For the family of curves rho^m=k^mcos(m theta), prove: 1°

For the family of curves $\rho^{m}=k^{m}\cos(m\theta)$, prove: 1° the angle $\varphi$ between the tangent and the radius vector satisfies $\varphi=\dfrac{(2n+1)\pi}{2}+m\theta$; 2° the angle $\psi$ between the polar axis and the normal satisfies $\psi=(m+1)\theta+n\pi$ ($n\in\mathbb Z$).

Step 1: Differentiate the curve equation

Differentiating $\rho^{m}=k^{m}\cos(m\theta)$ with respect to $\theta$ and dividing by $m$: \[\rho^{m-1}\rho' = -k^{m}\sin(m\theta). \tag{1}\]

Since $k^{m}=\rho^{m}/\cos(m\theta)$ from the curve equation itself, substituting into (1) and dividing by $\rho^{m-1}$: \[\rho' = -\rho\tan(m\theta). \tag{2}\]

Step 2: Part 1°: the tangent–radius angle

By the polar formula $\tan\varphi=\rho/\rho'$, \[\tan\varphi = \frac{\rho}{-\rho\tan(m\theta)} = -\cot(m\theta) = \tan\!\left(\frac\pi2+m\theta\right),\]

so, allowing for the periodicity of $\tan$, \[\varphi = \frac{(2n+1)\pi}{2} + m\theta, \qquad n\in\mathbb Z. \tag{$1^{\circ}$}\]

Step 3: Part 2°: the polar-axis–normal angle

The tangent makes angle $\theta+\varphi$ with the polar axis, so the normal (perpendicular to it) makes angle $\psi=\theta+\varphi-\tfrac\pi2$. Substituting (1°), \[\psi = \theta + \frac{(2n+1)\pi}{2} + m\theta - \frac\pi2 = (m+1)\theta + n\pi, \qquad n\in\mathbb Z. \tag{$2^{\circ}$}\]

Step 4: Verification

\[m=2,\ \theta=0.2\text{: } \tan\varphi \text{ from (1) matches } \tan\!\left(\tfrac\pi2+m\theta\right) \text{ (numerically, agreement to } 10^{-6})\ \checkmark\]