Question

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

Hint:

For trigonometric identities involving derivatives, simplify using tangent and secant relationships before differentiating.

Answer 1

Step 1: Rearrange the given equation
The given equation is: \[ x \cos(p + y) + \cos p \sin(p + y) = 0. \] Divide throughout by \( \cos(p + y) \): \[ x + \cos p \tan(p + y) = 0. \] This implies: \[ \tan(p + y) = -\frac{x}{\cos p}. \] 
Step 2: Differentiate with respect to \( x \)
Differentiate both sides with respect to \( x \): \[ \sec^2(p + y) \cdot \frac{d}{dx}(p + y) = -\frac{d}{dx}\left(\frac{x}{\cos p}\right). \] Simplify: \[ \sec^2(p + y) \frac{dy}{dx} = -\frac{1}{\cos p}. \] 
Step 3: Express \( \frac{dy}{dx} \) in terms of \( \cos^2(p + y) \)
Using \( \sec^2(p + y) = \frac{1}{\cos^2(p + y)} \), we get: \[ \frac{1}{\cos^2(p + y)} \cdot \frac{dy}{dx} = -\frac{1}{\cos p}. \] Multiply through by \( \cos^2(p + y) \): \[ \frac{dy}{dx} = -\cos^2(p + y) \cdot \cos p. \] 
Step 4: Conclude the result
\[ \cos p \frac{dy}{dx} = -\cos^2(p + y). \]

Answer 2

Given 

\[ x\cos(p+y)+\cos p\sin(p+y)=0,\qquad p\ \text{is constant}. \]

Differentiate implicitly w.r.t. \(x\)

Differentiate term-by-term (remember \(p\) is constant, \(y=y(x)\)): \[ \frac{d}{dx}\big[x\cos(p+y)\big] + \frac{d}{dx}\big[\cos p\sin(p+y)\big]=0. \] For the first term use product rule: \[ \frac{d}{dx}\big[x\cos(p+y)\big]=\cos(p+y)+x\big(-\sin(p+y)\big)\frac{dy}{dx}. \] For the second term (\(\cos p\) is constant): \[ \frac{d}{dx}\big[\cos p\sin(p+y)\big]=\cos p\cdot\cos(p+y)\frac{dy}{dx}. \] Putting these into the differentiated equation: \[ \cos(p+y)-x\sin(p+y)\frac{dy}{dx}+\cos p\cos(p+y)\frac{dy}{dx}=0. \] Collect the terms containing \(\dfrac{dy}{dx}\): \[ \cos(p+y)+\big[-x\sin(p+y)+\cos p\cos(p+y)\big]\frac{dy}{dx}=0. \] Hence \[ \big[-x\sin(p+y)+\cos p\cos(p+y)\big]\frac{dy}{dx}=-\cos(p+y). \] So \[ \frac{dy}{dx}=\frac{-\cos(p+y)}{-x\sin(p+y)+\cos p\cos(p+y)}. \]

Simplify using the original equation

From the given equation \[ x\cos(p+y) + \cos p\sin(p+y)=0 \] we get \[ x\cos(p+y)=-\cos p\sin(p+y)\quad\Rightarrow\quad x=\frac{-\cos p\sin(p+y)}{\cos(p+y)}. \] Substitute this \(x\) into the denominator: \[ -x\sin(p+y)+\cos p\cos(p+y) = -\Big(\frac{-\cos p\sin(p+y)}{\cos(p+y)}\Big)\sin(p+y)+\cos p\cos(p+y) \] \[ =\frac{\cos p\sin^2(p+y)}{\cos(p+y)}+\cos p\cos(p+y) =\cos p\;\frac{\sin^2(p+y)+\cos^2(p+y)}{\cos(p+y)} =\cos p\;\frac{1}{\cos(p+y)}. \] Therefore the denominator simplifies to \(\dfrac{\cos p}{\cos(p+y)}\). Substitute back into \(\dfrac{dy}{dx}\): \[ \frac{dy}{dx} =\frac{-\cos(p+y)}{\dfrac{\cos p}{\cos(p+y)}} =-\frac{\cos^2(p+y)}{\cos p}. \] Multiplying both sides by \(\cos p\) gives the required result: \[ \boxed{\;\cos p\;\frac{dy}{dx} = -\cos^2(p+y)\; }. \]