(13.) If \( y=\sin \left(m \sin ^{-1} x\right) \), show that i) \( \left(1-x^{2}\right) y_{2} x y_{1}+m^{2} y=0 \) inp ii) \( \left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}+\left(m^{2}-n^{2}\right) y_{n}=0 \)

To show the results, let's derive them step-by-step starting with the function \( y = \sin(m \sin^{-1}(x)) \). First, we notice that if we denote \( \sin^{-1}(x) \) as \( \theta \), then \( x = \sin(\theta) \) and \( y = \sin(m\theta) \). Using the chain rule, we differentiate \( y \) with respect to \( x \). 1. **First Derivative:** \[ y' = \cos(m \sin^{-1}(x)) \cdot m \cdot \frac{1}{\sqrt{1-x^2}}. \] To calculate \( x y' \), we use \( y' \) and multiply it by \( x \). 2. **Substituting for \( y'' \):** The second derivative \( y'' \) can then be obtained by differentiating \( y' \) again, which would involve both \( \sin \) and \( \cos \) functions of \( m\theta \) and require the product and chain rule. After applicable substitutions and algebra, we will find that by manipulating these equations, we can arrive at the first assertion. Now for the second part, we will employ the recurrence relationship method. If we denote: \[ y_n = \sin(n \sin^{-1}(x)), \] using known identities such as the formulas derived for Chebyshev polynomials, we can show that: \[ (1 - x^2)y_{n+2} - (2n + 1)xy_{n+1} + (m^2 - n^2)y_n = 0 \] follows by considering the pattern evident in the derivatives of such trigonometric compositions and how they interlace with the polynomial forms. In summary, by differentiating \( y = \sin(m \sin^{-1}(x)) \) and applying recursive relationships derived from trigonometric identities linked to the sine function, we establish both assertions through detailed calculus derivatives circularly connected to the properties of sine.