1. Simplify the two curves

Given $y_1=\sqrt{\frac{1+\sin x}{\cos x}},\qquad y_2=\sqrt{\frac{1-\sin x}{\cos x}}.$

Use the identity $\frac{1+\sin x}{\cos x}=\sec x+\tan x,$ $\frac{1-\sin x}{\cos x}=\sec x-\tan x.$

So, $y_1=\sqrt{\sec x+\tan x},\qquad y_2=\sqrt{\sec x-\tan x}.$

Also, $(\sec x+\tan x)(\sec x-\tan x)=1,$ so $y_1y_2=1 \implies y_2=\frac1{y_1}.$

2. Determine which curve is upper and which is lower

For $x\in[0,\pi/4]$, we have $\sin x\ge 0$, hence $1+\sin x \ge 1-\sin x,$ and since $\cos x>0$ on this interval, $\frac{1+\sin x}{\cos x} \ge \frac{1-\sin x}{\cos x}.$ Thus $y_1\ge y_2.$

Therefore the required area is

3. Rewrite the integrand conveniently

Let $u=\sqrt{\sec x+\tan x}.$ Then $y_1=u,\qquad y_2=\frac1u,$ so $y_1-y_2=u-\frac1u=\frac{u^2-1}{u}.$

But $u^2=\sec x+\tan x=\frac{1+\sin x}{\cos x}.$ A better standard substitution is $t=\tan\frac x2.$ Then $\sin x=\frac{2t}{1+t^2},\qquad \cos x=\frac{1-t^2}{1+t^2},\qquad dx=\frac{2}{1+t^2}\,dt.$

Now compute the two curves: $\frac{1+\sin x}{\cos x}=\frac{1+\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}=\frac{(1+t)^2}{1-t^2}=\frac{1+t}{1-t},$ so $y_1=\sqrt{\frac{1+t}{1-t}}.$

Similarly, $\frac{1-\sin x}{\cos x}=\frac{1-t}{1+t},$ so $y_2=\sqrt{\frac{1-t}{1+t}}.$

Hence

Take common form: $y_1-y_2=\frac{(1+t)-(1-t)}{\sqrt{1-t^2}}=\frac{2t}{\sqrt{1-t^2}}.$

Therefore