Question

If \(\cos y=x\cos(a+y)\) and \(\cos a\ne\pm1\), then prove that \(\dfrac{dy}{dx}=\dfrac{\cos^2(a+y)}{\sin a}\).

Hint:

Differentiate implicitly, then substitute x back from the original relation and use sin(A-B).

Answer 1

Step 1: Differentiate both sides implicitly w.r.t. \(x\):
\[ -\sin y\cdot\dfrac{dy}{dx}=\cos(a+y)+x\cdot\left(-\sin(a+y)\right)\dfrac{dy}{dx} \]

Step 2: Collect the \(dy/dx\) terms:
\[ -\sin y\cdot y'+x\sin(a+y)\cdot y'=\cos(a+y) \]
\[ y'\big[x\sin(a+y)-\sin y\big]=\cos(a+y) \]

Step 3: Substitute \(x=\dfrac{\cos y}{\cos(a+y)}\) from the original equation:
\[ x\sin(a+y)-\sin y=\dfrac{\cos y\sin(a+y)}{\cos(a+y)}-\sin y=\dfrac{\cos y\sin(a+y)-\sin y\cos(a+y)}{\cos(a+y)} \]
The numerator is \(\sin\big[(a+y)-y\big]=\sin a\) (sine subtraction formula), so this simplifies to \(\dfrac{\sin a}{\cos(a+y)}\).

Step 4: Solve for \(y'\):
\[ y'\cdot\dfrac{\sin a}{\cos(a+y)}=\cos(a+y) \implies y'=\dfrac{\cos^2(a+y)}{\sin a} \]

Final Answer:
\[ \boxed{\dfrac{dy}{dx}=\dfrac{\cos^2(a+y)}{\sin a}} \]