1. Given differential equation

$\frac{dy}{dx}=(\tan x-y)\sec^2 x, \qquad y(0)=0$

We need to find:

$y\left(-\frac{\pi}{4}\right)$

2. Rewrite in linear form

Expand the right side:

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

So,

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

This is a linear differential equation of the form

$\frac{dy}{dx}+P(x)y=Q(x)$

with

$P(x)=\sec^2 x, \qquad Q(x)=\tan x\sec^2 x$

3. Find the integrating factor

The integrating factor is

$\text{I.F.}=e^{\int \sec^2 x\,dx}=e^{\tan x}$

4. Multiply the equation by the integrating factor

Multiplying throughout by $e^{\tan x}$:

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

The left-hand side becomes:

$\frac{d}{dx}\left(ye^{\tan x}\right)=\tan x\sec^2 x\,e^{\tan x}$

5. Integrate both sides

$ye^{\tan x}=\int \tan x\sec^2 x\,e^{\tan x}\,dx + C$

Let

$t=\tan x \implies dt=\sec^2 x\,dx$

Then the integral becomes:

$\int t e^t\,dt$

Using integration by parts or the standard result,

$\int t e^t\,dt=e^t(t-1)+C$

So,

$ye^{\tan x}=e^{\tan x}(\tan x-1)+C$

Hence,

$y=\tan x-1+Ce^{-\tan x}$

6. Use the initial condition

Given $y(0)=0$ and $\tan 0=0$:

$0=0-1+Ce^0$

$0=-1+C$

$C=1$

Thus the solution is

$y=\tan x-1+e^{-\tan x}$

7. Evaluate at $x=-\frac{\pi}{4}$

Since

$\tan\left(-\frac{\pi}{4}\right)=-1$

we get

$y\left(-\frac{\pi}{4}\right)=(-1)-1+e^{-(-1)}$

$y\left(-\frac{\pi}{4}\right)=-2+e$

$y\left(-\frac{\pi}{4}\right)=e-2$

8. Compare with options

$e-2$

This matches Option B.