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Question
In ∆ABC, if a2, b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.
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Solution
\[Let \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k\]
Then,
\[\sin A = ka, \sin B = kb, \sin C = kc\]
a2, b2 and c2 are in A.P.
\[\Rightarrow 2 b^2 = a^2 + c^2 \]
\[ \Rightarrow 2\left( a^2 + c^2 - b^2 \right) = 2\left( 2 b^2 - b^2 \right) = 2 b^2 = b^2 + b^2 + c^2 - a^2 - c^2 + a^2 \]
\[ \Rightarrow 2\left( a^2 + c^2 - b^2 \right) = b^2 + c^2 - a^2 + a^2 + b^2 - c^2 \]
\[ \Rightarrow \frac{2\left( a^2 + c^2 - b^2 \right)}{2abc} = \frac{\left( b^2 + c^2 - a^2 \right)}{2abc} + \frac{\left( a^2 + b^2 - c^2 \right)}{2abc}\]
\[ \Rightarrow \frac{2\cos B}{kb} = \frac{\cos A}{ka} + \frac{\cos C}{kc}\]
\[ \Rightarrow \frac{2\cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}\]
\[ \Rightarrow 2\cot B = \cot A + \cot C\]
\[ \Rightarrow \cot A, \cot B and \cot C are in AP .\]
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