If b+c=3a,then cot frac{B}{2} cot frac{C}{2} | vedclass

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

Question

(D) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Multiplying these,we

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Multiplying these,we get $\cot \frac{B}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.Given $b+c=3a$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.Substituting $s=2a$ into the expression,we get $\frac{2a}{2a-a} = \frac{2a}{a} = 2$.

If $b+c=3a$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

Similar Questions

Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time $t_1$ as a function of time $t$ is given by:

For the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$,which of the following remains constant as $\alpha$ varies?

In the reaction,$2SO_2 + O_2 \xrightarrow[As_2O_3]{Pt} 2SO_3$,$As_2O_3$ acts as:

Phenol can be converted into salicylic acid by heating with

The solution of the equation $\cos^2 \theta + \sin \theta + 1 = 0$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module