IIT JEE , AIEEE Forum - Mathematics , Trignometry

ramkumar_november

$sinA=\frac{a}{2R}$

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cotA=\frac{cosA}{sinA}=\frac{2R(b^2+c^2-a^2)}{2abc}$

$cotA=\frac{b^2+c^2-a^2}{4\bigtriangleup}$

similarly

$cotB=\frac{a^2+c^2-b^2}{4\bigtriangleup}$

$cotC=\frac{a^2+b^2-c^2}{4\bigtriangleup}$

$\frac{cotC}{cotA+cotB}=\frac{a^2+b^2-c^2}{b^2+c^2-a^2+a^2+c^2-b^2}$

$\frac{cotC}{cotA+cotB}=\frac{a^2+b^2-c^2}{2c^2}$

$\frac{cotC}{cotA+cotB}=\frac{2007c^2-c^2}{2c^2}$

$\frac{cotC}{cotA+cotB}=1003$

PS: please type the question properly without any errors..... you have typed cotA/(cotB+cotA)