In triangle ABC,(r_1+r_2) operatorname{cosec} | vedclass

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

Question

(C) In $	riangle ABC$,$A+B+C=\pi$. We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ and $r_2 = 4R \cos \frac{A}{2} \sin \fr

Vedclass · Accepted Answer

$4 R \cot ^2 \frac{C}{2}$

Vedclass · Answer

(C) In $	riangle ABC$,$A+B+C=\pi$. We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ and $r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$. Adding these,$r_1+r_2 = 4R \cos \frac{C}{2} [\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}]$. Using the identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$,we get $r_1+r_2 = 4R \cos \frac{C}{2} \sin(\frac{A+B}{2})$. Since $A+B = \pi - C$,$\sin(\frac{A+B}{2}) = \sin(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2}$. Thus,$r_1+r_2 = 4R \cos^2 \frac{C}{2}$. Now,$(r_1+r_2) \operatorname{cosec}^2 \frac{C}{2} = \frac{4R \cos^2 \frac{C}{2}}{\sin^2 \frac{C}{2}} = 4R \cot^2 \frac{C}{2}$.

In $\triangle ABC$,$(r_1+r_2) \operatorname{cosec}^2 \frac{C}{2} =$

Similar Questions

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

The smallest angle of the triangle whose sides are $6 + \sqrt{12}$,$\sqrt{48}$,and $\sqrt{24}$ is:

If in a triangle $ABC$,$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$,then $\cos A$ is equal to

With usual notations, in $\triangle ABC$, if $\cos C = \frac{\sin A}{2 \sin B}$, then which of the following is true?

In a triangle $ABC$,if $2a^2b^2 + 2b^2c^2 = a^4 + b^4 + c^4$,then angle $B$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module