The value of $$\tan^{-1}\left(\frac{\cos\frac{15\pi}{4} - 1}{\sin\frac{\pi}{4}}\right)$$ is equal to

First evaluate the trigonometric numbers inside the inverse tangent.

Periodicity of cosine: $$\cos(\theta + 2\pi)=\cos\theta$$.
Write $$\frac{15\pi}{4}=2\pi+\frac{7\pi}{4}$$, so

$$\cos\frac{15\pi}{4}=\cos\frac{7\pi}{4}=\cos\left(-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}.$$

Since $$\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$$, the numerator becomes

$$\cos\frac{15\pi}{4}-1=\frac{\sqrt2}{2}-1.$$

The denominator is straightforward:

$$\sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$$

Form the required fraction:

$$\frac{\cos\frac{15\pi}{4}-1}{\sin\frac{\pi}{4}} =\frac{\frac{\sqrt2}{2}-1}{\frac{\sqrt2}{2}} =\left(\frac{\sqrt2}{2}\right)\!\Big/\!\left(\frac{\sqrt2}{2}\right) -\;1\!\Big/\!\left(\frac{\sqrt2}{2}\right) =1-\frac{2}{\sqrt2} =1-\sqrt2.$$

Thus we need $$\tan^{-1}(1-\sqrt2).$$ Recall the well-known exact value

$$\tan\frac{\pi}{8}=\sqrt2-1.$$

Therefore

$$1-\sqrt2=-(\sqrt2-1)=-\tan\frac{\pi}{8} =\tan\!\left(-\frac{\pi}{8}\right).$$

Since the principal value range of $$\tan^{-1}$$ is $$(-\frac{\pi}{2},\frac{\pi}{2})$$, we obtain

$$\tan^{-1}(1-\sqrt2)=-\frac{\pi}{8}.$$

Hence the correct option is Option B: $$-$$\frac{\pi}{8}$$.$