1. If $cos \frac{ \theta }{2}= \dfrac{4}{5}$, find the value of:

Solution:
Here,
$cos \theta = 2 cos^2 \frac{\theta}{2} -1$

$= 2 \left ( \dfrac{4}{5} \right )^2 - 1$

$= 2 × \dfrac{16}{25} -1$
$= \dfrac{32 - 25}{25}$
$= \dfrac{7}{25}$
Now,
a) $sin \theta$
Solution:

$= \sqrt{ 1 - cos^2 \theta}$
$= \sqrt{1 -  \left (  \dfrac{7}{25} \right )^2}$

$= \sqrt{\dfrac{25^2 - 49}{25^2}}$

$= \sqrt{\dfrac{625-49}{625}}$

$= \sqrt{\dfrac{576}{625}}$

$= \sqrt{ \left ( \dfrac{24}{25} \right )^2 }$

$= \dfrac{24}{25}$
b) $cos \theta$

Already mentioned above
c) $tan \theta$
Solution:
$= \dfrac{ sin \theta}{cos \theta}$

$= \dfrac{ \dfrac{24}{25} }{\dfrac{7}{25}}$

$= \dfrac{24}{25} × \dfrac{25}{7}$

$= \dfrac{24}{7}$