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: First isolate \(x\) explicitly from the given relation:
$x=\dfrac{\cos y}{\cos(a+y)}$ (valid since $\cos a\ne\pm1$ keeps things well-defined near $y=0$, ensuring $\cos(a+y)\ne0$ in the relevant range).

Step 2: Differentiate this explicit form using the quotient rule:
$\dfrac{dx}{dy}=\dfrac{-\sin y\cos(a+y)-\cos y\cdot(-\sin(a+y))}{\cos^2(a+y)}=\dfrac{-\sin y\cos(a+y)+\cos y\sin(a+y)}{\cos^2(a+y)}$.

Step 3: Simplify the numerator using the sine subtraction identity:
$\cos y\sin(a+y)-\sin y\cos(a+y)=\sin\big[(a+y)-y\big]=\sin a$. So $\dfrac{dx}{dy}=\dfrac{\sin a}{\cos^2(a+y)}$.

Step 4: Invert to get \(dy/dx\):
$\dfrac{dy}{dx}=\dfrac1{dx/dy}=\dfrac{\cos^2(a+y)}{\sin a}$.

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