A straight line through the vertex p of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then :

Let $${S_n} = \sum\limits_{k = 1}^n {{n \over {{n^2} + kn + {k^2}}}} $$ and $${T_n} = \sum\limits_{k = 0}^{n - 1} {{n \over {{n^2} + kn + {k^2}}}} $$ for $$n$$ $$=1, 2, 3, ............$$ Then,

If $$0 < x < 1$$, then

$$\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cot }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}}^2} - 1} \right]^{1/2}} = $$

A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.