If in any Delta ABC,cot frac{A}{2}, cot frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$This implies $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}

Vedclass · Accepted Answer

$\cot \frac{A}{2} \cot \frac{C}{2} = 3$

Vedclass · Answer

(B) Given that $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$This implies $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2}$.Using the identity $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$,we have $\cot \frac{A}{2} = \frac{s-a}{r}$,where $r$ is the inradius and $s$ is the semi-perimeter.Substituting this into the $A.P.$ condition: $2 \frac{s-b}{r} = \frac{s-a}{r} + \frac{s-c}{r}$.$2(s-b) = s-a + s-c \implies 2s - 2b = 2s - (a+c) \implies 2b = a+c$.This means $a, b, c$ are in $A.P.$For an equilateral triangle where $A=B=C=60^{\circ}$,$\cot 30^{\circ} = \sqrt{3}$.Then $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{3} 	imes \sqrt{3} = 3$. Thus,option $(b)$ is correct.

If in any $\Delta ABC$,$\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$,then:

Similar Questions

If in a triangle $ABC$,$a=2$,$b=3$ and $c=4$,then $\tan \left(\frac{A}{2}\right) = $

In $\triangle ABC$,if $A$ is an acute angle,$b=6, c=9$ and $\sin A=\frac{2 \sqrt{14}}{9}$,then $3a(\cos B+\cos C)=$

In triangle $ABC$,if $A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$,then $c =$

In triangle $ABC$,if $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$,then $\frac{a^2-b^2}{c^2}=$

The sides of a triangle are in the ratio $1 : \sqrt{3} : 2$. Then the angles are in the ratio

Vedclass Test Series

Exam Paper Generator

Online Exam Module