MHT CET · Maths · Properties of Triangles
In a triangle \(\mathrm{ABC}, l(\mathrm{AB})=\sqrt{23}\) units, \(l(\mathrm{BC})=3\) units, \(l(\mathrm{CA})=4\) units, then \(\frac{\cot A+\cot C}{\cot B}\) is
- A 1
- B 2
- C 4
- D 8
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{\cot A+\cot C}{\cot B} \\ = & \frac{\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}\end{aligned}\)
\(=\frac{\frac{b^2+c^2-a^2}{2 b c \sin A}+\frac{a^2+b^2-c^2}{2 a b \sin C}}{\frac{c^2+a^2-b^2}{2 c a \sin B}}\)
\(=\frac{\frac{b^2+c^2-a^2}{2(2 \Delta)}+\frac{a^2+b^2-c^2}{2(2 \Delta)}}{\frac{c^2+a^2-b^2}{2(2 \Delta)}}\)
\(\begin{aligned} & =\frac{2 b^2}{c^2+a^2-b^2} \\ & =\frac{2(4)^2}{(\sqrt{23})^2+(3)^2-(4)^2} \\ & =\frac{32}{16} \\ & =2\end{aligned}\)
\(=\frac{\frac{b^2+c^2-a^2}{2 b c \sin A}+\frac{a^2+b^2-c^2}{2 a b \sin C}}{\frac{c^2+a^2-b^2}{2 c a \sin B}}\)
\(=\frac{\frac{b^2+c^2-a^2}{2(2 \Delta)}+\frac{a^2+b^2-c^2}{2(2 \Delta)}}{\frac{c^2+a^2-b^2}{2(2 \Delta)}}\)
\(\begin{aligned} & =\frac{2 b^2}{c^2+a^2-b^2} \\ & =\frac{2(4)^2}{(\sqrt{23})^2+(3)^2-(4)^2} \\ & =\frac{32}{16} \\ & =2\end{aligned}\)
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