Question

(a) If \( x^{30} y^{20} = (x + y)^{50} \), prove that

\[ \frac{dy}{dx} = \frac{y}{x}. \]

Hint:

When working with matrices involving trigonometric functions, simplify calculations by substituting trigonometric identities, such as \( 1 - \cot^2 x = -\cos 2x \).

Answer 1

Step 1: Determine the transpose of \( A \).
The given matrix is: \[ A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix}. \] The transpose of \( A \), denoted \( A^T \), is obtained by interchanging rows and columns: \[ A^T = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Step 2: Calculate the determinant of \( A \).
The determinant of a \( 2 \times 2 \) matrix is given by: \[ \text{det}(A) = (1)(1) - (-\cot x)(\cot x). \] Simplify: \[ \text{det}(A) = 1 + \cot^2 x. \] Using the identity \( 1 + \cot^2 x = \csc^2 x \): \[ \text{det}(A) = \csc^2 x. \] Step 3: Compute the inverse of \( A \).
The inverse of a \( 2 \times 2 \) matrix is: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \] For \( A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix} \), this becomes: \[ A^{-1} = \frac{1}{\csc^2 x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Using \( \csc^2 x = 1 + \cot^2 x \), the inverse simplifies to: \[ A^{-1} = \frac{1}{1 + \cot^2 x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Step 4: Multiply \( A^T \) and \( A^{-1} \).
Substitute \( A^T \) and \( A^{-1} \): \[ A^T = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}, \quad A^{-1} = \frac{1}{1 + \cot^2 x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] The product is: \[ A^T A^{-1} = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix} \cdot \frac{1}{1 + \cot^2 x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Step 5: Simplify the product.
Multiply the matrices: \[ A^T A^{-1} = \frac{1}{1 + \cot^2 x} \begin{bmatrix} 1 - \cot^2 x & -\cot x - \cot x \\ \cot x + \cot x & 1 - \cot^2 x \end{bmatrix}. \] Simplify using \( 1 - \cot^2 x = -\cos 2x \) and \( -\cot x - \cot x = -2\cot x \): \[ A^T A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}. \] Conclusion:
The final result is: \[ \boxed{A^T A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}}. \]