In triangle ABC, if b2 + c 2 =2a2 , then value of cot⁡Acot⁡B+cot⁡C

cot⁡Acot⁡B+cot⁡C=Rb2+c2−a2abcRa2+c2−b2abc+Ra2+b2−c2abc=b2+c2−a22a2=2a2−a22a2=12

In triangle ABC, if b2 + c 2 =2a2 , then value of cot⁡Acot⁡B+cot⁡C

cot⁡Acot⁡B+cot⁡C=Rb2+c2−a2abcRa2+c2−b2abc+Ra2+b2−c2abc=b2+c2−a22a2=2a2−a22a2=12

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In triangle ABC, if b² + c ² =2a² , then value of $\frac{\cot A}{\cot B + \cot C}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{5}{2}$

$\frac{5}{3}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} \frac{\cot A}{\cot B + \cot C} = \frac{\frac{R (b^{2} + c^{2} - a^{2})}{a b c}}{\frac{R (a^{2} + c^{2} - b^{2})}{a b c} + \frac{R (a^{2} + b^{2} - c^{2})}{a b c}} \\ = \frac{b^{2} + c^{2} - a^{2}}{2 a^{2}} = \frac{2 a^{2} - a^{2}}{2 a^{2}} = \frac{1}{2} \end{aligned}$