Show that the value of $\sin 105^\circ$ can be expressed as $\tfrac{1}{4}(p+q)$, where $p$ and $q$ are constants. Given that $\dfrac{\cos(A+B)}{\cos(A-B)}=\dfrac{2}{7}$ and $\sin A\sin B\ne 0$, find the exact value of $\cot A\cot B$. Given that $B$ is a reflex angle and $\cot B = 3$, find the exact value of $\cos 2B$.

1. Express $105^\circ$ as a sum of special angles

$\sin 105^\circ = \sin(60^\circ + 45^\circ)$

2. Apply the sine addition formula

$\sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$

3. Substitute exact values and simplify to the required form

$= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)$
$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
$= \frac{1}{4}(\sqrt{6} + \sqrt{2})$

1. Expand the numerator and denominator using the addition and subtraction formulae for cosine.

$\frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B} = \frac{2}{7}$

2. Cross-multiply to form a linear equation in terms of $\cos A \cos B$ and $\sin A \sin B$.

$7(\cos A \cos B - \sin A \sin B) = 2(\cos A \cos B + \sin A \sin B)$
$7\cos A \cos B - 7\sin A \sin B = 2\cos A \cos B + 2\sin A \sin B$

3. Group like terms and solve for $\cot A \cot B$.

$5\cos A \cos B = 9\sin A \sin B$
Since $\sin A \sin B \neq 0$, we can divide both sides by $5\sin A \sin B$:
$\frac{\cos A \cos B}{\sin A \sin B} = \frac{9}{5}$
$\left(\frac{\cos A}{\sin A}\right)\left(\frac{\cos B}{\sin B}\right) = \frac{9}{5}$
$\cot A \cot B = \frac{9}{5}$

1. Find the value of $\tan B$ or $\sin^2 B$ from $\cot B$.

Since $\cot B = 3$, we have $\tan B = \frac{1}{3}$.
Alternatively, using $\operatorname{cosec}^2 B = 1 + \cot^2 B$:
$\operatorname{cosec}^2 B = 1 + 3^2 = 10 \implies \sin^2 B = \frac{1}{10}$.

2. Use a double angle formula to evaluate $\cos 2B$.

Using $\cos 2B = 1 - 2\sin^2 B$:
$\cos 2B = 1 - 2\left(\frac{1}{10}\right) = 1 - \frac{1}{5} = \frac{4}{5}$.

(Note: The quadrant of $B$ does not affect the value of $\cos 2B$ since it depends on squared trigonometric ratios.)