Given that $\cos(A+B)= -\dfrac{5}{13}$ and $\sin A\sin B=\dfrac{15}{26}$, where A and B are acute angles and angle A > angle B. Find $\cos A\cos B$. Using your result in (i), find $\tan(A-B)$. Give your answer in exact form. Show that $\displaystyle\frac{1+\sin A+\cos A}{1+\sin A-\cos A}=\cot\dfrac{A}{2}$.

1. Apply the formula for $\cos(A+B)$ and substitute the given values.

Using
$\cos(A+B)=\cos A\cos B-\sin A\sin B$,

$-\dfrac{5}{13}=\cos A\cos B-\dfrac{15}{26}$

$\cos A\cos B=-\dfrac{5}{13}+\dfrac{15}{26}$

$\cos A\cos B=-\dfrac{10}{26}+\dfrac{15}{26}=\dfrac{5}{26}$

1. Find $\sin A\cos B$ using the result from part (i).

From part (i),
$\cos A\cos B=\dfrac{5}{26}$.

Using
$\cos(A+B)=\cos A\cos B-\sin A\sin B$,
we have
$-\dfrac{5}{13}=\dfrac{5}{26}-\sin A\sin B$,
which is consistent with $\sin A\sin B=\dfrac{15}{26}$.

To find $\tan(A-B)$, use
$\tan A\tan B=\frac{\sin A\sin B}{\cos A\cos B}=\frac{15/26}{5/26}=3.$

2. Use the tangent subtraction identity and determine the exact value.

Also,
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$=\frac{5}{26}+\frac{15}{26}=\frac{20}{26}=\frac{10}{13}.$

Since $A>B$ and both are acute, $A-B$ is acute, so
$\sin(A-B)=\sqrt{1-\cos^2(A-B)}$
$=\sqrt{1-\left(\frac{10}{13}\right)^2}$
$=\sqrt{\frac{69}{169}}=\frac{\sqrt{69}}{13}.$

Hence
$\tan(A-B)=\frac{\sin(A-B)}{\cos(A-B)}=\frac{\sqrt{69}/13}{10/13}=\frac{\sqrt{69}}{10}.$

1. Express $1$, $\sin A$ and $\cos A$ using half-angle notation.

Let $x=\dfrac A2$.

Then
$\sin A=2\sin x\cos x,\qquad \cos A=\cos^2x-\sin^2x,$
and
$1=\sin^2x+\cos^2x.$

2. Simplify the numerator in terms of $\sin x$ and $\cos x$.

Numerator:

$$\begin{aligned} 1+\sin A+\cos A &=\sin^2x+\cos^2x+2\sin x\cos x+\cos^2x-\sin^2x\\ &=2\cos^2x+2\sin x\cos x\\ &=2\cos x(\cos x+\sin x). \end{aligned}$$

3. Simplify the denominator in terms of $\sin x$ and $\cos x$.

Denominator:

$$\begin{aligned} 1+\sin A-\cos A &=\sin^2x+\cos^2x+2\sin x\cos x-(\cos^2x-\sin^2x)\\ &=2\sin^2x+2\sin x\cos x\\ &=2\sin x(\sin x+\cos x). \end{aligned}$$

4. Cancel common factors and identify the result as $\cot \dfrac A2$.

Hence

$$\begin{aligned} \frac{1+\sin A+\cos A}{1+\sin A-\cos A} &=\frac{2\cos x(\cos x+\sin x)}{2\sin x(\sin x+\cos x)}\\ &=\frac{\cos x}{\sin x}\\ &=\cot x\\ &=\cot\frac A2. \end{aligned}$$