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Cosine Rule

In a ABC, s...

Question

In a $△ A B C, \frac{sin A}{sin C} = \frac{sin (A - B)}{sin (B - C)},$ then:

A

cot A, cot B, cot C

are in A.P

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B

sin 2 A, sin 2 B, sin 2 C

are in A.P

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C

cos 2 A, cos 2 B, cos 2 C

are in A.P

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D

a sin A, b sin B, c sin C

are in A.P.

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Solution

The correct options are
A $cot A, cot B, cot C$ are in A.P
C $cos 2 A, cos 2 B, cos 2 C$ are in A.P
D $a sin A, b sin B, c sin C$ are in A.P.
$∵ \frac{sin A}{sin C} = \frac{sin (A - B)}{sin (B - C)}$ ................ $(1)$
$\Rightarrow \frac{sin (π - (B + C))}{sin (π - (A + B))} = \frac{sin (A - B)}{sin (B - C)}$
$\Rightarrow sin (π - (B + C)) sin (B - C) = sin (π - (A + B)) sin (A - B)$
$\Rightarrow sin (B + C) sin (B - C) = sin (A + B) sin (A - B)$
$\Rightarrow {sin}^{2} B - {sin}^{2} C = {sin}^{2} A - {sin}^{2} B$
$\Rightarrow 2 {sin}^{2} B = {sin}^{2} A + {sin}^{2} C$
Hence, ${sin}^{2} A, {sin}^{2} B, {sin}^{2} C$ are in A.P. ............. $(2)$
or $- 2 {sin}^{2} A, - 2 {sin}^{2} B, - 2 {sin}^{2} C$ are in A.P.
or $1 - 2 {sin}^{2} A, 1 - 2 {sin}^{2} B, 1 - 2 {sin}^{2} C$ are in A.P.
or $cos 2 A, cos 2 B, cos 2 C$ are in A.P>
From $(2)$
$sin A sin A, sin B sin B, sin C sin C$ are in A.P.
$\Rightarrow \frac{a}{2 R} sin A, \frac{b}{2 R} sin B, \frac{c}{2 R} sin C$ are in A.P
$\Rightarrow a sin A, b sin B, c sin C$ are in A.P
From $(1)$
$\frac{sin (B - C)}{sin B sin C} = \frac{sin (A - B)}{sin A sin B}$
$\Rightarrow \frac{sin B cos C - cos B sin C}{sin B sin C} = \frac{sin A cos B - cos A sin B}{sin A sin B}$
$\Rightarrow cot C - cot B = cot B - cot A$ (on evaluation)
$\Rightarrow 2 cot B = cot A + cot C$
$∴ cot A, cot B, cot C$ are in A.P.

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