Question

In $\vartriangle ABC$ , if ${a^2}$ , ${b^2}$ , ${c^2}$ are in A.P. Hence prove that $CotA$ , $CotB$ , $CotC$ are also in A.P.

Answer 1

Hint : In arithmetic progression, each term differs from the preceding term by a constant called the common difference of the progression. $a$ , $b$ and $c$ are the sides of the triangle ABC, and $\angle A$ , $\angle B$ and $\angle C$ are the angles of the triangle. By using the cosine and sine rule we can find out the relation between the angles and sides and thus prove the required relation.

Complete step-by-step answer :
We are given that
${a^2}\,,\,{b^2}\,,\,{c^2}$ are in A.P.
Multiplying all the three terms by -2, we get –
$- 2{a^2}\,,\, - 2{b^2}\,,\, - 2{c^2}$ are in A.P.
Add ${a^2} + {b^2} + {c^2}$ with the three terms –
${a^2} + {b^2} + {c^2} - 2{a^2}\,,\,{a^2} + {b^2} + {c^2} - 2{b^2}\,,\,{a^2} + {b^2} + {c^2} - 2{c^2}$ are in A.P.
${b^2} + {c^2} - {a^2}\,,\,{a^2} - {b^2} + {c^2}\,,\,{a^2} + {b^2} - {c^2}$ are in A.P.
Now divide all the three terms by $2abc$ -
$\dfrac{{{b^2} + {c^2} - {a^2}}}{{2abc}}\,,\,\dfrac{{{a^2} - {b^2} + {c^2}}}{{2abc}}\,,\,\dfrac{{{a^2} + {b^2} - {c^2}}}{{2abc}}$ are in A.P.
Rearranging the terms in the denominator, we get –
$\dfrac{1}{a}(\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}})\,,\,\dfrac{1}{b}(\dfrac{{{a^2} - {b^2} + {c^2}}}{{2ac}})\,,\,\dfrac{1}{c}(\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}})$ are in A.P.
Now, by cosine rule, we have - $\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \operatorname{Cos} A\,,\,\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \operatorname{Cos} B\,,\,\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \operatorname{Cos} C$
Substituting these values in the above equation, we get –
$\dfrac{1}{a}\operatorname{Cos} A\,,\,\dfrac{1}{b}\operatorname{Cos} B\,,\,\dfrac{1}{c}\operatorname{Cos} C$ are in A.P.
Multiplying the three terms by a constant k, we get –
$\dfrac{k}{a}\operatorname{Cos} A\,,\,\dfrac{k}{b}\operatorname{Cos} B\,,\,\dfrac{k}{c}\operatorname{Cos} C$ are in A.P.
Now, by sine rule we have –
$\dfrac{a}{{\operatorname{Sin} A}} = \dfrac{b}{{\operatorname{Sin} B}} = \dfrac{c}{{\operatorname{Sin} C}} = k \\\ \Rightarrow \dfrac{k}{a} = \dfrac{1}{{\operatorname{Sin} A}}\,,\,\dfrac{k}{b} = \dfrac{1}{{\operatorname{Sin} B}}\,,\,\dfrac{k}{c} = \dfrac{1}{{\operatorname{Sin} C}} \\\$
Putting these values in the obtained equation, we get –
$\dfrac{{\operatorname{Cos} A}}{{\operatorname{Sin} A}}\,,\,\dfrac{{\operatorname{Cos} B}}{{\operatorname{Sin} B}}\,,\,\dfrac{{\operatorname{Cos} C}}{{\operatorname{Sin} C}}$ are in A.P.
$\Rightarrow CotA\,,\,CotB\,,\,CotC$ are in A.P.
Hence Proved.
So, the correct answer is “ $CotA\,,\,CotB\,,\,CotC$ are in A.P”.

Note : Sin, Cos, Tan, Sec, Cosec are Cot are the trigonometric functions that are expressed as the ratio of sides of a right-angled triangle. Sine rule and Cosine rule relate the sides of the triangle with the angles in the triangle so that if one of the quantities is known, the other one can be found out easily.