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 ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Properties of Triangle

In a triangle ABC, let

A B = 23, B C = 3

and

C A = 4

. Then the value of

\frac{c o t A + c o t C}{c o t B}

is ____.

Answer: 2

Solution

Given

c = 23, a = 3, b = 4

. Using the formula

cot A = \frac{c o s A}{s i n A} = \frac{b ^{2} + c ^{2} - a ^{2}}{4Δ}

where

Δ

is the area of the triangle. Similarly,

cot B = \frac{a ^{2} + c ^{2} - b ^{2}}{4Δ}

and

cot C = \frac{a ^{2} + b ^{2} - c ^{2}}{4Δ}

. Therefore,

\frac{c o t A + c o t C}{c o t B} = \frac{b ^{2} + c ^{2} - a ^{2} + a ^{2} + b ^{2} - c ^{2}}{a ^{2} + c ^{2} - b ^{2}} = \frac{2 b ^{2}}{a ^{2} + c ^{2} - b ^{2}} = \frac{2 ( 16 )}{9 + 23 - 16} = \frac{32}{16} = 2

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Integer answer

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