In a $Δ A B C$ , $\cot A + \cot B + \cot C =$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{a^{2} + b^{2} + c^{2}}{Δ}$

$\frac{a + b + c}{4 Δ}$

$\frac{a^{2} + b^{2} + c^{2}}{4 Δ}$

$\frac{a^{2} + b^{2} + c^{2}}{2 Δ}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let a triangle ABC of sides $a, b, c$ having area $Δ$
Question Image
Area $= Δ$
$\frac{1}{2} b c \sin A = Δ - - - - - - (i)$
and $\frac{1}{2} a c \sin B = Δ - - - - - - (i i)$
and $\frac{1}{2} a b \sin C = Δ - - - - - - (i i i)$
Using cosine rule
$a^{2} = b^{2} + c^{2} - 2 b c \cos A$
and $b^{2} = a^{2} + c^{2} - 2 a c \cos B$
and $c^{2} = a^{2} + b^{2} - 2 a b \cos C$
On adding, we get
$a^{2} + b^{2} + c^{2} = 2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b \cos C - 2 a c \cos B - 2 b c \cos A$
or $a^{2} + b^{2} + c^{2} = 2 (a b \cos C + a c \cos B + b c \cos A) - - - - - (i v)$
Now, from Equation (i)
$b c = \frac{2 Δ}{\sin A}$
Equation (ii), $a c = \frac{2 Δ}{\sin B}$
Equation (iii), $a b = \frac{2 Δ}{\sin C}$
Putting these values in equation (iv), we get
$a^{2} + b^{2} + c^{2} = 2$
$(\frac{2 Δ}{\sin C} \cos C + \frac{2 Δ}{\sin B} \cos B + \frac{2 Δ}{\sin A} \cos A)$
$a^{2} + b^{2} + c^{2} = 4 Δ (\cot C + \cot B + \cot A)$
or $\cot C + \cot B + \cot A = \frac{a^{2} + b^{2} + c^{2}}{4 Δ}$
or $\cot A + \cot B + \cot C = \frac{a^{2} + b^{2} + c^{2}}{4 Δ}$