Question

If $\cot(\theta - \alpha), 3\cot\theta, \cot(\theta + \alpha)$ are in AP (where, $\theta \neq \frac{n\pi}{2}, \alpha \neq k\pi, n, k \in I)$, then $\frac{2\sin^2\theta}{\sin^2\alpha}$ equal to

Answer 1

3

Answer 2

The problem states that $\cot(\theta - \alpha)$, $3\cot\theta$, and $\cot(\theta + \alpha)$ are in Arithmetic Progression (AP).

For three terms $a, b, c$ to be in AP, the condition is $2b = a + c$.

Applying this condition: $2(3\cot\theta) = \cot(\theta - \alpha) + \cot(\theta + \alpha)$ $6\cot\theta = \frac{\cos(\theta - \alpha)}{\sin(\theta - \alpha)} + \frac{\cos(\theta + \alpha)}{\sin(\theta + \alpha)}$

Combine the terms on the right-hand side: $6\cot\theta = \frac{\cos(\theta - \alpha)\sin(\theta + \alpha) + \cos(\theta + \alpha)\sin(\theta - \alpha)}{\sin(\theta - \alpha)\sin(\theta + \alpha)}$

The numerator is in the form $\sin A \cos B + \cos A \sin B = \sin(A+B)$. Here, $A = \theta + \alpha$ and $B = \theta - \alpha$. So, the numerator becomes $\sin((\theta + \alpha) + (\theta - \alpha)) = \sin(2\theta)$.

The equation now is: $6\cot\theta = \frac{\sin(2\theta)}{\sin(\theta - \alpha)\sin(\theta + \alpha)}$

Substitute $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and $\sin(2\theta) = 2\sin\theta\cos\theta$: $6\frac{\cos\theta}{\sin\theta} = \frac{2\sin\theta\cos\theta}{\sin(\theta - \alpha)\sin(\theta + \alpha)}$

Given $\theta \neq \frac{n\pi}{2}$, we know that $\sin\theta \neq 0$ and $\cos\theta \neq 0$. Therefore, we can cancel $\cos\theta$ from both sides and multiply by $\sin\theta$: $6 = \frac{2\sin^2\theta}{\sin(\theta - \alpha)\sin(\theta + \alpha)}$

Rearrange the equation: $6\sin(\theta - \alpha)\sin(\theta + \alpha) = 2\sin^2\theta$

Now, use the trigonometric identity: $\sin(A-B)\sin(A+B) = \sin^2 A - \sin^2 B$. Here, $A = \theta$ and $B = \alpha$. So, $\sin(\theta - \alpha)\sin(\theta + \alpha) = \sin^2\theta - \sin^2\alpha$.

Substitute this identity back into the equation: $6(\sin^2\theta - \sin^2\alpha) = 2\sin^2\theta$ $6\sin^2\theta - 6\sin^2\alpha = 2\sin^2\theta$

Rearrange the terms to solve for the desired expression: $6\sin^2\theta - 2\sin^2\theta = 6\sin^2\alpha$ $4\sin^2\theta = 6\sin^2\alpha$

We need to find the value of $\frac{2\sin^2\theta}{\sin^2\alpha}$. Divide both sides of the equation $4\sin^2\theta = 6\sin^2\alpha$ by $2\sin^2\alpha$. $\frac{4\sin^2\theta}{2\sin^2\alpha} = \frac{6\sin^2\alpha}{2\sin^2\alpha}$ $\frac{2\sin^2\theta}{\sin^2\alpha} = 3$

The second part of the question "The expression $\frac{\tan A}{1-\cot A}$ (1) $\sin A \cos A + 1$" appears to be an incomplete or unrelated query and is ignored in the solution of the main problem.

The final answer is $\boxed{3}$.

Explanation of the solution:

1. Apply the AP condition $2b=a+c$ to the given cotangent terms.
2. Combine the cotangent terms on the RHS using a common denominator.
3. Recognize the numerator as $\sin(A+B)$ and simplify to $\sin(2\theta)$.
4. Substitute $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and $\sin(2\theta) = 2\sin\theta\cos\theta$.
5. Simplify the equation by canceling common terms, noting the given conditions $\theta \neq \frac{n\pi}{2}$.
6. Use the identity $\sin(A-B)\sin(A+B) = \sin^2 A - \sin^2 B$.
7. Rearrange the resulting equation to solve for the required expression $\frac{2\sin^2\theta}{\sin^2\alpha}$.