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

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

(B) Given $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$.
We know that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - (\frac{1}{2} \times \frac{1}{3})} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$.
Since $\tan(A + B) = 1$,we have $A + B = 45^\circ$.
Therefore,$2A = 90^\circ - 2B$.
Taking cosine on both sides,$\cos 2A = \cos(90^\circ - 2B) = \sin 2B$.

(D) Given $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$.
Since $A$ and $B$ are acute,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Similarly,$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$\sin(A + B) = \left(\frac{1}{\sqrt{10}}\right)\left(\frac{2}{\sqrt{5}}\right) + \left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{5}}\right)$
$\sin(A + B) = \frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\sin(A + B) = \frac{1}{\sqrt{2}}$ and $A, B$ are acute,$A + B = \frac{\pi}{4}$.

(D) Given that $\sin A + \sin B = C$ and $\cos A + \cos B = D$.
Dividing the two equations:
$\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{C}{D}$
Using the sum-to-product formulas:
$\frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}} = \frac{C}{D}$
$\tan \frac{A+B}{2} = \frac{C}{D}$
Using the identity $\sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}$ where $\theta = A+B$:
$\sin (A+B) = \frac{2 \tan \frac{A+B}{2}}{1 + \tan^2 \frac{A+B}{2}}$
Substituting $\tan \frac{A+B}{2} = \frac{C}{D}$:
$\sin (A+B) = \frac{2(\frac{C}{D})}{1 + (\frac{C}{D})^2} = \frac{\frac{2C}{D}}{\frac{D^2 + C^2}{D^2}} = \frac{2C}{D} \times \frac{D^2}{C^2 + D^2} = \frac{2CD}{C^2 + D^2}$.

(B) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)$.
Substituting $A = 48^\circ$ and $B = 12^\circ$:
$\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ) \cos(48^\circ - 12^\circ)$
$= \cos 60^\circ \cos 36^\circ$
We know that $\cos 60^\circ = \frac{1}{2}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.
Therefore,the expression becomes:
$= \frac{1}{2} \times \left( \frac{\sqrt{5} + 1}{4} \right) = \frac{\sqrt{5} + 1}{8}$.

(B) We know that $\sin(A + B) = \sin A \cos B + \cos A \sin B$.
$\sin 75^\circ = \sin(45^\circ + 30^\circ)$
$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
$= (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} \times \frac{1}{2})$
$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$
$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$.

(B) We have $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$:
$\tan(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \left(\frac{m}{m + 1}\right) \left(\frac{1}{2m + 1}\right)}$
$= \frac{\frac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1)}}{\frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)}}$
$= \frac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}$
$= \frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$
Since $\tan(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

(B) We know that the formula for $\tan(A + B)$ is:
$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
Substituting $A = 20^\circ$ and $B = 40^\circ$:
$\tan(20^\circ + 40^\circ) = \frac{\tan 20^\circ + \tan 40^\circ}{1 - \tan 20^\circ \tan 40^\circ}$
Since $\tan(60^\circ) = \sqrt{3}$,we have:
$\sqrt{3} = \frac{\tan 20^\circ + \tan 40^\circ}{1 - \tan 20^\circ \tan 40^\circ}$
Multiplying both sides by $(1 - \tan 20^\circ \tan 40^\circ)$:
$\sqrt{3} - \sqrt{3} \tan 20^\circ \tan 40^\circ = \tan 20^\circ + \tan 40^\circ$
Rearranging the terms:
$\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = \sqrt{3}$

(D) Given expression: $\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ]$
Multiply and divide by $2$ inside the bracket:
$= \frac{1}{2} [\frac{\sqrt{3}}{2} \cos 23^\circ - \frac{1}{2} \sin 23^\circ]$
Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$:
$= \frac{1}{2} [\cos 30^\circ \cos 23^\circ - \sin 30^\circ \sin 23^\circ]$
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$= \frac{1}{2} \cos(30^\circ + 23^\circ)$
$= \frac{1}{2} \cos 53^\circ$
Since $\frac{1}{2} \cos 53^\circ$ is not equal to any of the given options,the correct answer is $D$.

(A) We know that $\tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.
Since $\cot 75^\circ = \frac{1}{\tan 75^\circ} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.
Therefore,$\tan 75^\circ - \cot 75^\circ = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}$.

(B) We are given $\tan A = -\frac{1}{2}$ and $\tan B = -\frac{1}{3}.$
Using 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 values:
$\tan(A + B) = \frac{-\frac{1}{2} - \frac{1}{3}}{1 - (-\frac{1}{2})(-\frac{1}{3})}$
$\tan(A + B) = \frac{-\frac{5}{6}}{1 - \frac{1}{6}} = \frac{-\frac{5}{6}}{\frac{5}{6}} = -1$
Since $\tan(A + B) = -1,$ and knowing that $\tan(\frac{3\pi}{4}) = -1,$
$A + B = \frac{3\pi}{4}.$

(D) Given $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$.
Since $A$ is in the first quadrant,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$.
Since $B$ is in the third quadrant,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$\cos(A + B) = \left(\frac{3}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)$
$= -\frac{36}{65} + \frac{20}{65}$
$= -\frac{16}{65}$.

(B) Given that $A + B = \frac{\pi}{4}.$
Taking tangent on both sides,we get $\tan(A + B) = \tan\left(\frac{\pi}{4}\right).$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},$ we have $\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1.$
This implies $\tan A + \tan B = 1 - \tan A \tan B.$
Rearranging the terms,we get $\tan A + \tan B + \tan A \tan B = 1.$
Now,consider the expression $(1 + \tan A)(1 + \tan B) = 1 + \tan B + \tan A + \tan A \tan B.$
Substituting the value $\tan A + \tan B + \tan A \tan B = 1$ into the expression,we get $1 + 1 = 2.$
Thus,$(1 + \tan A)(1 + \tan B) = 2.$

(C) We know the standard trigonometric identity for the cosine of a sum of two angles is:
$\cos (A + B) = \cos A \cos B - \sin A \sin B$
Comparing this with the given equation $\cos (A + B) = \alpha \cos A \cos B + \beta \sin A \sin B$,we get:
$\alpha = 1$
$\beta = -1$
Therefore,$(\alpha, \beta) = (1, -1)$.

(B) Given expression: $\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B}$
Using the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$,the numerator becomes $2 \sin(A + B) \sin(A - B)$ if we multiply by $2/2$.
Alternatively,multiply numerator and denominator by $2$:
$= \frac{2(\sin^2 A - \sin^2 B)}{2 \sin A \cos A - 2 \sin B \cos B}$
Using $2 \sin^2 \theta = 1 - \cos 2\theta$ and $2 \sin \theta \cos \theta = \sin 2\theta$:
$= \frac{(1 - \cos 2A) - (1 - \cos 2B)}{\sin 2A - \sin 2B} = \frac{\cos 2B - \cos 2A}{\sin 2A - \sin 2B}$
Using sum-to-product formulas $\cos C - \cos D = 2 \sin(\frac{C+D}{2}) \sin(\frac{D-C}{2})$ and $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$:
$= \frac{2 \sin(A + B) \sin(A - B)}{2 \cos(A + B) \sin(A - B)}$
$= \frac{\sin(A + B)}{\cos(A + B)} = \tan(A + B)$.

(B) Given $\cos (\alpha + \beta ) = \frac{4}{5}$ and $\sin (\alpha - \beta ) = \frac{5}{13}$.
Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha + \beta < \frac{\pi}{2}$ and $-\frac{\pi}{4} < \alpha - \beta < \frac{\pi}{4}$.
Thus,$\sin (\alpha + \beta ) = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$ and $\cos (\alpha - \beta ) = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$.
We know that $2\alpha = (\alpha + \beta ) + (\alpha - \beta )$.
Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$\sin 2\alpha = \sin ((\alpha + \beta ) + (\alpha - \beta )) = \sin (\alpha + \beta ) \cos (\alpha - \beta ) + \cos (\alpha + \beta ) \sin (\alpha - \beta )$.
$\sin 2\alpha = (\frac{3}{5} \times \frac{12}{13}) + (\frac{4}{5} \times \frac{5}{13}) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}$.
Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$\cos 2\alpha = \cos ((\alpha + \beta ) + (\alpha - \beta )) = \cos (\alpha + \beta ) \cos (\alpha - \beta ) - \sin (\alpha + \beta ) \sin (\alpha - \beta )$.
$\cos 2\alpha = (\frac{4}{5} \times \frac{12}{13}) - (\frac{3}{5} \times \frac{5}{13}) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$.
Therefore,$\tan 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{56/65}{33/65} = \frac{56}{33}$.

(A) Given $\cos \theta = \frac{8}{17}$ and $\theta$ is in the $1^{st}$ quadrant,we have $\sin \theta = \sqrt{1 - \left( \frac{8}{17} \right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.
Expanding the expression $\cos (30^\circ + \theta) + \cos (45^\circ - \theta) + \cos (120^\circ - \theta)$:
$= (\cos 30^\circ \cos \theta - \sin 30^\circ \sin \theta) + (\cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta) + (\cos 120^\circ \cos \theta + \sin 120^\circ \sin \theta)$
$= \cos \theta (\cos 30^\circ + \cos 45^\circ + \cos 120^\circ) + \sin \theta (-\sin 30^\circ + \sin 45^\circ + \sin 120^\circ)$
$= \cos \theta \left( \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \right) + \sin \theta \left( -\frac{1}{2} + \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \right)$
$= \frac{8}{17} \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right) + \frac{15}{17} \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)$
$= \left( \frac{8}{17} + \frac{15}{17} \right) \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)$
$= \frac{23}{17} \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)$.

(D) Given expression: $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ$
Group the terms: $(\sin 61^\circ + \sin 47^\circ) - (\sin 25^\circ + \sin 11^\circ)$
Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
$= 2 \sin \frac{61^\circ + 47^\circ}{2} \cos \frac{61^\circ - 47^\circ}{2} - 2 \sin \frac{25^\circ + 11^\circ}{2} \cos \frac{25^\circ - 11^\circ}{2}$
$= 2 \sin 54^\circ \cos 7^\circ - 2 \sin 18^\circ \cos 7^\circ$
$= 2 \cos 7^\circ (\sin 54^\circ - \sin 18^\circ)$
Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= 2 \cos 7^\circ \times 2 \cos \frac{54^\circ + 18^\circ}{2} \sin \frac{54^\circ - 18^\circ}{2}$
$= 4 \cos 7^\circ \cos 36^\circ \sin 18^\circ$
Substitute values $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$ and $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$:
$= 4 \cos 7^\circ \left( \frac{\sqrt{5}+1}{4} \right) \left( \frac{\sqrt{5}-1}{4} \right)$
$= 4 \cos 7^\circ \left( \frac{5-1}{16} \right) = 4 \cos 7^\circ \left( \frac{4}{16} \right) = \cos 7^\circ$

(A) We know that $\cos A - \sin A = \sqrt{2} \cos(A + 45^\circ)$.
Substituting $A = 15^\circ$:
$\cos 15^\circ - \sin 15^\circ = \sqrt{2} \cos(15^\circ + 45^\circ)$
$= \sqrt{2} \cos 60^\circ$
$= \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.

(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:
$\tan 5x \tan 3x \tan 2x = \tan 5x - \tan 3x - \tan 2x$.

(D) We know the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Let $A = \frac{2\pi}{5} = \frac{6\pi}{15}$ and $B = \frac{\pi}{15}$.
Then $A - B = \frac{6\pi}{15} - \frac{\pi}{15} = \frac{5\pi}{15} = \frac{\pi}{3}$.
Using the formula,$\tan\left(\frac{6\pi}{15} - \frac{\pi}{15}\right) = \frac{\tan \frac{6\pi}{15} - \tan \frac{\pi}{15}}{1 + \tan \frac{6\pi}{15} \tan \frac{\pi}{15}} = \tan \frac{\pi}{3}$.
Since $\tan \frac{\pi}{3} = \sqrt{3}$,we have $\frac{\tan \frac{2\pi}{5} - \tan \frac{\pi}{15}}{1 + \tan \frac{2\pi}{5} \tan \frac{\pi}{15}} = \sqrt{3}$.
Multiplying both sides by the denominator: $\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} = \sqrt{3} (1 + \tan \frac{2\pi}{5} \tan \frac{\pi}{15})$.
$\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} = \sqrt{3} + \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$.
Rearranging the terms: $\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15} = \sqrt{3}$.

(C) Given expression: $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$
Group the terms: $(\cos 132^\circ + \cos 12^\circ) + (\cos 156^\circ + \cos 84^\circ)$
Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$= 2 \cos \frac{132^\circ + 12^\circ}{2} \cos \frac{132^\circ - 12^\circ}{2} + 2 \cos \frac{156^\circ + 84^\circ}{2} \cos \frac{156^\circ - 84^\circ}{2}$
$= 2 \cos 72^\circ \cos 60^\circ + 2 \cos 120^\circ \cos 36^\circ$
Since $\cos 60^\circ = \frac{1}{2}$ and $\cos 120^\circ = -\frac{1}{2}$:
$= 2 \cos 72^\circ \left(\frac{1}{2}\right) + 2 \left(-\frac{1}{2}\right) \cos 36^\circ$
$= \cos 72^\circ - \cos 36^\circ$
Using values $\cos 72^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:
$= \frac{\sqrt{5}-1}{4} - \frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}-1-\sqrt{5}-1}{4} = \frac{-2}{4} = -\frac{1}{2}$.

(A) Dividing the numerator and the denominator by $\cos 17^\circ$,we get:
$\frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} = \frac{1 + \tan 17^\circ}{1 - \tan 17^\circ}$
Since $\tan 45^\circ = 1$,we can write this as:
$= \frac{\tan 45^\circ + \tan 17^\circ}{1 - \tan 45^\circ \tan 17^\circ}$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = 45^\circ$ and $B = 17^\circ$:
$= \tan(45^\circ + 17^\circ) = \tan 62^\circ$.

(A) Divide the numerator and denominator by $\cos 9^\circ$:
$\frac{\frac{\cos 9^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 9^\circ}}{\frac{\cos 9^\circ}{\cos 9^\circ} - \frac{\sin 9^\circ}{\cos 9^\circ}} = \frac{1 + \tan 9^\circ}{1 - \tan 9^\circ}$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = 45^\circ$ and $B = 9^\circ$:
$\frac{\tan 45^\circ + \tan 9^\circ}{1 - \tan 45^\circ \tan 9^\circ} = \tan(45^\circ + 9^\circ) = \tan 54^\circ$.

(A) Given $\cos (A - B) = \frac{3}{5}$.
Using the expansion formula,$\cos A \cos B + \sin A \sin B = \frac{3}{5}$ ..... $(i)$.
Given $\tan A \tan B = 2$,which implies $\frac{\sin A \sin B}{\cos A \cos B} = 2$,so $\sin A \sin B = 2 \cos A \cos B$ ..... $(ii)$.
Substitute $(ii)$ into $(i)$:
$\cos A \cos B + 2 \cos A \cos B = \frac{3}{5}$
$3 \cos A \cos B = \frac{3}{5}$
$\cos A \cos B = \frac{1}{5}$.
Now,substitute $\cos A \cos B = \frac{1}{5}$ into $(ii)$:
$\sin A \sin B = 2 \times \frac{1}{5} = \frac{2}{5}$.
Thus,the correct option is $\cos A \cos B = \frac{1}{5}$.

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

(D) Given $\sin \alpha = \frac{15}{17}$ and $\frac{\pi}{2} < \alpha < \pi$ (second quadrant),so $\cos \alpha = -\sqrt{1 - (\frac{15}{17})^2} = -\frac{8}{17}$.
Given $\tan \beta = \frac{12}{5}$ and $\pi < \beta < \frac{3\pi}{2}$ (third quadrant),so $\sin \beta = -\frac{12}{13}$ and $\cos \beta = -\frac{5}{13}$.
Using the formula $\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha$:
$\sin(\beta - \alpha) = (-\frac{12}{13})(-\frac{8}{17}) - (-\frac{5}{13})(\frac{15}{17})$
$= \frac{96}{221} + \frac{75}{221} = \frac{171}{221}$.

(A) Divide the numerator and denominator by $\cos 10^o$:
$\frac{1 + \tan 10^o}{1 - \tan 10^o}$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = 45^o$ and $B = 10^o$:
$\tan(45^o + 10^o) = \tan 55^o$

(B) Given,$\cos P = \frac{1}{7}$ and $\cos Q = \frac{13}{14}$.
Since $P$ and $Q$ are acute,$\sin P = \sqrt{1 - \cos^2 P} = \sqrt{1 - (\frac{1}{7})^2} = \sqrt{\frac{48}{49}} = \frac{4\sqrt{3}}{7}$.
Similarly,$\sin Q = \sqrt{1 - \cos^2 Q} = \sqrt{1 - (\frac{13}{14})^2} = \sqrt{\frac{196 - 169}{196}} = \sqrt{\frac{27}{196}} = \frac{3\sqrt{3}}{14}$.
Using the formula $\cos(P - Q) = \cos P \cos Q + \sin P \sin Q$:
$\cos(P - Q) = (\frac{1}{7})(\frac{13}{14}) + (\frac{4\sqrt{3}}{7})(\frac{3\sqrt{3}}{14})$
$= \frac{13}{98} + \frac{12 \times 3}{98} = \frac{13 + 36}{98} = \frac{49}{98} = \frac{1}{2}$.
Since $\cos(P - Q) = \frac{1}{2}$,we have $P - Q = 60^o$.

(B) Given $\tan \alpha = \frac{1}{1 + 2^{-x}} = \frac{2^x}{2^x + 1}$ and $\tan \beta = \frac{1}{1 + 2^{x+1}}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$:
$\tan(\alpha + \beta) = \frac{\frac{2^x}{2^x + 1} + \frac{1}{1 + 2^{x+1}}}{1 - \left(\frac{2^x}{2^x + 1}\right) \left(\frac{1}{1 + 2^{x+1}}\right)}$
Multiply numerator and denominator by $(2^x + 1)(1 + 2^{x+1})$:
$\tan(\alpha + \beta) = \frac{2^x(1 + 2^{x+1}) + (2^x + 1)}{(2^x + 1)(1 + 2^{x+1}) - 2^x}$
$= \frac{2^x + 2^x \cdot 2^{x+1} + 2^x + 1}{2^x + 2^x \cdot 2^{x+1} + 1 + 2^{x+1} - 2^x}$
$= \frac{2 \cdot 2^x + 2 \cdot 2^{2x} + 1}{2 \cdot 2^{2x} + 2^{x+1} + 1}$
Since $2^{x+1} = 2 \cdot 2^x$,the numerator is $2 \cdot 2^x + 2 \cdot 2^{2x} + 1$ and the denominator is $2 \cdot 2^{2x} + 2 \cdot 2^x + 1$.
Thus,$\tan(\alpha + \beta) = 1$.
Therefore,$\alpha + \beta = \frac{\pi}{4}$.

(B) Given $\sin \theta = \frac{12}{13}$ and $0 < \theta < \frac{\pi}{2}$ (first quadrant),$\cos \theta$ is positive.
$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{12}{13})^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.
Given $\cos \phi = -\frac{3}{5}$ and $\pi < \phi < \frac{3\pi}{2}$ (third quadrant),$\sin \phi$ is negative.
$\sin \phi = -\sqrt{1 - \cos^2 \phi} = -\sqrt{1 - (-\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.
Using the identity $\sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi$:
$\sin(\theta + \phi) = (\frac{12}{13})(-\frac{3}{5}) + (\frac{5}{13})(-\frac{4}{5})$
$= -\frac{36}{65} - \frac{20}{65} = -\frac{56}{65}$.

(C) Given expression: $\frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta}$
Group the terms: $\frac{(\sin 9\theta + \sin 3\theta) + (\sin 7\theta + \sin 5\theta)}{(\cos 9\theta + \cos 3\theta) + (\cos 7\theta + \cos 5\theta)}$
Using the sum-to-product formulas $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$:
Numerator: $2 \sin 6\theta \cos 3\theta + 2 \sin 6\theta \cos \theta = 2 \sin 6\theta (\cos 3\theta + \cos \theta)$
Denominator: $2 \cos 6\theta \cos 3\theta + 2 \cos 6\theta \cos \theta = 2 \cos 6\theta (\cos 3\theta + \cos \theta)$
Dividing the numerator by the denominator: $\frac{2 \sin 6\theta (\cos 3\theta + \cos \theta)}{2 \cos 6\theta (\cos 3\theta + \cos \theta)} = \frac{\sin 6\theta}{\cos 6\theta} = \tan 6\theta$.

(B) Given expression: $\sin {163^\circ} \cos {347^\circ} + \sin {73^\circ} \sin {167^\circ}$
Using the reduction formulas:
$\sin {163^\circ} = \sin (180^\circ - 17^\circ) = \sin {17^\circ}$
$\cos {347^\circ} = \cos (360^\circ - 13^\circ) = \cos {13^\circ}$
$\sin {73^\circ} = \sin (90^\circ - 17^circ) = \cos {17^\circ}$
$\sin {167^\circ} = \sin (180^\circ - 13^\circ) = \sin {13^\circ}$
Substituting these values into the expression:
$= \sin {17^\circ} \cos {13^\circ} + \cos {17^\circ} \sin {13^\circ}$
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$= \sin (17^\circ + 13^\circ) = \sin {30^\circ}$
Since $\sin {30^\circ} = 1/2$,the final answer is $1/2$.

(B) Using the sum-to-product formula $\cos(X + Y) + \cos(X - Y) = 2 \cos X \cos Y$,we have:
$\cos A + \cos(240^\circ + A) + \cos(240^\circ - A)$
$= \cos A + 2 \cos 240^\circ \cos A$
$= \cos A(1 + 2 \cos 240^\circ)$
Since $\cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$,
$= \cos A(1 + 2(-\frac{1}{2}))$
$= \cos A(1 - 1)$
$= \cos A(0) = 0$.

(A) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)$.
Let $A = \frac{\pi}{6} + \theta$ and $B = \frac{\pi}{6} - \theta$.
Then,$A + B = \left( \frac{\pi}{6} + \theta \right) + \left( \frac{\pi}{6} - \theta \right) = \frac{2\pi}{6} = \frac{\pi}{3}$.
And,$A - B = \left( \frac{\pi}{6} + \theta \right) - \left( \frac{\pi}{6} - \theta \right) = 2\theta$.
Substituting these into the identity:
$\cos^2 \left( \frac{\pi}{6} + \theta \right) - \sin^2 \left( \frac{\pi}{6} - \theta \right) = \cos \left( \frac{\pi}{3} \right) \cos(2\theta)$.
Since $\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$,the expression becomes $\frac{1}{2} \cos 2\theta$.

(C) Given $b \sin \alpha = a \sin (\alpha + 2\beta)$.
Dividing both sides,we get $\frac{a}{b} = \frac{\sin \alpha}{\sin (\alpha + 2\beta)}$.
Using the componendo and dividendo rule,$\frac{a + b}{a - b} = \frac{\sin \alpha + \sin (\alpha + 2\beta)}{\sin \alpha - \sin (\alpha + 2\beta)}$.
Applying the sum-to-product formulas $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\frac{a + b}{a - b} = \frac{2 \sin (\alpha + \beta) \cos (-\beta)}{2 \cos (\alpha + \beta) \sin (-\beta)}$.
Since $\cos(-\beta) = \cos \beta$ and $\sin(-\beta) = -\sin \beta$:
$\frac{a + b}{a - b} = \frac{2 \sin (\alpha + \beta) \cos \beta}{-2 \cos (\alpha + \beta) \sin \beta} = -\tan (\alpha + \beta) \cot \beta$.
This can be rewritten as $-\frac{\cot \beta}{\cot (\alpha + \beta)}$.

(B) Given expression: $E = \frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)}$
Using the identities $\cos \theta = \sin(90^\circ - \theta)$:
$E = \frac{\sin(B + A) + \sin(90^\circ - (B - A))}{\sin(B - A) + \sin(90^\circ - (B + A))}$
$E = \frac{\sin(B + A) + \sin(90^\circ - B + A)}{\sin(B - A) + \sin(90^\circ - B - A)}$
Using $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
Numerator: $\sin(B+A) + \sin(90^\circ - B + A) = 2 \sin(\frac{B+A+90^\circ-B+A}{2}) \cos(\frac{B+A-90^\circ+B-A}{2}) = 2 \sin(45^\circ + A) \cos(B - 45^\circ)$
Denominator: $\sin(B-A) + \sin(90^\circ - B - A) = 2 \sin(\frac{B-A+90^\circ-B-A}{2}) \cos(\frac{B-A-90^\circ+B+A}{2}) = 2 \sin(45^\circ - A) \cos(B - 45^\circ)$
$E = \frac{2 \sin(45^\circ + A) \cos(B - 45^\circ)}{2 \sin(45^\circ - A) \cos(B - 45^\circ)} = \frac{\sin(45^\circ + A)}{\sin(45^\circ - A)}$
Using $\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$:
$E = \frac{\sin 45^\circ \cos A + \cos 45^\circ \sin A}{\sin 45^\circ \cos A - \cos 45^\circ \sin A} = \frac{\frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A}{\frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A} = \frac{\cos A + \sin A}{\cos A - \sin A}$.

(B) Given,$\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$.
Applying the Componendo and Dividendo rule:
$\frac{\sin(x + y) + \sin(x - y)}{\sin(x + y) - \sin(x - y)} = \frac{(a + b) + (a - b)}{(a + b) - (a - b)}$.
Using the identities $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$ and $\sin(A+B) - \sin(A-B) = 2\cos A \sin B$:
$\frac{2\sin x \cos y}{2\cos x \sin y} = \frac{2a}{2b}$.
$\frac{\sin x}{\cos x} \cdot \frac{\cos y}{\sin y} = \frac{a}{b}$.
$\tan x \cdot \frac{1}{\tan y} = \frac{a}{b}$.
Therefore,$\frac{\tan x}{\tan y} = \frac{a}{b}$.

(A) We have the expression: $E = \cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta )$.
Expanding each term using the identity $\sin(x - y) = \sin x \cos y - \cos x \sin y$:
$E = \cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$.
Distributing the terms:
$E = \cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$.
Observing the terms:
$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$.
Thus,the correct option is $A$.

(B) Let the expression be $E = \sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma )$.
Using the sum-to-product formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
Group the first two terms: $\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) = 2 \sin \gamma \cos (\beta - \alpha )$.
Group the last two terms: $\sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma ) = 2 \cos (\alpha + \beta ) \sin (-\gamma) = -2 \sin \gamma \cos (\alpha + \beta )$.
Combining these,$E = 2 \sin \gamma [\cos (\beta - \alpha ) - \cos (\alpha + \beta )]$.
Using the identity $\cos (A-B) - \cos (A+B) = 2 \sin A \sin B$:
$E = 2 \sin \gamma [2 \sin \alpha \sin \beta] = 4 \sin \alpha \sin \beta \sin \gamma $.

(A) Given,$m \tan (\theta - 30^\circ) = n \tan (\theta + 120^\circ)$.
$\frac{m}{n} = \frac{\tan (\theta + 120^\circ)}{\tan (\theta - 30^\circ)}$.
Applying componendo and dividendo:
$\frac{m + n}{m - n} = \frac{\tan (\theta + 120^\circ) + \tan (\theta - 30^\circ)}{\tan (\theta + 120^\circ) - \tan (\theta - 30^\circ)}$.
Using $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$ and $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$:
$\frac{m + n}{m - n} = \frac{\sin((\theta + 120^\circ) + (\theta - 30^\circ))}{\sin((\theta + 120^\circ) - (\theta - 30^\circ))} = \frac{\sin(2\theta + 90^\circ)}{\sin(150^\circ)}$.
Since $\sin(2\theta + 90^\circ) = \cos 2\theta$ and $\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$:
$\frac{m + n}{m - n} = \frac{\cos 2\theta}{1/2} = 2 \cos 2\theta$.

(C) Given expression: $1 + \cos 2x + \cos 4x + \cos 6x$
Group the terms: $(1 + \cos 6x) + (\cos 2x + \cos 4x)$
Using the identities $1 + \cos 2\theta = 2\cos^2 \theta$ and $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$:
$= 2\cos^2 3x + 2\cos(\frac{4x+2x}{2})\cos(\frac{4x-2x}{2})$
$= 2\cos^2 3x + 2\cos 3x \cos x$
Factor out $2\cos 3x$:
$= 2\cos 3x(\cos 3x + \cos x)$
Using $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$ again:
$= 2\cos 3x \cdot 2\cos(\frac{3x+x}{2})\cos(\frac{3x-x}{2})$
$= 4\cos 3x \cos 2x \cos x$
Thus,the correct option is $C$.

(B) We have $\frac{\tan 70^o - \tan 20^o}{\tan 50^o}$.
Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$= \frac{\frac{\sin 70^o}{\cos 70^o} - \frac{\sin 20^o}{\cos 20^o}}{\tan 50^o}$
$= \frac{\sin 70^o \cos 20^o - \cos 70^o \sin 20^o}{\cos 70^o \cos 20^o \tan 50^o}$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$,the numerator becomes $\sin(70^o - 20^o) = \sin 50^o$.
$= \frac{\sin 50^o}{\cos 70^o \cos 20^o \tan 50^o}$
Since $\tan 50^o = \frac{\sin 50^o}{\cos 50^o}$,we have:
$= \frac{\sin 50^o \cos 50^o}{\cos 70^o \cos 20^o \sin 50^o} = \frac{\cos 50^o}{\cos 70^o \cos 20^o}$
Multiply numerator and denominator by $2$:
$= \frac{2 \cos 50^o}{2 \cos 70^o \cos 20^o}$
Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$:
$= \frac{2 \cos 50^o}{\cos(70^o + 20^o) + \cos(70^o - 20^o)}$
$= \frac{2 \cos 50^o}{\cos 90^o + \cos 50^o}$
Since $\cos 90^o = 0$:
$= \frac{2 \cos 50^o}{0 + \cos 50^o} = \frac{2 \cos 50^o}{\cos 50^o} = 2$.

(B) Given $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5}$.
Since $\theta$ and $\phi$ are acute angles,$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ and $\sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$.
Using the formula $\cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi$,we get:
$\cos(\theta - \phi) = \left(\frac{3}{5}\right)\left(\frac{4}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{3}{5}\right) = \frac{12}{25} + \frac{12}{25} = \frac{24}{25}$.
We know that $2\cos^2\left(\frac{\theta - \phi}{2}\right) = 1 + \cos(\theta - \phi)$.
Substituting the value,$2\cos^2\left(\frac{\theta - \phi}{2}\right) = 1 + \frac{24}{25} = \frac{49}{25}$.
Therefore,$\cos^2\left(\frac{\theta - \phi}{2}\right) = \frac{49}{50}$.
Taking the square root,$\cos\left(\frac{\theta - \phi}{2}\right) = \sqrt{\frac{49}{50}} = \frac{7}{5\sqrt{2}}$.

(A) Expanding the expression:
$(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = \cos ^2 \alpha + \cos ^2 \beta + 2\cos \alpha \cos \beta + \sin ^2 \alpha + \sin ^2 \beta + 2\sin \alpha \sin \beta$
Using the identity $\sin ^2 \theta + \cos ^2 \theta = 1$:
$= (\cos ^2 \alpha + \sin ^2 \alpha) + (\cos ^2 \beta + \sin ^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)$
$= 1 + 1 + 2\cos (\alpha - \beta)$
$= 2 + 2\cos (\alpha - \beta)$
$= 2(1 + \cos (\alpha - \beta))$
Using the identity $1 + \cos \theta = 2\cos ^2 \frac{\theta}{2}$:
$= 2 \times 2\cos ^2 \left( \frac{\alpha - \beta }{2} \right)$
$= 4\cos ^2 \left( \frac{\alpha - \beta }{2} \right)$.

(D) We know the trigonometric identity $\cos(2A) = 1 - 2\sin^2(A)$.
Substituting $A = \frac{\pi}{4} + \theta$,we get:
$1 - 2\sin^2\left( \frac{\pi}{4} + \theta \right) = \cos\left( 2\left( \frac{\pi}{4} + \theta \right) \right)$
$= \cos\left( \frac{\pi}{2} + 2\theta \right)$
Using the identity $\cos\left( \frac{\pi}{2} + x \right) = -\sin(x)$,we have:
$= -\sin(2\theta)$.

(C) We know that $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
$\tan 15^\circ = \tan(45^\circ - 30^\circ)$
$= \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$
$= \frac{1 - 1/\sqrt{3}}{1 + 1 \times 1/\sqrt{3}}$
$= \frac{(\sqrt{3} - 1)/\sqrt{3}}{(\sqrt{3} + 1)/\sqrt{3}}$
$= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
To rationalize the denominator,multiply the numerator and denominator by $(\sqrt{3} - 1)$:
$= \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$
$= \frac{3 + 1 - 2\sqrt{3}}{3 - 1}$
$= \frac{4 - 2\sqrt{3}}{2}$
$= 2 - \sqrt{3}$.

(D) Given,$\tan \theta = \frac{1}{7}$ and $\sin \phi = \frac{1}{\sqrt{10}}$.
Since $\theta$ is in the $1^{st}$ quadrant,$\sin \theta = \frac{1}{\sqrt{50}}$ and $\cos \theta = \frac{7}{\sqrt{50}}$.
Since $\phi$ is in the $1^{st}$ quadrant,$\cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - \frac{1}{10}} = \frac{3}{\sqrt{10}}$.
Now,$\cos 2\phi = 2\cos^2 \phi - 1 = 2(\frac{9}{10}) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$.
And $\sin 2\phi = 2\sin \phi \cos \phi = 2(\frac{1}{\sqrt{10}})(\frac{3}{\sqrt{10}}) = \frac{6}{10} = \frac{3}{5}$.
Now,consider $\cos(\theta + 2\phi) = \cos \theta \cos 2\phi - \sin \theta \sin 2\phi$.
$\cos(\theta + 2\phi) = (\frac{7}{\sqrt{50}})(\frac{8}{10}) - (\frac{1}{\sqrt{50}})(\frac{6}{10}) = \frac{56 - 6}{10\sqrt{50}} = \frac{50}{10(5\sqrt{2})} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\theta$ and $\phi$ are in the $1^{st}$ quadrant,$\theta + 2\phi = 45^\circ$.

(B) Given $\sin A = n \sin B,$ we have $\frac{n}{1} = \frac{\sin A}{\sin B}.$
Applying componendo and dividendo,we get $\frac{n - 1}{n + 1} = \frac{\sin A - \sin B}{\sin A + \sin B}.$
Using the sum-to-product formulas $\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$ and $\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2},$
$\frac{n - 1}{n + 1} = \frac{2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}}{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}} = \cot \frac{A + B}{2} \tan \frac{A - B}{2}.$
Therefore,$\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = \tan \frac{A - B}{2}.$

(A) Given $\sin \alpha = 1/\sqrt{5}$,then $\cos \alpha = \sqrt{1 - (1/\sqrt{5})^2} = 2/\sqrt{5}$.
Given $\sin \beta = 3/5$,then $\cos \beta = \sqrt{1 - (3/5)^2} = 4/5$.
Using the formula $\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha$:
$\sin(\beta - \alpha) = (3/5)(2/\sqrt{5}) - (4/5)(1/\sqrt{5}) = 6/(5\sqrt{5}) - 4/(5\sqrt{5}) = 2/(5\sqrt{5})$.
Since $5\sqrt{5} \approx 11.18$,$\sin(\beta - \alpha) = 2/11.18 \approx 0.1789$.
We know $\sin 0 = 0$ and $\sin(\pi/4) = 1/\sqrt{2} \approx 0.707$.
Since $0 < 0.1789 < 0.707$,it follows that $0 < \beta - \alpha < \pi/4$.
Thus,$\beta - \alpha$ lies in the interval $[0, \pi/4]$.