0.96

It is given that: $\space \sin{\theta} = \frac{24}{25}$

$\Rightarrow \cos{\theta} = \sqrt{1- \left( \frac{24}{25} \right)^2} = \sqrt{ \frac{25^2 - 24^2}{25^2}} = \sqrt{ \frac{(25 - 24)(25+24)}{25^2}} = \sqrt{ \frac{49}{25^2}} = \frac{7}{25}$

$\begin{aligned} \frac {\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}} & = \frac {\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}} \times \frac{2\cos{\frac{\theta}{2}}}{2\cos{\frac{\theta}{2}}} \\ & = \frac {2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}+2\cos^2{\frac{\theta}{2}}} \\ & = \frac {2\cos^2{\frac{\theta}{2}}-1+1}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}+2\cos^2{\frac{\theta}{2}}-1+1} \\ & = \frac {\cos{\theta}+1}{\sin{\theta}+\cos{\theta}+1} = \frac {\frac{7}{25}}{\frac{24}{25}+\frac{7}{25}+1} = \frac{32}{56} = \boxed{\frac{4}{7}}\end{aligned}$

sin(theta)=24/25 so (theta)=sin inverse of 24/25 which is 73.74 nw substitute in place of theta as 73.74

Let x= cos (theta), y= sin(theta). Let alpha= cos(theta/2), beta=sin(theta/2).

Let the value of the given expression be V. Then.

V=alpha/(alpha+beta)

=alpha(alpha-beta)/(alpha^2-beta^2)

= (2alpha^2-2alpha beta)/[2(alpha^2-beta^2)]

From the definitions of alpha and beta,

2 alpha beta= sin(theta)=y

2 alpha^2-1=alpha^2-beta^2=cos(theta)=x, i.e., 2 alpha^2=1+x

Therefore,

V=(1+x-y)/(2x)

y=24/25 implies x= 7/25 (since x^2+y^2=1)

Therefore, V= (25+7-24)/14=8/14=4/7