Question

If A, B and C are vertices of a triangle and $ 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: We start solving the problem by using the fact that if p, q, r are in H.P, then $ \dfrac{1}{p} $ , $ \dfrac{1}{q} $ and $ \dfrac{1}{r} $ are in A.P. We then make use of the fact $ \dfrac{1}{\sec \theta }=\cos \theta $ to proceed through the problem. We then make use of the fact that if p, q, r are in A.P, then $ 2q=p+r $ to proceed further through the problem. We then apply the cosine rule of triangle $ \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} $ , $ \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} $ , $ \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} $ and make necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are given that A, B and C are vertices and $ a\sec A $ , $ b\sec B $ , $ c\sec C $ are in H.P (Harmonic Progression). We need to find the progression that $ {{a}^{2}} $ , $ {{b}^{2}} $ , $ {{c}^{2}} $ are in.
We are given that $ a\sec A $ , $ b\sec B $ , $ c\sec C $ . We know that if p, q, r are in H.P (Harmonic Progression), then $ \dfrac{1}{p} $ , $ \dfrac{1}{q} $ and $ \dfrac{1}{r} $ are in A.P (Arithmetic mean).
So, we get $ \dfrac{1}{a\sec A} $ , $ \dfrac{1}{b\sec B} $ , $ \dfrac{1}{c\sec C} $ are in A.P.
We know that $ \dfrac{1}{\sec \theta }=\cos \theta $ .
So, we get $ \dfrac{\cos A}{a} $ , $ \dfrac{\cos B}{b} $ , $ \dfrac{\cos C}{c} $ are in A.P.
We know that if p, q, r are in A.P, then $ 2q=p+r $ .
 $ \Rightarrow \dfrac{2\cos B}{b}=\dfrac{\cos A}{a}+\dfrac{\cos C}{c} $ .
We know that from cosine rule of triangle $ \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} $ , $ \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} $ and $ \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} $ .
 $ \Rightarrow 2\times \left( \dfrac{\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}}{b} \right)=\dfrac{\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}}{a}+\dfrac{\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}}{c} $ .
 $ \Rightarrow 2\times \left( \dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2abc} \right)=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2abc}+\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2abc} $ .
 $ \Rightarrow 2\times \left( \dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2abc} \right)=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}+{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2abc} $ .
 $ \Rightarrow 2\times \left( \dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2abc} \right)=\dfrac{2{{b}^{2}}}{2abc} $ .
 $ \Rightarrow {{a}^{2}}+{{c}^{2}}-{{b}^{2}}={{b}^{2}} $ .
 $ \Rightarrow {{a}^{2}}+{{c}^{2}}=2{{b}^{2}} $ .
We can see that $ {{a}^{2}} $ , $ {{b}^{2}} $ , $ {{c}^{2}} $ are in A.P (Arithmetic Progression).
 $ \, therefore, $ The correct option for the given problem is (a).

Note:
We should know that if A, B, and C are not the vertices of a triangle, then the obtained result is not true. We should perform each step carefully in order to avoid confusion and calculation mistakes. We can make use of the fact that if p, q, r are in H.P (Harmonic Progression), then $ q=\dfrac{2pr}{p+r} $ to solve the given problem. Similarly, we can expect the problem to find the values of angles A, B, and C using the obtained result.