Skip to main content

In a triangle ABC, let $AB=\sqrt {23}$, $BC=3$ and ....

In a triangle ABC, let $AB=\sqrt {23}$, $BC=3$ and $CA=4$. Then the value of

$$\frac {cot A + cot C}{cot B}$$

is _______ .

Solution

$\frac {cot A + cot C}{cot B}=\frac {sin (A+C).sin B}{sin A sin C.cos B}$

$= \frac {sin^2 B}{sin A. sin C. cos B}$

$= \frac {b^2}{a.c. \frac {c^2+a^2-b^2}{2ac}}$

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

Putting $AB = c = \sqrt {23},BC=a=3,CA=b=4$ we have, $ \frac {2.4^2}{23+3^2-4^2}$

$= \frac {2\times 16}{23+9-16}=\frac {2\times 16}{16}=2$

Popular posts from this blog