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

Step 1: Given Data
We are given a triangle \( ABC \) with the following sides:
- \( AB = \sqrt{23} \)
- \( BC = 3 \)
- \( CA = 4 \)
We are asked to find the value of \( \frac{\cot A + \cot C}{\cot B} \).

Step 2: Applying the Cotangent Formula
The cotangent of an angle in a triangle can be expressed using the law of cosines and the sides of the triangle. Specifically:
\[ \frac{\cot A + \cot C}{\cot B} = \frac{\frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}} \] We can rewrite this expression using the formula for cosine in terms of the sides of the triangle:
\[ \frac{\cot A + \cot C}{\cot B} = \frac{\frac{b^2 + c^2 - a^2}{2bc} + \frac{a^2 + b^2 - c^2}{2ab}}{\frac{c^2 + a^2 - b^2}{2ac}} \]

Step 3: Simplifying the Expression
Now, simplifying the above expression, we have:
\[ \frac{\cot A + \cot C}{\cot B} = \frac{\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta}}{\frac{c^2 + a^2 - b^2}{4\Delta}} \] where \( \Delta \) is the area of the triangle.
Simplifying further, we get: \[ \frac{\cot A + \cot C}{\cot B} = \frac{2b^2}{a^2 + c^2 - b^2} \]

Step 4: Final Calculation
Substituting the given values of \( a = 3 \), \( b = 4 \), and \( c = \sqrt{23} \), we get:
\[ \frac{\cot A + \cot C}{\cot B} = \frac{2 \times 4^2}{3^2 + (\sqrt{23})^2 - 4^2} \] Simplifying the expression: \[ = \frac{2 \times 16}{9 + 23 - 16} \] \[ = \frac{32}{16} = 2 \]

Final Answer:
The value of \( \frac{\cot A + \cot C}{\cot B} \) is \( 2 \).