Question

If $\cos y=x\cos (a+y)$ with $\cos a\ne \pm 1$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{\sin a}{{\cos }^2(a+y)}$
• B. $\frac{{\cos }^2(a+y)}{\sin a}$
• C. $\frac{\cos a}{{\sin }^2(a+y)}$
• D. $\frac{{\cos }^2(a+y)}{\cos a}$

Answer 1

The correct option is B $\frac{{\cos }^2(a+y)}{\sin a}$

$\cos y=x\cos (a+y)$
$\Rightarrow \frac{\cos y}{\cos (a+y)}=x$
Differentiating both sides wrt $x$, we get
$\frac{-\sin \left(y\right)y^{\prime }.\cos (a+y)+\sin (a+y)y^{\prime }.\cos y}{{\cos }^2(a+y)}=1$
$\Rightarrow \frac{y^{\prime }\sin (a+y-y)}{{\cos }^2(a+y)}=1$
$\text{i.e.},y^{\prime }=\frac{dy}{dx}=\frac{{\cos }^2(a+y)}{\sin a}$