In ΔABC,r1.r2[4R−r1−r2r1+r2]=

r1r24R=r1−r2r1+r2=r1r2r3−rr1+r2=Δ2(s−a)(s−b)cs(s−c)2s−a−b(s−a)(s−b)=ss−ccs(s−c)c(s−a)(s−c)=s(s−c)tanc2=s.r=Δr.r=Δ

In ΔABC,r1.r2[4R−r1−r2r1+r2]=

r1r24R=r1−r2r1+r2=r1r2r3−rr1+r2=Δ2(s−a)(s−b)cs(s−c)2s−a−b(s−a)(s−b)=ss−ccs(s−c)c(s−a)(s−c)=s(s−c)tanc2=s.r=Δr.r=Δ

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In $Δ A B C, r_{1} . r_{2} \sqrt{[\frac{4 R - r_{1} - r_{2}}{r_{1} + r_{2}}] =}$

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$r_{1} r_{2} r_{3}$

$(r_{1} + r_{2}) \sec^{2} \frac{A}{2}$

$Δ$

answer is C.

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Detailed Solution

$r_{1} r_{2} \sqrt{\frac{4 R = r_{1} - r_{2}}{r_{1} + r_{2}}} = r_{1} r_{2} \sqrt{\frac{r_{3} - r}{r_{1} + r_{2}}}$

$= \frac{Δ^{2}}{(s - a) (s - b)} \sqrt{\frac{\frac{c}{s (s - c)}}{\frac{2 s - a - b}{(s - a) (s - b)}}}$

$= s (s - c \sqrt{\frac{\frac{c}{s (s - c)}}{\frac{c}{(s - a) (s - c)}}})$

$\begin{array}{l} = s (s - c) \tan \frac{c}{2} \\ = s . r = \frac{Δ}{r} . r = Δ \end{array}$

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In ΔABC,r1.r2[4R−r1−r2r1+r2]=

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In $Δ A B C, r_{1} . r_{2} \sqrt{[\frac{4 R - r_{1} - r_{2}}{r_{1} + r_{2}}] =}$