If y = y (x) is the solution of the differential equation

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

Now, put $\tan \mathrm{x}=\mathrm{t} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\sec ^{2} \mathrm{x}$

So $\frac{d y}{d t}+y=t$

On solving, we get $y e^{t}=e^{t}(t-1)+c$

$\Rightarrow \mathrm{y}=(\tan \mathrm{x}-1)+\mathrm{ce}^{-\tan x}$

$\Rightarrow \mathrm{y}(0)=0 \Rightarrow \mathrm{c}=1$

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

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