The value of $$\frac{1}{\sec x - \tan x} - \frac{1}{\cos x}, 0^\circ < x < 90^\circ,$$ is equal to:

$$\frac{1}{\sec x - \tan x} - \frac{1}{\cos x}$$

simplifying the sum secx-tanx

use the identity tanx=$$\frac{sinx}{cosx}$$

then use the identity secx=$$\frac{1}{cosx}$$

$$\frac{1}{1÷cosx-sinx÷cosx}$$ - $$\frac{1}{cosx}$$

$$\frac{1}{1-sinx÷cosx}$$ - $$\frac{1}{cosx}$$

$$\frac{cosx}{1-sinx}$$ - $$\frac{-1}{cosx}$$

order factor for (1-sinx)cosx

$$\frac{cos^2x-1+sinx}{cosx(1-sinx)}$$

use the identity 

$$\cos^2 \theta +\sin^2 \theta$$=1

$$\sin^2 \theta$$=1-$$\cos^2 \theta$$

-$$\sin^2 \theta$$ = -(1-$$\cos^2 \theta$$)

-$$\sin^2 \theta$$= -1+$$\cos^2 \theta$$

where θ is replaced by x

Hence =  $$\frac{-1+cos^2x+sinx}{cosx(1-sinx)}$$

           = $$\frac{-sin^2x+sinx}{cosx(1-sinx)}$$

.          = $$\frac{sinx(1-sinx)}{cosx(1-sinx)}$$

cancel (1-sinx)(1-sinx)

then =$$\frac{sinx}{cosx}$$

by using the identity we know that tanx=$$\frac{sinx}{cosx}$$