1. Given differential equation

$\frac{dy}{dx}+\frac{y}{2}\sec x=\frac{\tan x}{2y}, \qquad 0\le x<\frac{\pi}{2}, \qquad y(0)=1$

We want to solve this initial value problem.

2. Convert into an equation in $y^2$

Multiply both sides by $2y$:

$2y\frac{dy}{dx}+y^2\sec x=\tan x$

Now note that

$\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

So the equation becomes

$\frac{d}{dx}(y^2)+y^2\sec x=\tan x$

Let

$v=y^2$

Then we get the linear differential equation

$\frac{dv}{dx}+v\sec x=\tan x$

3. Solve the linear equation

The integrating factor is

$\text{I.F.}=e^{\int \sec x\,dx}$

We know

$\int \sec x\,dx=\ln(\sec x+\tan x)$

Hence

$\text{I.F.}=e^{\ln(\sec x+\tan x)}=\sec x+\tan x$

Multiplying the equation by the integrating factor:

$(\sec x+\tan x)\frac{dv}{dx}+v\sec x(\sec x+\tan x)=\tan x(\sec x+\tan x)$

The left side becomes

$\frac{d}{dx}\left[v(\sec x+\tan x)\right]$

Therefore,

$\frac{d}{dx}\left[v(\sec x+\tan x)\right]=\tan x(\sec x+\tan x)$

4. Integrate the right-hand side

We need

$\int \tan x(\sec x+\tan x)\,dx$

Expand:

$\int (\sec x\tan x+\tan^2 x)\,dx$

Using

$\tan^2 x=\sec^2 x-1$

we get

$\int (\sec x\tan x+\sec^2 x-1)\,dx$

Now integrate termwise:

$\int \sec x\tan x\,dx=\sec x$ $\int \sec^2 x\,dx=\tan x$ $\int 1\,dx=x$

So,

$\int \tan x(\sec x+\tan x)\,dx=\sec x+\tan x-x+C$

Hence

$v(\sec x+\tan x)=\sec x+\tan x-x+C$

Thus

$v=1-\frac{x}{\sec x+\tan x}+\frac{C}{\sec x+\tan x}$

Since $v=y^2$,

$y^2=1-\frac{x}{\sec x+\tan x}+\frac{C}{\sec x+\tan x}$

5. Use the initial condition

Given $y(0)=1$, so

$y^2(0)=1$

At $x=0$,

$\sec 0+\tan 0=1+0=1$

So

$1=1-\frac{0}{1}+\frac{C}{1}$

This gives

$C=0$

Therefore,

$y^2=1-\frac{x}{\sec x+\tan x}$

6. Match with the options

This is exactly:

$\boxed{y^2=1-\frac{x}{\sec x+\tan x}}$

So the correct option is C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They match.