If `a^(2),b^(2),c^(2)` are in A.P.,then which of the following is also in A.P.?

Options

A. sin A , sin B, sin C
B. tan A , tan B, tan C
C. cot A , cot B , cot C
D. none of these

Answer

To solve the problem, we need to show that if \( a^2, b^2, c^2 \) are in arithmetic progression (A.P.), then \( \cot A, \cot B, \cot C \) are also in A.P. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the condition of A.P.
We know that \( a^2, b^2, c^2 \) are in A.P. This means that:
\[ 2b^2 = a^2 + c^2 \]

### Step 2: Using the sine rule
According to the sine rule in triangles, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
where \( k \) is a constant. From this, we can express \( a, b, c \) in terms of \( k \):
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

### Step 3: Substituting into the A.P. condition
Substituting \( a, b, c \) into the A.P. condition:
\[ 2(k \sin B)^2 = (k \sin A)^2 + (k \sin C)^2 \]
This simplifies to:
\[ 2k^2 \sin^2 B = k^2 \sin^2 A + k^2 \sin^2 C \]
Dividing through by \( k^2 \) (assuming \( k \neq 0 \)):
\[ 2 \sin^2 B = \sin^2 A + \sin^2 C \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \sin^2 B - \sin^2 A = \sin^2 C - \sin^2 B \]

### Step 5: Factoring the equation
We can factor both sides:
\[ (\sin B + \sin A)(\sin B - \sin A) = (\sin C + \sin B)(\sin C - \sin B) \]

### Step 6: Using the sine addition formula
Using the sine addition formula, we can express:
\[ \sin(A + B) = \sin(180^\circ - C) = \sin C \]
This leads to:
\[ \sin A \cos B - \cos A \sin B = \sin C \cos B - \cos C \sin B \]

### Step 7: Arriving at the conclusion
From the derived equations, we can conclude that:
\[ \cot A, \cot B, \cot C \text{ are in A.P.} \]

Thus, we have shown that if \( a^2, b^2, c^2 \) are in A.P., then \( \cot A, \cot B, \cot C \) are also in A.P.

### Final Answer:
The correct option is that \( \cot A, \cot B, \cot C \) are in A.P.

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