Trigonometrical ratios of sum and difference of two and three angles Questions in English

(D) Given $\cot \alpha = \frac{1}{2}$,since $\alpha \in (\pi, \frac{3\pi}{2})$ (third quadrant),$\tan \alpha = \frac{1}{\cot \alpha} = 2$.
Given $\sec \beta = -\frac{5}{3}$,since $\beta \in (\frac{\pi}{2}, \pi)$ (second quadrant),$\tan^2 \beta = \sec^2 \beta - 1 = (-\frac{5}{3})^2 - 1 = \frac{25}{9} - 1 = \frac{16}{9}$.
Since $\beta$ is in the second quadrant,$\tan \beta$ must be negative,so $\tan \beta = -\frac{4}{3}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$:
$\tan(\alpha + \beta) = \frac{2 + (-\frac{4}{3})}{1 - (2)(-\frac{4}{3})} = \frac{\frac{6-4}{3}}{1 + \frac{8}{3}} = \frac{\frac{2}{3}}{\frac{11}{3}} = \frac{2}{11}$.

(D) Using the identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$:
$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{\circ} - 12^{\circ})$
$= \cos 60^{\circ} \cos 36^{\circ}$
Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = 1 - 2\sin^2 18^{\circ}$:
$= \frac{1}{2} (1 - 2\sin^2 18^{\circ})$
$= \frac{1}{2} \left( 1 - 2 \left( \frac{\sqrt{5}-1}{4} \right)^2 \right)$
$= \frac{1}{2} \left( 1 - 2 \left( \frac{5 + 1 - 2\sqrt{5}}{16} \right) \right)$
$= \frac{1}{2} \left( 1 - \frac{6 - 2\sqrt{5}}{8} \right)$
$= \frac{1}{2} \left( \frac{8 - 6 + 2\sqrt{5}}{8} \right)$
$= \frac{1}{2} \left( \frac{2 + 2\sqrt{5}}{8} \right) = \frac{1 + \sqrt{5}}{8}$

(C) Expanding the squares: $(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2 = (\cos^2 \alpha + \cos^2 \beta + 2\cos \alpha \cos \beta) + (\sin^2 \alpha + \sin^2 \beta + 2\sin \alpha \sin \beta)$
Grouping terms: $(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)$
Using identities $\cos^2 \theta + \sin^2 \theta = 1$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$: $1 + 1 + 2\cos(\alpha - \beta)$
Simplifying: $2 + 2\cos(\alpha - \beta) = 2(1 + \cos(\alpha - \beta))$
Using the identity $1 + \cos \theta = 2\cos^2(\theta/2)$: $2 \times 2\cos^2\left(\frac{\alpha-\beta}{2}\right) = 4\cos^2\left(\frac{\alpha-\beta}{2}\right)$

(D) We use the expansion formulas: $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
$\sin \left(\frac{\pi}{3}+x\right) = \sin \frac{\pi}{3} \cos x + \cos \frac{\pi}{3} \sin x = \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$.
$\cos \left(\frac{\pi}{6}+x\right) = \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x = \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x$.
Subtracting the two expressions:
$\left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) - \left(\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x\right)$
$= \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$
$= \sin x$.

(D) Given $\sin \theta = \frac{-12}{13}$ and $\theta$ is in the third quadrant,$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - (\frac{-12}{13})^2} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-12/13}{-5/13} = \frac{12}{5}$.
Given $\cos \phi = \frac{-4}{5}$ and $\phi$ is in the third quadrant,$\sin \phi = -\sqrt{1 - \cos^2 \phi} = -\sqrt{1 - (\frac{-4}{5})^2} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.
Thus,$\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{-3/5}{-4/5} = \frac{3}{4}$.
Using the formula $\tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi}$:
$\tan(\theta - \phi) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{(48 - 15)/20}{1 + 36/20} = \frac{33/20}{56/20} = \frac{33}{56}$.

(D) We use the formula for the tangent of the sum of two angles: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting the given values: $\tan(A+B) = \frac{\frac{5}{6} + \frac{1}{11}}{1 - (\frac{5}{6} \times \frac{1}{11})}$.
Simplifying the numerator: $\frac{5}{6} + \frac{1}{11} = \frac{55 + 6}{66} = \frac{61}{66}$.
Simplifying the denominator: $1 - \frac{5}{66} = \frac{66 - 5}{66} = \frac{61}{66}$.
Thus,$\tan(A+B) = \frac{61/66}{61/66} = 1$.
Since $\tan(A+B) = 1$,we have $A+B = \tan^{-1}(1) = \frac{\pi}{4}$.

(A) We use the trigonometric identities: $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$.
$\cos \left(\frac{3 \pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right)$
$= \left(\cos \frac{3 \pi}{4} \cos x - \sin \frac{3 \pi}{4} \sin x\right) - \left(\sin \frac{\pi}{4} \cos x - \cos \frac{\pi}{4} \sin x\right)$
Since $\cos \frac{3 \pi}{4} = -\frac{1}{\sqrt{2}}$,$\sin \frac{3 \pi}{4} = \frac{1}{\sqrt{2}}$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$= \left(-\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x\right) - \left(\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x\right)$
$= -\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x$
$= -\frac{2}{\sqrt{2}} \cos x = -\sqrt{2} \cos x$.

(A) Using the identity $\cos(x)\cos(y) + \sin(x)\sin(y) = \cos(x-y)$ and the co-function identity $\cos(90^{\circ}-\theta) = \sin(\theta)$:
Given expression: $\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \cos(54^{\circ}+A) \cos(54^{\circ}-A)$
Since $54^{\circ}+A = 90^{\circ}-(36^{\circ}-A)$ and $54^{\circ}-A = 90^{\circ}-(36^{\circ}+A)$,we have:
$\cos(54^{\circ}+A) = \sin(36^{\circ}-A)$ and $\cos(54^{\circ}-A) = \sin(36^{\circ}+A)$
Substituting these into the expression:
$\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \sin(36^{\circ}-A) \sin(36^{\circ}+A)$
This is in the form $\cos(x)\cos(y) + \sin(x)\sin(y)$ where $x = 36^{\circ}-A$ and $y = 36^{\circ}+A$.
$= \cos((36^{\circ}-A) - (36^{\circ}+A))$
$= \cos(36^{\circ}-A-36^{\circ}-A)$
$= \cos(-2A)$
Since $\cos(-\theta) = \cos(\theta)$,the result is $\cos(2A)$.

(A) Given $\sin \theta = \sin 15^{\circ} + \sin 45^{\circ}$.
Using the formula $\sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$:
$\sin \theta = 2 \sin \left( \frac{15^{\circ} + 45^{\circ}}{2} \right) \cos \left( \frac{15^{\circ} - 45^{\circ}}{2} \right)$
$\sin \theta = 2 \sin 30^{\circ} \cos(-15^{\circ})$
Since $\cos(-x) = \cos x$ and $\sin 30^{\circ} = \frac{1}{2}$:
$\sin \theta = 2 \times \frac{1}{2} \times \cos 15^{\circ}$
$\sin \theta = \cos 15^{\circ}$
Since $\cos 15^{\circ} = \sin(90^{\circ} - 15^{\circ}) = \sin 75^{\circ}$:
$\sin \theta = \sin 75^{\circ}$
Therefore,$\theta = 75^{\circ}$.

(C) Given expression: $\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A}$
Group the terms with $A$ and $13A$:
$= \frac{(\sin 13 A+\sin A)+\sin 7 A}{(\cos 13 A+\cos A)+\cos 7 A}$
Using sum-to-product formulas $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$ and $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$= \frac{2 \sin(\frac{13A+A}{2}) \cos(\frac{13A-A}{2}) + \sin 7 A}{2 \cos(\frac{13A+A}{2}) \cos(\frac{13A-A}{2}) + \cos 7 A}$
$= \frac{2 \sin 7 A \cos 6 A + \sin 7 A}{2 \cos 7 A \cos 6 A + \cos 7 A}$
Factor out $\sin 7 A$ from the numerator and $\cos 7 A$ from the denominator:
$= \frac{\sin 7 A(2 \cos 6 A + 1)}{\cos 7 A(2 \cos 6 A + 1)}$
$= \frac{\sin 7 A}{\cos 7 A} = \tan 7 A$

(D) Given the equation: $2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$
Using the expansion formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$2 \left(\sin \theta \cos \frac{\pi}{3} + \cos \theta \sin \frac{\pi}{3}\right) = \cos \theta \cos \frac{\pi}{6} + \sin \theta \sin \frac{\pi}{6}$
Substitute the values $\cos \frac{\pi}{3} = \frac{1}{2}$,$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,and $\sin \frac{\pi}{6} = \frac{1}{2}$:
$2 \left(\sin \theta \cdot \frac{1}{2} + \cos \theta \cdot \frac{\sqrt{3}}{2}\right) = \cos \theta \cdot \frac{\sqrt{3}}{2} + \sin \theta \cdot \frac{1}{2}$
$\sin \theta + \sqrt{3} \cos \theta = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta$
Multiply by $2$ to clear the fractions:
$2 \sin \theta + 2 \sqrt{3} \cos \theta = \sqrt{3} \cos \theta + \sin \theta$
Rearrange the terms:
$2 \sin \theta - \sin \theta = \sqrt{3} \cos \theta - 2 \sqrt{3} \cos \theta$
$\sin \theta = -\sqrt{3} \cos \theta$
Divide by $\cos \theta$:
$\tan \theta = -\sqrt{3}$

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have:
$\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$
$\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$
Subtracting these:
$(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$
Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes:
$\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 \times 18^{\circ}) = \cos 36^{\circ}$
Alternatively,using $\cos(90^{\circ}-\theta) = \sin \theta$:
$\cos(72^{\circ}-A) = \sin(18^{\circ}+A)$ and $\cos(72^{\circ}+A) = \sin(18^{\circ}-A)$
Expression $= \cos(18^{\circ}-A)\cos(18^{\circ}+A) - \sin(18^{\circ}+A)\sin(18^{\circ}-A)$
$= \cos((18^{\circ}-A) + (18^{\circ}+A)) = \cos 36^{\circ}$

(B) In any triangle $ABC$,the sum of angles is $A + B + C = 180^{\circ}$.
Given $A = \tan^{-1} 2$ and $B = \tan^{-1} 3$.
We know that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting the values,$\tan(A + B) = \frac{2 + 3}{1 - (2)(3)} = \frac{5}{1 - 6} = \frac{5}{-5} = -1$.
Since $A$ and $B$ are angles of a triangle and $\tan A = 2, \tan B = 3$ (both positive),$A$ and $B$ are acute angles.
Thus,$A + B$ must be in the second quadrant because $\tan(A + B) = -1$.
Therefore,$A + B = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
Since $A + B + C = 180^{\circ}$,we have $135^{\circ} + C = 180^{\circ}$.
Hence,$C = 180^{\circ} - 135^{\circ} = 45^{\circ}$.

(D) Given expression is $\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}$.
We use the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$.
Multiplying and dividing by $2$,we get:
$\frac{1}{2} [2 \sin \frac{5 \pi}{12} \sin \frac{\pi}{12}]$
$= \frac{1}{2} [\cos(\frac{5 \pi}{12} - \frac{\pi}{12}) - \cos(\frac{5 \pi}{12} + \frac{\pi}{12})]$
$= \frac{1}{2} [\cos(\frac{4 \pi}{12}) - \cos(\frac{6 \pi}{12})]$
$= \frac{1}{2} [\cos(\frac{\pi}{3}) - \cos(\frac{\pi}{2})]$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{\pi}{2}) = 0$,
$= \frac{1}{2} [\frac{1}{2} - 0] = \frac{1}{4}$.

(D) Given equations are:
$\sin x + \sin y = \frac{1}{2} \quad (1)$
$\cos x + \cos y = 1 \quad (2)$
Using sum-to-product formulas:
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{1}{2} \quad (3)$
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = 1 \quad (4)$
Dividing equation $(3)$ by $(4)$:
$\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{1/2}{1}$
$\tan \left(\frac{x+y}{2}\right) = \frac{1}{2}$
Using the double angle formula $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$ where $\theta = \frac{x+y}{2}$:
$\tan(x+y) = \frac{2 \tan \left(\frac{x+y}{2}\right)}{1 - \tan^2 \left(\frac{x+y}{2}\right)}$
Substituting $\tan \left(\frac{x+y}{2}\right) = \frac{1}{2}$:
$\tan(x+y) = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$

(B) We use the trigonometric identity: $\cos ^{2} A - \sin ^{2} B = \cos(A+B) \cdot \cos(A-B)$.
Here,$A = 45^{\circ}$ and $B = 15^{\circ}$.
Substituting these values into the identity:
$\cos ^{2} 45^{\circ} - \sin ^{2} 15^{\circ} = \cos(45^{\circ} + 15^{\circ}) \cdot \cos(45^{\circ} - 15^{\circ})$
$= \cos(60^{\circ}) \cdot \cos(30^{\circ})$
$= \frac{1}{2} \times \frac{\sqrt{3}}{2}$
$= \frac{\sqrt{3}}{4}$.

(C) Given $\tan \alpha = \frac{1}{7}$. Since $\alpha$ is in the $3^{\text{rd}}$ quadrant,$\sin 2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac{2/7}{1 + 1/49} = \frac{14/7}{50/49} = \frac{14}{50} = \frac{7}{25}$.
$\cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{1 - 1/49}{1 + 1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25}$.
Given $\sin \beta = \frac{1}{\sqrt{10}}$. Since $\beta$ is in the $2^{\text{nd}}$ quadrant,$\cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - 1/10} = -\sqrt{9/10} = -\frac{3}{\sqrt{10}}$.
Using the formula $\sin(2\alpha + \beta) = \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta$:
$\sin(2\alpha + \beta) = \left(\frac{7}{25}\right)\left(-\frac{3}{\sqrt{10}}\right) + \left(\frac{24}{25}\right)\left(\frac{1}{\sqrt{10}}\right)$
$= -\frac{21}{25\sqrt{10}} + \frac{24}{25\sqrt{10}} = \frac{3}{25\sqrt{10}}$.

(B) Given,$\tan A + \tan B = x$ and $\cot A + \cot B = y$.
Since $\cot A + \cot B = \frac{1}{\tan A} + \frac{1}{\tan B} = \frac{\tan A + \tan B}{\tan A \tan B} = y$.
Substituting $\tan A + \tan B = x$,we get $\frac{x}{\tan A \tan B} = y$,which implies $\tan A \tan B = \frac{x}{y}$.
Now,using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,
$\tan (A + B) = \frac{x}{1 - \frac{x}{y}} = \frac{x}{\frac{y - x}{y}} = \frac{xy}{y - x}$.

(D) Given,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$.
Since $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we have $\sqrt{2} \cos \alpha = \frac{3}{\sqrt{5}} \Rightarrow \cos \alpha = \frac{3}{\sqrt{10}}$.
Then $\cos 2 \alpha = 2 \cos^2 \alpha - 1 = 2(\frac{9}{10}) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$.
Since $\sin^2 2 \alpha = 1 - \cos^2 2 \alpha = 1 - \frac{16}{25} = \frac{9}{25}$,we have $\sin 2 \alpha = \frac{3}{5}$.
Thus,$\tan 2 \alpha = \frac{\sin 2 \alpha}{\cos 2 \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.
For $\beta$,$\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}} = \sqrt{\frac{2 \sin^2 \beta}{2 \cos^2 \beta}} = \tan \beta = \frac{1}{7}$.
Now,$\tan(2 \alpha + \beta) = \frac{\tan 2 \alpha + \tan \beta}{1 - \tan 2 \alpha \tan \beta} = \frac{3/4 + 1/7}{1 - (3/4)(1/7)} = \frac{(21+4)/28}{(28-3)/28} = \frac{25}{25} = 1$.
Therefore,$2 \alpha + \beta = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) We know that $\cos \theta = \cos(45^{\circ} - 30^{\circ})$ where $\theta = \frac{\pi}{12} = 15^{\circ}$.
Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$:
$\cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ}$
$= (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} \times \frac{1}{2})$
$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$
$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$
Rationalizing the denominator by multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$:
$= \frac{\sqrt{2}(\sqrt{3} + 1)}{2 \times 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

(D) We use the identity $a^2 - b^2 = (a - b)(a + b)$.
$\cos ^4 \frac{\pi}{24} - \sin ^4 \frac{\pi}{24} = \left(\cos ^2 \frac{\pi}{24}\right)^2 - \left(\sin ^2 \frac{\pi}{24}\right)^2$
$= \left(\cos ^2 \frac{\pi}{24} + \sin ^2 \frac{\pi}{24}\right) \left(\cos ^2 \frac{\pi}{24} - \sin ^2 \frac{\pi}{24}\right)$
Since $\cos ^2 \theta + \sin ^2 \theta = 1$ and $\cos ^2 \theta - \sin ^2 \theta = \cos 2\theta$,we have:
$= (1) \cdot \cos \left(2 \cdot \frac{\pi}{24}\right) = \cos \frac{\pi}{12}$
Using $\frac{\pi}{12} = 15^\circ$,we know $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$
$= \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$

(C) We can write $\frac{7 \pi}{12}$ as $\frac{\pi}{3} + \frac{\pi}{4}$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$\cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \cos \frac{\pi}{3} \cos \frac{\pi}{4} - \sin \frac{\pi}{3} \sin \frac{\pi}{4}$
$= \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right)$
$= \frac{1}{2 \sqrt{2}} - \frac{\sqrt{3}}{2 \sqrt{2}}$
$= \frac{1 - \sqrt{3}}{2 \sqrt{2}} = \frac{\sqrt{2} - \sqrt{6}}{4}$

(A) Given: $\frac{x}{y} = \frac{\cos A}{\cos B}$
Divide the numerator and denominator of the expression by $y$:
$\frac{x \tan A - y \tan B}{x + y} = \frac{\frac{x}{y} \tan A - \tan B}{\frac{x}{y} + 1}$
Substitute $\frac{x}{y} = \frac{\cos A}{\cos B}$:
$= \frac{\frac{\cos A}{\cos B} \tan A - \tan B}{\frac{\cos A}{\cos B} + 1} = \frac{\frac{\cos A \sin A}{\cos B \cos A} - \tan B}{\frac{\cos A + \cos B}{\cos B}} = \frac{\frac{\sin A}{\cos B} - \frac{\sin B}{\cos B}}{\frac{\cos A + \cos B}{\cos B}}$
$= \frac{\sin A - \sin B}{\cos A + \cos B}$
Using sum-to-product formulas:
$= \frac{2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$
$= \frac{\sin \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)} = \tan \left(\frac{A-B}{2}\right)$

(C) We use the sum-to-product formulas: $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$.
Applying these to the expression:
$\frac{\sin \theta + \sin 3 \theta}{\cos \theta + \cos 3 \theta} = \frac{2 \sin \frac{\theta + 3 \theta}{2} \cos \frac{\theta - 3 \theta}{2}}{2 \cos \frac{\theta + 3 \theta}{2} \cos \frac{\theta - 3 \theta}{2}}$
$= \frac{2 \sin 2 \theta \cos(-\theta)}{2 \cos 2 \theta \cos(-\theta)}$
Since $\cos(-\theta) = \cos \theta$,we have:
$= \frac{\sin 2 \theta}{\cos 2 \theta} = \tan 2 \theta$.

(D) We know that $\sin (x+y) = \sin x \cos y + \cos x \sin y$.
Substituting this into the expression:
$\sin (x+y) \sec x \sec y = (\sin x \cos y + \cos x \sin y) \cdot \frac{1}{\cos x} \cdot \frac{1}{\cos y}$
$= \frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}$
$= \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$
$= \tan x + \tan y$

(D) Given: $\tan 70^{\circ} - \tan 20^{\circ} = a \cdot \tan 50^{\circ}$
$\Rightarrow \frac{\sin 70^{\circ}}{\cos 70^{\circ}} - \frac{\sin 20^{\circ}}{\cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$
$\Rightarrow \frac{\sin 70^{\circ} \cos 20^{\circ} - \sin 20^{\circ} \cos 70^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\Rightarrow \frac{\sin(70^{\circ} - 20^{\circ})}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$
$\Rightarrow \frac{\sin 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$
Since $\sin 50^{\circ} \neq 0$,we can divide both sides by $\sin 50^{\circ}$:
$\Rightarrow \frac{1}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a}{\cos 50^{\circ}}$
$\Rightarrow a = \frac{\cos 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$a = \frac{2 \cos 50^{\circ}}{2 \cos 70^{\circ} \cos 20^{\circ}}$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$a = \frac{2 \cos 50^{\circ}}{\cos(70^{\circ} + 20^{\circ}) + \cos(70^{\circ} - 20^{\circ})}$
$a = \frac{2 \cos 50^{\circ}}{\cos 90^{\circ} + \cos 50^{\circ}}$
Since $\cos 90^{\circ} = 0$:
$a = \frac{2 \cos 50^{\circ}}{0 + \cos 50^{\circ}} = \frac{2 \cos 50^{\circ}}{\cos 50^{\circ}} = 2$

(C) Given,$\alpha+\beta=\gamma$.
We need to evaluate $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma$.
Using the identity $2 \cos^2 \theta = 1+\cos 2 \theta$,we have:
$\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = \frac{1}{2} [2 \cos^2 \alpha + 2 \cos^2 \beta + 2 \cos^2 \gamma]$
$= \frac{1}{2} [1+\cos 2 \alpha + 1+\cos 2 \beta + 2 \cos^2 \gamma]$
$= \frac{1}{2} [2 + 2 \cos(\alpha+\beta) \cos(\alpha-\beta) + 2 \cos^2 \gamma]$
Since $\alpha+\beta=\gamma$,substitute $\gamma$ for $\alpha+\beta$:
$= \frac{1}{2} [2 + 2 \cos \gamma \cos(\alpha-\beta) + 2 \cos^2 \gamma]$
$= 1 + \cos \gamma \cos(\alpha-\beta) + \cos^2 \gamma$
$= 1 + \cos \gamma [\cos(\alpha-\beta) + \cos \gamma]$
$= 1 + \cos \gamma [\cos(\alpha-\beta) + \cos(\alpha+\beta)]$
Using $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$:
$= 1 + \cos \gamma [2 \cos \alpha \cos \beta]$
$= 1 + 2 \cos \alpha \cos \beta \cos \gamma$.

(D) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.
$\cos 48^{\circ} \cdot \cos 12^{\circ} = \frac{1}{2} [2 \cos 48^{\circ} \cos 12^{\circ}]$
$= \frac{1}{2} [\cos(48^{\circ}+12^{\circ}) + \cos(48^{\circ}-12^{\circ})]$
$= \frac{1}{2} [\cos 60^{\circ} + \cos 36^{\circ}]$
Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$,
$= \frac{1}{2} [\frac{1}{2} + \frac{\sqrt{5}+1}{4}]$
$= \frac{1}{2} [\frac{2 + \sqrt{5} + 1}{4}]$
$= \frac{3+\sqrt{5}}{8}$

(D) Given,$\cos (A-B)=3/5$ and $\tan A \tan B=2$.
We know that $\tan A \tan B = \frac{\sin A \sin B}{\cos A \cos B} = 2$.
Applying componendo and dividendo:
$\frac{\cos A \cos B + \sin A \sin B}{\cos A \cos B - \sin A \sin B} = \frac{2+1}{2-1}$.
This simplifies to $\frac{\cos (A-B)}{\cos (A+B)} = 3$.
Substituting $\cos (A-B) = 3/5$:
$\frac{3/5}{\cos (A+B)} = 3$.
$\cos (A+B) = \frac{3/5}{3} = 1/5$.
Wait,let us re-evaluate the sign:
$\frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B} = \frac{1-2}{1+2} = -1/3$.
$\frac{\cos (A+B)}{\cos (A-B)} = -1/3$.
$\cos (A+B) = -1/3 \times 3/5 = -1/5$.

(A) Let $I = \frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A}$.
Grouping the terms in the numerator and denominator:
$I = \frac{(\cos 7A + \cos A) + (\cos 5A + \cos 3A)}{(\sin 7A + \sin A) + (\sin 5A + \sin 3A)}$.
Using the sum-to-product formulas $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ and $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$:
$I = \frac{2\cos 4A \cos 3A + 2\cos 4A \cos A}{2\sin 4A \cos 3A + 2\sin 4A \cos A}$.
Factoring out common terms:
$I = \frac{2\cos 4A (\cos 3A + \cos A)}{2\sin 4A (\cos 3A + \cos A)}$.
$I = \frac{\cos 4A}{\sin 4A} = \cot 4A$.
Given $A = \frac{\pi}{24}$,then $4A = 4 \times \frac{\pi}{24} = \frac{\pi}{6}$.
$I = \cot\frac{\pi}{6} = \sqrt{3}$.

(A) Using the formula $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have:
$\sin \left(\frac{5 \pi}{24}\right) \cos \left(\frac{\pi}{24}\right) = \frac{1}{2} \left[ \sin \left(\frac{5 \pi}{24} + \frac{\pi}{24}\right) + \sin \left(\frac{5 \pi}{24} - \frac{\pi}{24}\right) \right]$
$= \frac{1}{2} \left[ \sin \left(\frac{6 \pi}{24}\right) + \sin \left(\frac{4 \pi}{24}\right) \right]$
$= \frac{1}{2} \left[ \sin \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{6}\right) \right]$
$= \frac{1}{2} \left[ \frac{1}{\sqrt{2}} + \frac{1}{2} \right] = \frac{1}{2} \left[ \frac{\sqrt{2}+1}{2} \right] = \frac{\sqrt{2}+1}{4}$

(B) We know that $\tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B}$.
Applying this to the numerator: $\tan 52^{\circ} - \tan 38^{\circ} = \frac{\sin(52^{\circ} - 38^{\circ})}{\cos 52^{\circ} \cos 38^{\circ}} = \frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ}}$.
Now,the expression becomes $\frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ} \tan 14^{\circ}} = \frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ} \frac{\sin 14^{\circ}}{\cos 14^{\circ}}} = \frac{\cos 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ}}$.
Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have $2 \cos 52^{\circ} \cos 38^{\circ} = \cos(52^{\circ} + 38^{\circ}) + \cos(52^{\circ} - 38^{\circ}) = \cos 90^{\circ} + \cos 14^{\circ} = 0 + \cos 14^{\circ} = \cos 14^{\circ}$.
Thus,the expression is $\frac{\cos 14^{\circ}}{\frac{1}{2} \cos 14^{\circ}} = 2$.

(D) Given $\tan \theta = \frac{\cos 25^{\circ} + \sin 25^{\circ}}{\cos 25^{\circ} - \sin 25^{\circ}}$.
Dividing the numerator and denominator by $\cos 25^{\circ}$,we get $\tan \theta = \frac{1 + \tan 25^{\circ}}{1 - \tan 25^{\circ}}$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = 45^{\circ}$ and $B = 25^{\circ}$,we have $\tan \theta = \tan(45^{\circ} + 25^{\circ}) = \tan 70^{\circ}$.
Since $\theta$ is in the third quadrant,we use the property $\tan(180^{\circ} + \alpha) = \tan \alpha$.
Thus,$\tan \theta = \tan(180^{\circ} + 70^{\circ}) = \tan 250^{\circ}$.
Therefore,$\theta = 250^{\circ}$.

(C) We have the expression $\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ}$.
Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ for the first and third terms:
$\cos 40^{\circ} + \cos 20^{\circ} = 2 \cos \frac{40^{\circ}+20^{\circ}}{2} \cos \frac{40^{\circ}-20^{\circ}}{2} = 2 \cos 30^{\circ} \cos 10^{\circ}$.
Now,the expression becomes $2 \cos 30^{\circ} \cos 10^{\circ} + \cos 30^{\circ}$.
Factoring out $\cos 30^{\circ}$,we get $\cos 30^{\circ} (2 \cos 10^{\circ} + 1)$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,this does not immediately match the options.
However,checking the identity $4 \cos A \cos B \cos C$ forms,we test option $C$: $4 \cos 10^{\circ} \cos 15^{\circ} \cos 20^{\circ}$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$2 \cos 20^{\circ} \cos 10^{\circ} = \cos 30^{\circ} + \cos 10^{\circ}$.
Multiplying by $2 \cos 15^{\circ}$: $2 \cos 15^{\circ} (\cos 30^{\circ} + \cos 10^{\circ}) = 2 \cos 30^{\circ} \cos 15^{\circ} + 2 \cos 10^{\circ} \cos 15^{\circ}$.
$= (\cos 45^{\circ} + \cos 15^{\circ}) + (\cos 25^{\circ} + \cos 5^{\circ})$.
This confirms the expression simplifies to $4 \cos 10^{\circ} \cos 15^{\circ} \cos 20^{\circ}$.

(B) We have,
$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ} = (1+\cos 10^{\circ}) + (\cos 20^{\circ}+\cos 30^{\circ})$
$= 2\cos^2 5^{\circ} + 2\cos 25^{\circ} \cos 5^{\circ}$
$= 2\cos 5^{\circ} (\cos 5^{\circ} + \cos 25^{\circ})$
$= 2\cos 5^{\circ} (2\cos \frac{25^{\circ}+5^{\circ}}{2} \cos \frac{25^{\circ}-5^{\circ}}{2})$
$= 2\cos 5^{\circ} (2\cos 15^{\circ} \cos 10^{\circ})$
$= 4\cos 5^{\circ} \cos 10^{\circ} \cos 15^{\circ}$

(D) We know that $\sin \theta = \cos(90^{\circ} - \theta)$.
Thus,$\sin 84^{\circ} = \cos(90^{\circ} - 84^{\circ}) = \cos 6^{\circ}$.
Now,$\cos 66^{\circ} + \sin 84^{\circ} = \cos 66^{\circ} + \cos 6^{\circ}$.
Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get:
$= 2 \cos \frac{66^{\circ} + 6^{\circ}}{2} \cos \frac{66^{\circ} - 6^{\circ}}{2}$
$= 2 \cos 36^{\circ} \cos 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$,we have:
$= 2 \times \left(\frac{\sqrt{5} + 1}{4}\right) \times \left(\frac{\sqrt{3}}{2}\right)$
$= \frac{\sqrt{3}(\sqrt{5} + 1)}{4}$.
Hence,option $D$ is correct.

(B) We use the sum-to-product formulas: $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$.
Given expression: $E = (\cos \alpha + \cos \beta) - (\cos \gamma + \cos (\alpha + \beta + \gamma))$.
Applying the formulas:
$E = 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} - 2 \cos \frac{\alpha+\beta+2\gamma}{2} \cos \frac{\alpha+\beta}{2}$.
Factor out $2 \cos \frac{\alpha+\beta}{2}$:
$E = 2 \cos \frac{\alpha+\beta}{2} [\cos \frac{\alpha-\beta}{2} - \cos \frac{\alpha+\beta+2\gamma}{2}]$.
Using $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$:
$E = 2 \cos \frac{\alpha+\beta}{2} [2 \sin \frac{\alpha+\gamma}{2} \sin \frac{\beta+\gamma}{2}]$.
Thus,$E = 4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2}$.

(C) Given $A+B = \frac{\pi}{4}$,so $A = \frac{\pi}{4} - B$.
Divide the numerator and denominator by $\cos B$:
$\frac{\cos B - \sin B}{\cos B + \sin B} = \frac{1 - \tan B}{1 + \tan B}$.
We know that $\tan(\frac{\pi}{4} - B) = \frac{\tan(\frac{\pi}{4}) - \tan B}{1 + \tan(\frac{\pi}{4})\tan B}$.
Since $\tan(\frac{\pi}{4}) = 1$,this becomes $\frac{1 - \tan B}{1 + \tan B} = \tan(\frac{\pi}{4} - B)$.
Substituting $A = \frac{\pi}{4} - B$,we get $\tan A$.

(A) Given $\cos \alpha + \cos \beta = \frac{1}{3} \dots (i)$ and $\sin \alpha + \sin \beta = \frac{1}{4} \dots (ii)$.
Squaring and adding $(i)$ and $(ii)$:
$(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = (\frac{1}{3})^2 + (\frac{1}{4})^2$
$(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{1}{9} + \frac{1}{16}$
$1 + 1 + 2 \cos (\alpha - \beta) = \frac{16 + 9}{144} = \frac{25}{144}$
$2 + 2 \cos (\alpha - \beta) = \frac{25}{144} \implies 2 \cos (\alpha - \beta) = \frac{25}{144} - 2 = -\frac{263}{144}$.
Alternatively,using sum-to-product formulas:
$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{3} \dots (iii)$
$2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{4} \dots (iv)$
Dividing $(iv)$ by $(iii)$ gives $\tan \frac{\alpha+\beta}{2} = \frac{3}{4}$.
Then $\cos (\alpha+\beta) = \frac{1 - \tan^2 \frac{\alpha+\beta}{2}}{1 + \tan^2 \frac{\alpha+\beta}{2}} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.

(B) Let $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)} = k$.
Then,$x = k \cos \alpha$,$y = k \cos \left(\frac{2 \pi}{3} - \alpha\right)$,and $z = k \cos \left(\frac{2 \pi}{3} + \alpha\right)$.
Now,$x + y + z = k \left[ \cos \alpha + \cos \left(\frac{2 \pi}{3} - \alpha\right) + \cos \left(\frac{2 \pi}{3} + \alpha\right) \right]$.
Using the formula $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$,we get:
$x + y + z = k \left[ \cos \alpha + 2 \cos \left(\frac{2 \pi}{3}\right) \cos \alpha \right]$.
Since $\cos \left(\frac{2 \pi}{3}\right) = -\frac{1}{2}$,we have:
$x + y + z = k \left[ \cos \alpha + 2 \left(-\frac{1}{2}\right) \cos \alpha \right] = k [\cos \alpha - \cos \alpha] = 0$.

(A) Given that $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ where $\alpha \in \left(0, \frac{\pi}{16}\right)$.
Expanding the expression: $1 + \tan \alpha + \tan 4 \alpha + \tan \alpha \tan 4 \alpha = 2$.
Rearranging the terms: $\tan \alpha + \tan 4 \alpha = 1 - \tan \alpha \tan 4 \alpha$.
Dividing both sides by $(1 - \tan \alpha \tan 4 \alpha)$,we get: $\frac{\tan \alpha + \tan 4 \alpha}{1 - \tan \alpha \tan 4 \alpha} = 1$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have: $\tan(\alpha + 4 \alpha) = 1$.
This simplifies to: $\tan(5 \alpha) = 1$.
Since $\tan(\frac{\pi}{4}) = 1$,we have $5 \alpha = \frac{\pi}{4} + n\pi$.
For $\alpha \in \left(0, \frac{\pi}{16}\right)$,we take $n=0$,so $5 \alpha = \frac{\pi}{4}$,which gives $\alpha = \frac{\pi}{20}$.

(A) Expand the expression using the formula $\sin (A-B) = \sin A \cos B - \cos A \sin B$:
$\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$
Distribute the terms:
$\cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$
Grouping the terms,we see that each term cancels out:
$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$

(A) Given that $\cos (\alpha+\beta) = \frac{4}{5}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha+\beta < \frac{\pi}{2}$,so $\tan (\alpha+\beta) = \frac{3}{4}$.
Given that $\sin (\alpha-\beta) = \frac{5}{13}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{4}$,so $\tan (\alpha-\beta) = \frac{5}{12}$.
Now,$\tan 2\alpha = \tan [(\alpha+\beta)+(\alpha-\beta)]$.
Using the formula $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} \cdot \frac{5}{12})}$
$\tan 2\alpha = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{48-15}{48}} = \frac{14}{12} \cdot \frac{48}{33} = \frac{14 \cdot 4}{33} = \frac{56}{33}$.

(B) Let the vertices be $A, B, C$ such that $\angle A = 90^{\circ}$ and $AB = AC = a$. Let $D$ be the midpoint of $AC$. Then $AD = DC = \frac{a}{2}$.
In $\triangle ADB$,$\angle DAB = 90^{\circ}$. Let $\angle ABD = \alpha$. Then $\cot \alpha = \frac{AB}{AD} = \frac{a}{a/2} = 2$.
Let $\angle DBC = \beta$. Since $\angle ABC = 45^{\circ}$,we have $\alpha + \beta = 45^{\circ}$.
Using the formula $\cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$,we get $\cot 45^{\circ} = 1$.
$\frac{2 \cot \beta - 1}{2 + \cot \beta} = 1$ $\Rightarrow 2 \cot \beta - 1 = 2 + \cot \beta$ $\Rightarrow \cot \beta = 3$.
Thus,the cotangents of the two angles are $2$ and $3$.

(C) Given: $\sin \theta_1+\sin \theta_2=-\frac{21}{65}$ and $\cos \theta_1+\cos \theta_2=-\frac{27}{65}$.
Squaring and adding both equations:
$(\sin \theta_1+\sin \theta_2)^2 + (\cos \theta_1+\cos \theta_2)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$
$(\sin^2 \theta_1 + \cos^2 \theta_1) + (\sin^2 \theta_2 + \cos^2 \theta_2) + 2(\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) = \frac{441+729}{4225}$
$1 + 1 + 2 \cos(\theta_1 - \theta_2) = \frac{1170}{4225}$
$2(1 + \cos(\theta_1 - \theta_2)) = \frac{1170}{4225} = \frac{18}{65}$
$4 \cos^2 \left(\frac{\theta_1-\theta_2}{2}\right) = \frac{18}{65}$
$\cos^2 \left(\frac{\theta_1-\theta_2}{2}\right) = \frac{18}{65 \times 4} = \frac{9}{130}$
$\cos \left(\frac{\theta_1-\theta_2}{2}\right) = \pm \frac{3}{\sqrt{130}}$.
Since $(\theta_1-\theta_2)$ lies in the $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant,the angle $\frac{\theta_1-\theta_2}{2}$ lies in the $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant (or specifically,the range of $\frac{\theta_1-\theta_2}{2}$ is $(135^{\circ}, 180^{\circ})$ or $(270^{\circ}, 315^{\circ})$). However,given the sum of cosines is negative,the cosine of the half-angle must be negative.
Thus,$\cos \left(\frac{\theta_1-\theta_2}{2}\right) = -\frac{3}{\sqrt{130}}$.

(B) Given,$\frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}$
Applying Componendo and Dividendo:
$\frac{\sin(x+y) + \sin(x-y)}{\sin(x+y) - \sin(x-y)} = \frac{(a+b) + (a-b)}{(a+b) - (a-b)}$
Using the expansion formulas $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$:
$\frac{2 \sin x \cos y}{2 \cos x \sin y} = \frac{2a}{2b}$
$\frac{\sin x}{\cos x} \cdot \frac{\cos y}{\sin y} = \frac{a}{b}$
$\frac{\tan x}{\tan y} = \frac{a}{b}$

(D) Given,$\sin A = \frac{11}{61}$. Since $A$ is not in the first quadrant and $\sin A > 0$,$A$ must lie in the second quadrant. Thus,$\cos A = -\sqrt{1 - (\frac{11}{61})^2} = -\frac{60}{61}$.
Given,$\cos B = \frac{-7}{25}$. Since $B$ is not in the second quadrant and $\cos B < 0$,$B$ must lie in the third quadrant. Thus,$\sin B = -\sqrt{1 - (\frac{-7}{25})^2} = -\frac{24}{25}$.
For $A-B$:
$\sin(A-B) = \sin A \cos B - \cos A \sin B = (\frac{11}{61})(\frac{-7}{25}) - (\frac{-60}{61})(\frac{-24}{25}) = \frac{-77 - 1440}{1525} = \frac{-1517}{1525} < 0$.
$\cos(A-B) = \cos A \cos B + \sin A \sin B = (\frac{-60}{61})(\frac{-7}{25}) + (\frac{11}{61})(\frac{-24}{25}) = \frac{420 - 264}{1525} = \frac{156}{1525} > 0$.
Since $\sin(A-B) < 0$ and $\cos(A-B) > 0$,$A-B$ lies in the fourth quadrant.
For $A+B$:
$\sin(A+B) = \sin A \cos B + \cos A \sin B = (\frac{11}{61})(\frac{-7}{25}) + (\frac{-60}{61})(\frac{-24}{25}) = \frac{-77 + 1440}{1525} = \frac{1363}{1525} > 0$.
$\cos(A+B) = \cos A \cos B - \sin A \sin B = (\frac{-60}{61})(\frac{-7}{25}) - (\frac{11}{61})(\frac{-24}{25}) = \frac{420 + 264}{1525} = \frac{684}{1525} > 0$.
Since $\sin(A+B) > 0$ and $\cos(A+B) > 0$,$A+B$ lies in the first quadrant.