Let $$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}$$. Then $f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)$ is equal to

$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$ <br/><br/>$$ \begin{aligned} & =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\\\ & =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\ & =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\\\ & \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\\\ & =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\\\ & =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\\\ & =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\\\ & =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9} \end{aligned} $$