Question

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 ______

Answer 1

Correct Answer: 2

We have, \(c = \sqrt{23}\), a = 3 and b = 4

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

= \(\frac{\left(\frac{b^2 + c^2 - a^2}{2bc} \sin A + \frac{a^2 + b^2 - c^2}{2ab} \sin C\right)}{\frac{c^2 + a^2 - b^2}{2ac} \sin B}\)

=\(\frac{\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta}}{\frac{c^2 + a^2 - b^2}{4\Delta}}\)

= \(\frac{2b^2}{a^2+c^2-b^2}\)

\(= 2\)