Question

find the angle between the radius vector and tangent for r=a. sec^2 theta/2​

Answer 1

The angle between the radius vector and tangent is Ф = tan-1 ($\frac{1}{tan(theta/2)\\}$)

The formula for calculating the angle between the radius vector and tangent is:

tan Ф = $\frac{r}{dr/d(theta)}$

Thereforre, differentiating r with respect to theta

$\frac{dr}{d(theta)}$ = d ( a × sec² (theta/2))

= a × 2 × sec $\frac{theta}{2\\}$ × sec $\frac{theta}{2\\}$ × tan$\frac{theta}{2}$ × $\frac{1}{2}$

= a × sec² $\frac{theta}{2}$ × tan$\frac{theta}{2}$

Therefore,

tan Ф = $\frac{asec2(theta/2)}{asec2(theta/2)tan(theta/2)}$

= $\frac{1}{tan(theta/2)}$

Therefore,

Ф = tan-1 ($\frac{1}{tan(theta/2)\\}$)