In a triangle ABC, with the usual notations, $\angle B = \dfrac{\pi}{3}$, $\angle C = \dfrac{\pi}{4}$. If D divides BC internally in the ratio 1:3, then $\dfrac{\sin \angle BAD}{\sin \angle CAD} =$

Let $P_1$, $P_2$, $P_3$ be the altitudes of a triangle ABC from the vertices A, B, C respectively. If $\triangle$ denotes the area of the triangle and s is the semi-perimeter of the triangle, then $\dfrac{\cos A}{P_1} + \dfrac{\cos B}{P_2} + \dfrac{\cos C}{P_3} =$

In $\triangle ABC$, with the usual notations, $\angle C = 90^\circ$, then $\sin(A - B)$ is equal to....

In $\triangle ABC$ with usual notations, if $1 + \tan\left(\dfrac{A}{2}\right)\tan\left(\dfrac{B}{2}\right) = \dfrac{k}{s}$ (where $s$ is the semi-perimeter), then the value of $k$ is...