Let $$f(x) = \dfrac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$$, $$x \in [0, \pi] - \{\dfrac{\pi}{4}\}$$, then $$f\left(\dfrac{7\pi}{12}\right) f''\left(\dfrac{7\pi}{12}\right)$$ is equal to

To solve this, we first simplify $$f(x)$$ using trigonometric identities.

1. Simplify $$f(x)$$

Recall that $$\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})$$ and $$\sin x - \cos x = -\sqrt{2}\cos(x + \frac{\pi}{4})$$.

$$f(x) = \frac{\sqrt{2}\sin(x + \frac{\pi}{4}) - \sqrt{2}}{-\sqrt{2}\cos(x + \frac{\pi}{4})} = \frac{1 - \sin(x + \frac{\pi}{4})}{\cos(x + \frac{\pi}{4})}$$

Using half-angle identities (let $$\theta = x + \frac{\pi}{4}$$), we get:

$$f(x) = \frac{1 - \cos(\frac{\pi}{2} - \theta)}{\sin(\frac{\pi}{2} - \theta)} = \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \tan\left(\frac{\pi}{8} - \frac{x}{2}\right)$$

2. Find the Derivatives

Let $$u = \frac{\pi}{8} - \frac{x}{2}$$, so $$f(x) = \tan u$$.

• $$f'(x) = \sec^2 u \cdot (-\frac{1}{2})$$
• $$f''(x) = -\frac{1}{2} [2\sec u \cdot \sec u \tan u \cdot (-\frac{1}{2})] = \frac{1}{2} \sec^2 u \tan u$$

3. Evaluate at $$x = \frac{7\pi}{12}$$

Calculate $$u$$ first:

$$u = \frac{\pi}{8} - \frac{7\pi}{24} = \frac{3\pi - 7\pi}{24} = -\frac{4\pi}{24} = -\frac{\pi}{6}$$

Now find the values:

• $$f(-\frac{\pi}{6}) = \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$$
• $$f''(-\frac{\pi}{6}) = \frac{1}{2} \sec^2(-\frac{\pi}{6}) \tan(-\frac{\pi}{6}) = \frac{1}{2} \cdot (\frac{2}{\sqrt{3}})^2 \cdot (-\frac{1}{\sqrt{3}}) = \frac{1}{2} \cdot \frac{4}{3} \cdot (-\frac{1}{\sqrt{3}}) = -\frac{2}{3\sqrt{3}}$$

4. Final Calculation

$$f\left(\frac{7\pi}{12}\right) f''\left(\frac{7\pi}{12}\right) = \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{2}{3\sqrt{3}}\right) = \frac{2}{3 \cdot 3} = \mathbf{\frac{2}{9}}$$