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

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(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

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

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

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