Question

If the sides a,b,c of the triangle ABC are in harmonic progression, then \( \text{cosec}^2 A/2, \text{cosec}^2 B/2, \text{cosec}^2 C/2 \) are in

• A. Arithmetic-geometric progression
• B. Arithmetic progression
• C. Geometric progression
• D. Harmonic progression Correct Answer

Hint:

Recall properties relating sides and angles in a triangle, especially half-angle formulas. Key half-angle formulas: \( \sin^2(A/2) = \frac{(s-b)(s-c)}{bc} \), \( \cos^2(A/2) = \frac{s(s-a)}{bc} \), \( \tan^2(A/2) = \frac{(s-b)(s-c)}{s(s-a)} \). If \(a,b,c\) are in H.P., then \(1/a, 1/b, 1/c\) are in A.P. A standard result often useful: If sides \(a,b,c\) are in H.P., then \( \sin^2(A/2), \sin^2(B/2), \sin^2(C/2) \) are in H.P. (and consequently their reciprocals are in A.P.).

Answer 1

The Correct Option is B

Step 1: Relate \( \text{cosec}^2 (X/2) \) to sides of the triangle.
We know \( \sin^2(A/2) = \frac{(s-b)(s-c)}{bc} \), where \( s = (a+b+c)/2 \) is the semi-perimeter.
So, \( \text{cosec}^2(A/2) = \frac{1}{\sin^2(A/2)} = \frac{bc}{(s-b)(s-c)} \).
Similarly, \( \text{cosec}^2(B/2) = \frac{ac}{(s-a)(s-c)} \) and \( \text{cosec}^2(C/2) = \frac{ab}{(s-a)(s-b)} \).

Step 2: Use the condition that a, b, c are in Harmonic Progression (H.
P.
).
If a, b, c are in H.
P.
, then \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.
P.
).
This means \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \implies \frac{2}{b} = \frac{a+c}{ac} \implies b = \frac{2ac}{a+c} \).
This also implies \( \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \).
A known result is that if a, b, c are in H.
P.
, then \( s-a, s-b, s-c \) are also in H.
P.
If a, b, c are in H.
P, then \( \frac{s-a}{a}, \frac{s-b}{b}, \frac{s-c}{c} \) are in A.
P.
Also, \( \sin^2(A/2), \sin^2(B/2), \sin^2(C/2) \) are in H.
P.
if a,b,c are in H.
P.
If \( x, y, z \) are in H.
P.
, then \( 1/x, 1/y, 1/z \) are in A.
P.
So, if \( \sin^2(A/2), \sin^2(B/2), \sin^2(C/2) \) are in H.
P.
, then \( \text{cosec}^2(A/2), \text{cosec}^2(B/2), \text{cosec}^2(C/2) \) are in A.
P.
Let's prove that \( \sin^2(A/2), \sin^2(B/2), \sin^2(C/2) \) are in H.
P.
This means we need to show \( \frac{bc}{(s-b)(s-c)}, \frac{ac}{(s-a)(s-c)}, \frac{ab}{(s-a)(s-b)} \) are in A.
P.

Step 3: Alternative approach using \( \cot(X/2) \).
We know \( \cot(A/2) = \frac{s-a}{\Delta/s} = \frac{s(s-a)}{\Delta} \), \( \cot(B/2) = \frac{s(s-b)}{\Delta} \), \( \cot(C/2) = \frac{s(s-c)}{\Delta} \).
If a, b, c are in H.
P.
, then \( s-a, s-b, s-c \) are in H.
P.
So \( \frac{1}{s-a}, \frac{1}{s-b}, \frac{1}{s-c} \) are in A.
P.
Consider \( \tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \).
This seems more complex.
A known property: If a,b,c are in H.
P.
, then \( \cot(A/2), \cot(B/2), \cot(C/2) \) are in A.
P.
Proof: a,b,c in H.
P \( \implies \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) in A.
P.
\( \implies \frac{s}{a}, \frac{s}{b}, \frac{s}{c} \) in A.
P.
\( \implies \frac{s}{a}-1, \frac{s}{b}-1, \frac{s}{c}-1 \) in A.
P.
\( \implies \frac{s-a}{a}, \frac{s-b}{b}, \frac{s-c}{c} \) in A.
P.
We want to show \( \frac{s(s-a)}{\Delta}, \frac{s(s-b)}{\Delta}, \frac{s(s-c)}{\Delta} \) are in A.
P.
This is equivalent to showing \( s(s-a), s(s-b), s(s-c) \) are in A.
P.
This is not directly obvious from a,b,c being in H.
P.
Let's use the result: If a, b, c are in H.
P.
, then \( \sin^2(A/2), \sin^2(B/2), \sin^2(C/2) \) are in H.
P.
(This is often stated as a standard result).
If \( x, y, z \) are in H.
P.
, their reciprocals \( 1/x, 1/y, 1/z \) are in A.
P.
Let \( x = \sin^2(A/2), y = \sin^2(B/2), z = \sin^2(C/2) \).
Then \( 1/x = \text{cosec}^2(A/2), 1/y = \text{cosec}^2(B/2), 1/z = \text{cosec}^2(C/2) \).
If \( \sin^2(A/2), \sin^2(B/2), \sin^2(C/2) \) are in H.
P.
, then \( \text{cosec}^2(A/2), \text{cosec}^2(B/2), \text{cosec}^2(C/2) \) are in A.
P.
This matches option (2).