Question

In a $\Delta ABC$. with usual notation, match the items in List - I with the items in List - II and choose the correct option.List I	List II
(A)	$r_1r_2\sqrt{\bigg(\frac{4R-r_1-r_2}{r_1+r_2}\bigg)}$
(B)	$\frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$
(C)	$\frac{a}{c}=\frac{sin(A-B)}{sin(B-C)}$
(D)	$bc\;cos^2\frac{A}{2}$
	5

• A. (A)-(4), (B)-(3), (C)-(1), (D)-(5)
• B. (A)-(5), (B)-(4), (C)-(3), (D)-(2)
• C. (A)-(3) , (B)-(1), (C)-(2) , (D)-(5)
• D. (A)-(4) , (B)-(5), (C)-(2) , (D)-(1)

Answer 1

Answer: (C)

(A)-(3) , (B)-(1), (C)-(2) , (D)-(5)

Answer 2

(A) $r_{1} r_{2} \sqrt{\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}=\left[\frac{\Delta^{2}}{(s-a)(s-b)}\right.$ $\left.\sqrt{\frac{4 R-4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right)}{4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right)}}\right]$

$=\frac{\Delta^{2}}{(s-a)(s-b)} \sqrt{\frac{4 R\left(1-\cos ^{2} \frac{C}{2}\right)}{4 R \cos ^{2} \frac{C}{2}}}$

$=\frac{\Delta^{2}}{(s-a)(s-b)} \tan \frac{C}{2}$

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

$=\frac{\Delta^{2}}{\Delta}=\Delta$

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(B) $\frac{r_{2}\left(r_{3}+r_{1}\right)}{\sqrt{r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}}}$

$=\frac{\frac{\Delta}{s-b}\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-a}\right)}{\sqrt{\frac{(s-a)+(s-b)+(s-c)}{(s-a)(s-b)(s-c)}}}$

$=\frac{\frac{\Delta}{(s-b)} \times \frac{2 S-a-c}{(s-a)(s-c)}}{\sqrt{\frac{3 S-a-b-c}{(s-a)(s-b)(s-c)}}}=\frac{\Delta(b)}{\sqrt{s(s-a)(s-b)(s-c)}}=b$

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(C) $\frac{a}{c}=\frac{\sin (A-B)}{\sin (B-C)}$

$\Rightarrow \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$
$\Rightarrow \sin A \sin (B-C)=\sin (A-B) \sin C$
$\Rightarrow \sin A(\sin B \cos C-\sin C \cos B)$ $=(\sin A \cos B-\cos A \sin B) \sin C$
$\Rightarrow 2 \sin A \cos B \sin C=\sin A \sin B \cos C$ $+\sin B \cos A \sin C$

$\Rightarrow 2 \frac{a}{2 R} \times \frac{a^{2}+c^{2}-b^{2}}{2 a c} \times \frac{c}{2 R}$ $=\left(\frac{a}{2 R} \times \frac{b}{2 R} \times \frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)+$ $\left(\frac{b}{2 R} \times \frac{b^{2}+c^{2}-a^{2}}{2 b c} \times \frac{c}{2 R}\right)$

$\Rightarrow 2\left(a^{2}+c^{2}-b^{2}\right)=a^{2}+b^{2}-c^{2}+b^{2}+c^{2}-a^{2}$
$\Rightarrow 2 a^{2}+2 c^{2}-2 b^{2}=2 b^{2}$
$\Rightarrow 2 b^{2}=a^{2}+c^{2}$
$\Rightarrow a^{2}, b^{2}, c^{2}$ are in AP.

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(D) $b c \cos ^{2} \frac{A}{2}=b c \times \frac{s(s-a)}{b c}=s(s-a)$

Therefore, the correct option is (C) : (A)-(3) , (B)-(1), (C)-(2) , (D)-(5)