A sector OAP of a circle …

Given, Radius = r
{tex} \triangle{/tex}AOB is a right triangle.
In the right {tex} \triangle{/tex}OAB,
{tex} \tan \theta = \frac { \mathrm { AB } } { \mathrm { OA } }{/tex}
{tex} \Rightarrow A B = O A \times \tan \theta{/tex}
{tex} = r \tan \theta{/tex}
Now, area of {tex} \triangle{/tex}OAB ={tex}\frac12 b \times h{/tex}={tex} \frac { 1 } { 2 } \mathrm { OA } \times \mathrm { AB }{/tex}
{tex} = \frac { 1 } { 2 } \times r \times r \tan \theta{/tex}
{tex} = \frac { 1 } { 2 } r ^ { 2 } \tan \theta{/tex}
and area of sector OAC ={tex} \pi r ^ { 2 } \times \frac { \theta } { 360 ^ { \circ } }{/tex}

 length of arc AC = {tex} 2 \pi r \times \frac { \theta } { 360 ^ { \circ } }{/tex}
{tex} = \frac { 2 \pi r \theta } { 360 ^ { \circ } } = \frac { \pi r \theta } { 180 ^ { \circ } }{/tex}
{tex} \therefore{/tex}Perimeter of the shaded region
= Arc AC + AB + BC
{tex} = \frac { \pi r\theta } { 180 } + r \tan \theta + ( \mathrm { OB } - \mathrm { OC } ){/tex}
{tex} = \frac { \pi r\theta } { 180 } + r \tan \theta + ( r \sec \theta - r ){/tex}
{tex} = r \left( \frac { \pi \theta } { 180 } + \tan \theta + \sec \theta - 1 \right){/tex}
{tex} = r \left( \tan \theta + \sec \theta + \frac { \pi \theta } { 180 } - 1 \right){/tex}