Question

In a $\Delta ABC$, if ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in A.P. , then show that $\cot A,\cot B,\cot C$ are also in AP.

Answer 1

Hint: From the question we know that ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in A.P. so the common difference between ${{a}^{2}}$ and ${{b}^{2}}$ is equal to common difference between ${{b}^{2}}$and ${{c}^{2}}$. So, it can be written as ${{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}$. Then, we will apply the sine rule of triangle $\left( i.e.\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k \right)$ and replace a, b, c by $k\sin A, k\sin B, k\sin C$ and then try to convert all of them into $\cot A,\cot B,\cot C$ and find relation among them using the property
${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right)\sin \left( A-B \right)$ and $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$

Complete step by step answer:

From the figure, we can see that in $\Delta ABC$ a, b, and c are opposite sides of the angle A, angle B, angle C.
Since, it in the question that in a $\Delta ABC$ ,${{a}^{2}}, {{b}^{2}}, {{c}^{2}}$ are in A.P., so the common difference between ${{a}^{2}}$ and ${{b}^{2}}$is equal to common difference between ${{b}^{2}}$ and ${{c}^{2}}$.
Hence, it can be written as:
${{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}...................(1)$
Now, from the sine rule of a triangle we know that:
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$
$\therefore a=k\sin A$, $b=k\sin B$ and $c=k\sin C$
Now, after putting the value of a, b, c in equation (1) we will get:
${{k}^{2}}{{\sin }^{2}}B-{{k}^{2}}{{\sin }^{2}}A={{k}^{2}}{{\sin }^{2}}C-{{k}^{2}}{{\sin }^{2}}B$
$\Rightarrow {{k}^{2}}\left( {{\sin }^{2}}B-{{\sin }^{2}}A \right)={{k}^{2}}\left( {{\sin }^{2}}C-{{\sin }^{2}}B \right)$
$\therefore \left( {{\sin }^{2}}B-{{\sin }^{2}}A \right)=\left( {{\sin }^{2}}C-{{\sin }^{2}}B \right)$
We know that ${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right)\sin \left( A-B \right)$:
Hence, the above equation can be rewritten as:
$\Rightarrow \sin \left( B+A \right)\sin \left( B-A \right)=\sin \left( C+B \right)\sin \left( C-B \right)................(2)$
Now, for every \[\Delta ABC\], we know that:
$A+B+C=\pi =180{}^\circ $
$\therefore \left( B+A \right)=\pi -C$and $\left( C+B \right)=\pi -A$
Now, put $\left( B+A \right)=\pi -C$and $\left( C+B \right)=\pi -A$in equation (2).
$\Rightarrow \sin \left( \pi -C \right)\sin \left( B-A \right)=\sin \left( \pi -A \right)\sin \left( C-B \right)$
We know that $\sin \left( \pi -C \right)=\sin C$and $\sin \left( \pi -A \right)=\sin A$, we also know that $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$
$\Rightarrow \sin C\sin \left( B-A \right)=\sin A\sin \left( C-B \right)$
$\Rightarrow \sin C\left( sinB\cos A-\sin A\cos B \right)=\sin A\left( \sin C\cos B-\sin B\cos C \right)$
$\Rightarrow \sin C\sin B\cos A-\sin C\sin A\cos B=\sin A\sin C\cos B-\sin A\sin B\cos C$
Now, after dividing both the side of the equation by $\sin A\sin B\sin C$,we will get:
$\Rightarrow \dfrac{\sin C\sin B\cos A}{\sin A\sin B\sin C}-\dfrac{\sin C\sin A\cos B}{\sin A\sin B\sin C}=\dfrac{\sin A\sin C\cos B}{\sin A\sin B\sin C}-\dfrac{\sin A\sin B\cos C}{\sin A\sin B\sin C}$
$\Rightarrow \dfrac{\cos A}{\sin A}-\dfrac{\cos B}{\sin B}=\dfrac{\cos B}{\sin B}-\dfrac{\cos C}{\sin C}$
$\Rightarrow \cot A-\cot B=\cot B-\cot C$
$\Rightarrow 2\cot B=\cot A+\cot C................(3)$
We know that if $a,b,c$ are A.P. then $2b=a+c$
So, from equation (3) we can say that $\cot A,\cot B,\cot C$ are in A.P.
Hence, it is proved that $\cot A,\cot B,\cot C$ are in A.P.

Note: The above question uses so many trigonometric formulas, so students are required to memorize them, and use them carefully, and so that chance of making mistakes reduces. We usually
used to write sine rule for $\vartriangle ABC$ as $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k$, but it is wrong , the correct form of it is $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$. So, students should avoid these mistakes.