If (α+β) = (3)/(5) and α = (5)/(13) , where α+β lies in the IV^ th...

Given $\cos(\alpha+\beta) = \dfrac{3}{5}$ and $\alpha+\beta$ is in the $\text{IV}^{\text{th}}$ quadrant.
So, $\sin(\alpha+\beta) = -\dfrac{4}{5}$ and $\tan(\alpha+\beta) = -\dfrac{4}{3}$.

Also given $\sin\alpha = \dfrac{5}{13}$ and $\alpha$ is in the $\text{II}^{\text{nd}}$ quadrant.
So, $\cos\alpha = -\dfrac{12}{13}$ and $\tan\alpha = -\dfrac{5}{12}$.

We can write $\beta = (\alpha+\beta) - \alpha$.
Using the formula $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$, we get:
$\tan\beta = \dfrac{\tan(\alpha+\beta) - \tan\alpha}{1 + \tan(\alpha+\beta)\tan\alpha}$
$\Rightarrow \tan\beta = \dfrac{-\dfrac{4}{3} - \left(-\dfrac{5}{12}\right)}{1 + \left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{12}\right)} = \dfrac{\dfrac{-16 + 5}{12}}{1 + \dfrac{20}{36}} = \dfrac{-\dfrac{11}{12}}{\dfrac{56}{36}}$
$\Rightarrow \tan\beta = -\dfrac{11}{12} \times \dfrac{36}{56} = -\dfrac{33}{56}$.

To find the quadrant of $\beta$, we check the signs of $\sin\beta$ and $\cos\beta$:
$\sin\beta = \sin((\alpha+\beta) - \alpha) = \sin(\alpha+\beta)\cos\alpha - \cos(\alpha+\beta)\sin\alpha$
$\Rightarrow \sin\beta = \left(-\dfrac{4}{5}\right)\left(-\dfrac{12}{13}\right) - \left(\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right) = \dfrac{48}{65} - \dfrac{15}{65} = \dfrac{33}{65} > 0$.

$\cos\beta = \cos((\alpha+\beta) - \alpha) = \cos(\alpha+\beta)\cos\alpha + \sin(\alpha+\beta)\sin\alpha$
$\Rightarrow \cos\beta = \left(\dfrac{3}{5}\right)\left(-\dfrac{12}{13}\right) + \left(-\dfrac{4}{5}\right)\left(\dfrac{5}{13}\right) = -\dfrac{36}{65} - \dfrac{20}{65} = -\dfrac{56}{65} < 0$.

Since $\sin\beta$ is positive and $\cos\beta$ is negative, $\beta$ lies in the $\text{II}^{\text{nd}}$ quadrant.

Answer: $-\dfrac{33}{56}$ and $\text{II}^{\text{nd}}$ quadrant