In any $Δ A B C$ , if $a^{2}, b^{2}, c^{2}$ are in AP then prove that $c o t A, c o t B, c o t C$ , are in AP

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Solution

$G i v e a^{2}, b^{2}, c^{2} a r e i n A P$

$b^{2} - a^{2} = c^{2} - b^{2}$

$k^{2} {sin}^{2} B - k^{2} {sin}^{2} A = k^{2} {sin}^{2} C - k^{2} {sin}^{2} B$

$sin (B + A) sin (B - A) = sin (C + B) sin (C - B)$

$sin (π - C) sin (B - A) = sin (π - A) sin (C - B)$

$- sin C sin (B - A) = - sin A - sin (C - B)$

$\frac{sin B cos A - cos B sin A}{sin A} = \frac{sin C cos B - cos C sin B}{sin C}$

$sin B cot A - cos B = cos B - cot C sin B$

$cot A - cot B = cot B - cot C$

$cot A, cot B, cot C a r e i n A P$

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