Answer:
![sec(11\pi )/(3) =2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/pfa3tnj6fu57arryn3gi7q238f3yizmhlz.png)
Explanation:
Given
![sec(11\pi )/(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/6w5z5b9yv1ryqkc4kk8bg9aqfti1aopqys.png)
=
![sec( 4\pi -(\pi )/(3))](https://img.qammunity.org/2019/formulas/mathematics/middle-school/9dqdw7oiu0zt2wqbvim3vvpoqny7qcd7v4.png)
It lies in the IV quadrant .Therefore we can write as
![sec(11\pi )/(3) =sec(\pi )/(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/qaoulzn0qtivacomi5xodtvd5cwgvdt15n.png)
Because
![sec(2\pi -\theta)= sec\theta](https://img.qammunity.org/2019/formulas/mathematics/middle-school/rr14zq9osed50njhdf8u0rbmqmrh1qbla7.png)
Because
is positive in first quadrant and IV quadrant . There is
lies in the IV quadrant .
Hence,
is positive in IV quadrant .
![sec(11\pi )/(3) =sec(\pi )/(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/qaoulzn0qtivacomi5xodtvd5cwgvdt15n.png)
We know that value of
![sec(\pi )/(3) =sec60^(\circ)=2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/x1gflz0l31h4f3uanmujhcn8cf7fb3jcc6.png)
Therefore,
.