JEE Advanced 2021 Paper 1 · Q18 · Trigonometric Identities

Question

In a triangle $ABC$, let $AB = \sqrt{23}$, $BC = 3$ and $CA = 4$. Then the value of $\dfrac{\cot A + \cot C}{\cot B}$ is _____ .

SolveKar AI · Accepted Answer

Sides: $c=AB=\sqrt{23}$, $a=BC=3$, $b=CA=4$. Using $\cot A+\cot C=\dfrac{\sin(A+C)}{\sin A\sin C}=\dfrac{\sin B}{\sin A\sin C}$, and $\cot B=\dfrac{\cos B}{\sin B}$, we get $\dfrac{\cot A+\cot C}{\cot B}=\dfrac{\sin^2 B}{\sin A\sin C\cos B}$. Using law of sines $\sin A/a=\sin B/b=\sin C/c$ and law of cosines $\cos B=\dfrac{a^2+c^2-b^2}{2ac}$, this simplifies to $\dfrac{b^2}{a^2+c^2-b^2}\cdot 2=\dfrac{2b^2}{a^2+c^2-b^2}=\dfrac{2\cdot 16}{9+23-16}=\dfrac{32}{16}=2$.

JEE Advanced 2021 Paper 1 · Q18 · Trigonometric Identities

Solution