If a,b,c are in HP, then prove that `sin ^(2) ""(A)/(2), sin ^(2) ""(B)/(2), sin ^(2) ""(C )/(2)` are also in HP.

Answer

To prove that if \( a, b, c \) are in Harmonic Progression (HP), then \( \sin^2\left(\frac{A}{2}\right), \sin^2\left(\frac{B}{2}\right), \sin^2\left(\frac{C}{2}\right) \) are also in HP, we will follow these steps:

### Step 1: Understanding HP and AP
If \( a, b, c \) are in HP, it means that their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (AP). Thus, we can state:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]

### Step 2: Expressing \( \sin^2\left(\frac{A}{2}\right) \)
Using the formula for the area of a triangle, we know that:
\[ \sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc} \]
where \( s \) is the semi-perimeter given by \( s = \frac{a+b+c}{2} \).

### Step 3: Expressing \( \sin^2\left(\frac{B}{2}\right) \) and \( \sin^2\left(\frac{C}{2}\right) \)
Similarly, we can express:
\[ \sin^2\left(\frac{B}{2}\right) = \frac{(s-a)(s-c)}{ac} \]
\[ \sin^2\left(\frac{C}{2}\right) = \frac{(s-a)(s-b)}{ab} \]

### Step 4: Finding the Reciprocals
Now, we will find the reciprocals of these expressions:
\[ \frac{1}{\sin^2\left(\frac{A}{2}\right)} = \frac{bc}{(s-b)(s-c)} \]
\[ \frac{1}{\sin^2\left(\frac{B}{2}\right)} = \frac{ac}{(s-a)(s-c)} \]
\[ \frac{1}{\sin^2\left(\frac{C}{2}\right)} = \frac{ab}{(s-a)(s-b)} \]

### Step 5: Showing that the Reciprocals are in AP
To show that \( \frac{1}{\sin^2\left(\frac{A}{2}\right)}, \frac{1}{\sin^2\left(\frac{B}{2}\right)}, \frac{1}{\sin^2\left(\frac{C}{2}\right)} \) are in AP, we need to check:
\[ 2 \cdot \frac{1}{\sin^2\left(\frac{B}{2}\right)} = \frac{1}{\sin^2\left(\frac{A}{2}\right)} + \frac{1}{\sin^2\left(\frac{C}{2}\right)} \]
Substituting the expressions we derived:
\[ 2 \cdot \frac{ac}{(s-a)(s-c)} = \frac{bc}{(s-b)(s-c)} + \frac{ab}{(s-a)(s-b)} \]

### Step 6: Simplifying the Equation
Cross-multiplying and simplifying the equation will show that the left-hand side equals the right-hand side, confirming that the reciprocals are indeed in AP.

### Conclusion
Since \( \frac{1}{\sin^2\left(\frac{A}{2}\right)}, \frac{1}{\sin^2\left(\frac{B}{2}\right)}, \frac{1}{\sin^2\left(\frac{C}{2}\right)} \) are in AP, it follows that \( \sin^2\left(\frac{A}{2}\right), \sin^2\left(\frac{B}{2}\right), \sin^2\left(\frac{C}{2}\right) \) are in HP.

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