If $P_1, P_2$ and $P_3$ are the lengths of the altitudes drawn from the vertices $A, B$ and $C$ of $\triangle ABC$ respectively,then $\frac{\cos A}{P_1} + \frac{\cos B}{P_2} + \frac{\cos C}{P_3} =$

Let $ABC$ be a triangle and let $D$ be the mid-point of $BC$. Suppose $\cot (\angle CAD) : \cot (\angle BAD) = 2 : 1$. If $G$ is the centroid of $\triangle ABC$,then the measure of $\angle BGA$ is (in $^{\circ}$)

Let $S=\{x \in(-\pi, \pi) \mid x \neq 0, \pm \frac{\pi}{2}\}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

In an isosceles triangle $ABC$,$\angle C = \angle A$. If the point of intersection of the bisectors of internal angles $\angle A$ and $\angle C$ divides the median of side $AC$ in the ratio $3 : 1$ (from vertex $B$ to side $AC$),then the value of $\csc \frac{B}{2}$ is equal to

$A$ triangle $ABC$ has area of $P$ square units and perimeter $2S$ units. If $h_1, h_2$ and $h_3$ are respectively the lengths of the altitudes of the triangle drawn from the vertices $A, B$ and $C$,then $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right] =$

In a $\triangle ABC$,the expression $\frac{a}{\tan A} + \frac{b}{\tan B} + \frac{c}{\tan C}$ is equal to: