(a)OR(b)

If $\displaystyle x \cos (p+y)+\cos p \sin (p+y)=0$, prove that $\displaystyle \cos p \frac{d y}{d x}=-\cos ^{2}(p+y)$, where $\displaystyle p$ is a constant.

$$\begin{aligned} & x \cos (p+y)+\cos p \sin (p+y)=0 \\ & \Rightarrow x=\frac{-\cos p \sin (p+y)}{\cos (p+y)} \Rightarrow x=-\cos p \cdot \tan (p+y) \\ & \Rightarrow \frac{d x}{d y}=-\cos p \cdot \sec ^{2}(p+y) \\ & \Rightarrow \frac{d y}{d x}=\frac{-1}{\cos p \cdot \sec ^{2}(p+y)} \\ & \Rightarrow \cos p \frac{d y}{d x}=-\cos ^{2}(p+y) \end{aligned} \begin{aligned} & f(x)=\left\{\begin{array}{cl} \frac{x-2}{-(x-2)}+a & ; x<2 \\ a+b & ; x=2 \\ \frac{x-2}{(x-2)}+b & ; x>2 \end{array} \Rightarrow f(x)=\left\{\begin{array}{cc} -1+a & ; x<2 \\ a+b & ; x=2 \\ 1+b & ; x>2 \end{array}\right.\right. \\ & \lim _{x \rightarrow 2^{-}} f(x)=-1+a, \lim _{x \rightarrow 2^{+}} f(x)=1+b \text { and } f(2)=a+b \\ & \text { as } f \text { is continous at } x=2 \therefore-1+a=1+b=a+b \\ & \Rightarrow a=1, b=-1 \end{aligned} $$