3A$ is given by:
$\cot 3A = \frac{\cot^3 A - 3 \cot A}{3 \cot^2 A - 1}$
Alternatively,by multiplying the numerator and denominator by $-1$,we get:
$\cot 3A = \frac{3 \cot A - \cot^3 A}{1 - 3 \cot^2 A}$
Since it is given that $\cot 3A = x$,the value of $x$ is $\frac{3 \cot A - \cot^3 A}{1 - 3 \cot^2 A}$.

(D) Given expression: $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$
Factor out $\sin A$ from the numerator and $\cos A$ from the denominator:
$= \frac{\sin A (1 - 2 \sin^2 A)}{\cos A (2 \cos^2 A - 1)}$
We know that $\sin^2 A = 1 - \cos^2 A$:
$= \tan A \cdot \frac{1 - 2(1 - \cos^2 A)}{2 \cos^2 A - 1}$
$= \tan A \cdot \frac{1 - 2 + 2 \cos^2 A}{2 \cos^2 A - 1}$
$= \tan A \cdot \frac{2 \cos^2 A - 1}{2 \cos^2 A - 1}$
$= \tan A$