Question

If $a\sec A$, $b\sec B$, $c\sec C$ are in H.P. then ${{a}^{2}}$, ${{b}^{2}}$, ${{c}^{2}}$ are in
A. A.P.
B. G.P.
C. H.P.
D. A.G.P

Answer 1

Hint: In this problem we need to find the relation or progression that exists between the variables ${{a}^{2}}$, ${{b}^{2}}$, ${{c}^{2}}$. Where $a\sec A$, $b\sec B$, $c\sec C$ are in H.P. We will consider the given statement which is $a\sec A$, $b\sec B$, $c\sec C$ are in H.P, so we will use the H.P rule that is $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$. Here we will use the trigonometric formula $\dfrac{1}{\sec A}=\cos A$ and rewrite the obtained equation. Now we will use the property of triangle or cosine rule which is ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A$. From this formula we will calculate the required value which is useful in the obtained equation, similarly we can calculate the remaining values and substitute them in the equation and simplify it to get the required result.

Complete step by step solution:
Given that $a\sec A$, $b\sec B$, $c\sec C$ are in H.P. So, we are going to write
$\dfrac{2}{b\sec B}=\dfrac{1}{a\sec A}+\dfrac{1}{c\sec C}$
We have the trigonometric formula $\dfrac{1}{\sec A}=\cos A$. So, the above equation is modified as
$\dfrac{2\cos B}{b}=\dfrac{\cos A}{a}+\dfrac{\cos C}{c}$
From the cosine rule ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A$, we can write $\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$. Similarly, we may have $\cos B=\dfrac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac}$, $\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$. Substituting these values in the above equation, then we will get
$\dfrac{2\left( \dfrac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac} \right)}{b}=\dfrac{\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}}{a}+\dfrac{\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}}{c}$
Simplifying the above equation, then we will get
$\begin{align}
  & \dfrac{2\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2abc}=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2abc}+\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2abc} \\
 & \Rightarrow \dfrac{2\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2abc}=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}+{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2abc} \\
 & \Rightarrow \dfrac{2\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2abc}=\dfrac{2{{b}^{2}}}{2abc} \\
\end{align}$
Cancelling the $2abc$ which is in denominator and $2$ which is in numerator, then we will get
${{c}^{2}}+{{a}^{2}}-{{b}^{2}}={{b}^{2}}$
Adding ${{b}^{2}}$ on both sides of the above equation, then we will have
$2{{b}^{2}}={{a}^{2}}+{{c}^{2}}$
The above equation shows that the terms ${{a}^{2}}$, ${{b}^{2}}$, ${{c}^{2}}$ are in A.P.

So, the correct answer is “Option A”.

Note: In this problem students may do cancelations where there is possibility. But when you do cancelation everywhere then we need to do more steps for solving the equation like taking the LCM, addition of fractions likewise. So, don’t cancel any term immediately.