Question

In \[\Delta ABC\], if \[\cot A\], \[\cot B\], \[\cot C\] be in A.P., then \[{a^2},{b^2},{c^2}\] are in
A. H.P
B. G.P.
C. A.P.
D. None of these

Answer 1

Hint: First we will apply A.P condition on \[\cot A\], \[\cot B\], \[\cot C\]. Then break it into cos function and sin function. Using cosine law and sine law, we will simplify the given expression.

Formula used:
If a,b,c are in A.P., then a + c = 2b.
\[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
Sine law:
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
Cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
Given \[\cot A\], \[\cot B\], \[\cot C\] are in A.P.
Apply the condition of A.P
\[\cot A + \cot C = 2\cot B\]
Apply the formula \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
\[\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\cos C}}{{\sin C}} = 2\dfrac{{\cos B}}{{\sin B}}\] …..(i)
Assume that, \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {say} \right)\]
\[\sin A = ak\], \[\sin B = bk\], \[\sin C = ck\]
Putting \[\sin A = ak\], \[\sin B = bk\], \[\sin C = ck\] in equation (i)
\[\dfrac{{\cos A}}{{ak}} + \dfrac{{\cos C}}{{ck}} = 2\dfrac{{\cos B}}{{bk}}\]
Now applying cosine rule:
\[\dfrac{{\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{{ak}} + \dfrac{{\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}}{{ck}} = 2\dfrac{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}}{{bk}}\]
\[\dfrac{{{b^2} + {c^2} - {a^2}}}{{2abck}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2abck}} = 2\dfrac{{{a^2} + {c^2} - {b^2}}}{{2abck}}\]
Cancel out \[2abck\] from denominator
\[{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2} = 2{a^2} + 2{c^2} - 2{b^2}\]
\[2{b^2} = 2{a^2} + 2{c^2} - 2{b^2}\]
Cancel out 2 from both sides
\[{b^2} = {a^2} + {c^2} - {b^2}\]
\[{a^2} + {c^2} = 2{b^2}\]
Thus \[{a^2},{b^2},{c^2}\] are in AP.
Hence option C is the correct option.

Note: Students often confuse with condition of AP. They often take a + b = 2c, where a, b, c are in AP but the correct condition for AP is a + c = 2b.