Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 21

Hint:

We must know about the derivative of trigonometric function.

Given:

$x=f(t) \cos t-f^{\prime}(t) \sin t \\$

$y=f(t) \sin t-f^{\prime}(t) \cos t \\$

Explanation:

$x=f(t) \cos t-f^{\prime}(t) \sin t \\$

$y=f(t) \sin t-f^{\prime}(t) \cos t \\$

$\frac{d x}{d t}=f^{\prime}(t) \cos t-f(t) \sin t-f^{\prime \prime}(t) \sin t-f^{\prime}(t) \cos t \\$

\begin{aligned} & &\frac{d x}{d t}=-f(t) \sin t-f^{\prime \prime}(t) \sin t \end{aligned}

$\frac{d x}{d t}=-\sin t\left[f(t)+f^{\prime \prime}(t)\right] \\$

$\frac{d y}{d t}=f^{\prime}(t) \sin t-f(t) \cos t-f^{\prime \prime}(t) \cos t-f^{\prime}(t) \sin t \\$

\begin{aligned} &\frac{d y}{d t}=-f(t) \cos t-f^{\prime \prime}(t) \cos t \end{aligned}

$\frac{d y}{d t}=-\cos t\left[f(t)+f^{\prime \prime}(t)\right]$

$\\ \left(\frac{d x}{d t}\right)^{2}=\left[-\sin t\left[f(t)+f^{\prime \prime}(t)\right]\right]^{2} \\$

\begin{aligned} & &\left(\frac{d y}{d t}\right)^{2}=\left[-\cos t\left[f(t)+f^{\prime \prime}(t)\right]\right]^{2} \end{aligned}

$\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=(\sin t)^{2}\left[f(t)+f^{\prime \prime}(t)\right]^{2}+(\cos t)^{2}\left[f(t)+f^{\prime \prime}(t)\right]^{2} \\$

\begin{aligned} & &\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=\left[f(t)+f^{\prime \prime}(t)\right]^{2}\left[(\sin t)^{2}+(\cos t)^{2}\right] \end{aligned}

$=\left[f(t)+f^{\prime \prime}(t)\right]^{2}$                                                         $\left[\because(\sin t)^{2}+(\cos t)^{2}=1\right]$