QJEEAdv21Ma1456: JEE Advanced 2021 Mathematics - Class 11 JEEAdv21Ma1456 Solution: In a triangle ABC, let AB = , BC = 3 and CA = 4. Then the value …

Step-by-Step Verified Solution

Given,

$AB = \sqrt {23} = c$

$BC = 3 = a$

$CA = 4 = b$


Now,

${{\cot A + \cot C} \over {\cot B}} = {{{{\cos A} \over {\sin A}} + {{\cos C} \over {\sin C}}} \over {{{\cos B} \over {\sin B}}}}$

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

$= {{{{{b^2} + {c^2} - {a^2}} \over {4\Delta }} + {{{a^2} + {b^2} - {c^2}} \over {4\Delta }}} \over {{{{c^2} + {a^2} - {b^2}} \over {4\Delta }}}}$

$= {{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}} \over {{c^2} + {a^2} - {b^2}}}$

$= {{2{b^2}} \over {{a^2} + {c^2} - {b^2}}} = {{2 \times 16} \over {9 + 23 - 16}} = {{32} \over {16}} = 2$