Convert $y=x\tan \alpha$ in polar form.
This is quite a trivial case. However, the answer given in the book does not seem quite correct to me. Here goes my solution.
We know that in a polar system we can represent every point $P(x,y)$ as $P(r,\theta)$ where $\theta$ is the angle of $OP$ with the $x$ axis with origin $O$ and $r$ is the distance of $P$ from $O$. So, we can say $x=r\cos\theta $ and $y=r\sin\theta$ and hence $\tan \theta =y/x$. Now, we know that $\tan \alpha=y/x$ (given). Hence, $\theta=\alpha$. Now, substituting $x=r\cos\theta$ and $y=r\sin\theta$ in the given equation we get now a so called polar equation.
However, the answer given in the book is $\sin\theta=\theta \cos\theta$. How is this possible? I am not quite getting it.