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

In $\triangle ABC$,if $\angle A=90^{\circ}$,then $(r_2-r_1)(r_3-r_1)=$

If $p_1, p_2, p_3$ are the altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively,then with the usual notation,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=$

Assertion $(A)$: If $A=15^{\circ}, B=17^{\circ}$ and $C=13^{\circ}$,then $\cot 2A + \cot 2B + \cot 2C = \cot 2A \cot 2B \cot 2C$.
Reason $(R)$: In a $\triangle PQR$,$\tan \frac{P}{2} \tan \frac{Q}{2} + \tan \frac{Q}{2} \tan \frac{R}{2} + \tan \frac{P}{2} \tan \frac{R}{2} = 1$.
The correct option among the following is:

If the reciprocals of the lengths of the sides of a $\triangle ABC$ are in harmonic progression,then its ex-radii $r_1, r_2, r_3$ are in

In a right-angled triangle,the hypotenuse is $2\sqrt{2}$ times the length of the perpendicular drawn from the opposite vertex to the hypotenuse. Then,the other two angles are: