Show that :(i) ` sin^(-1) [ sin. (3pi)/4] != (3pi)/4`
(ii) `tan^(-1) [ tan. (5pi)/6] != (5pi)/6` . What is its value ?

Answer

To solve the given problems, we will analyze each part step by step.

### Part (i): Show that \( \sin^{-1} \left( \sin \left( \frac{3\pi}{4} \right) \right) \neq \frac{3\pi}{4} \)

**Step 1: Understand the range of the inverse sine function.**
The function \( \sin^{-1}(x) \) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

**Hint:** Remember that inverse trigonometric functions return values within a specific range.

**Step 2: Calculate \( \sin \left( \frac{3\pi}{4} \right) \).**
Using the sine function, we find:
\[ \sin \left( \frac{3\pi}{4} \right) = \sin \left( \pi - \frac{\pi}{4} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \]

**Hint:** Use the identity \( \sin(\pi - x) = \sin(x) \).

**Step 3: Apply the inverse sine function.**
Now, we need to find:
\[ \sin^{-1} \left( \sin \left( \frac{3\pi}{4} \right) \right) = \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) \]
Since \( \frac{\sqrt{2}}{2} \) corresponds to \( \frac{\pi}{4} \) in the range of \( \sin^{-1} \), we have:
\[ \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = \frac{\pi}{4} \]

**Hint:** The inverse sine function returns the angle in its defined range.

**Step 4: Conclusion.**
Thus, we conclude:
\[ \sin^{-1} \left( \sin \left( \frac{3\pi}{4} \right) \right) = \frac{\pi}{4} \neq \frac{3\pi}{4} \]

### Part (ii): Show that \( \tan^{-1} \left( \tan \left( \frac{5\pi}{6} \right) \right) \neq \frac{5\pi}{6} \)

**Step 1: Understand the range of the inverse tangent function.**
The function \( \tan^{-1}(x) \) has a range of \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

**Hint:** Similar to sine, the range of the inverse tangent function is limited.

**Step 2: Calculate \( \tan \left( \frac{5\pi}{6} \right) \).**
Using the tangent function, we find:
\[ \tan \left( \frac{5\pi}{6} \right) = \tan \left( \pi - \frac{\pi}{6} \right) = -\tan \left( \frac{\pi}{6} \right) = -\frac{1}{\sqrt{3}} \]

**Hint:** Use the identity \( \tan(\pi - x) = -\tan(x) \).

**Step 3: Apply the inverse tangent function.**
Now, we need to find:
\[ \tan^{-1} \left( \tan \left( \frac{5\pi}{6} \right) \right) = \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) \]
The angle corresponding to \( -\frac{1}{\sqrt{3}} \) in the range of \( \tan^{-1} \) is:
\[ \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) = -\frac{\pi}{6} \]

**Hint:** The inverse tangent function gives the angle in its defined range.

**Step 4: Conclusion.**
Thus, we conclude:
\[ \tan^{-1} \left( \tan \left( \frac{5\pi}{6} \right) \right) = -\frac{\pi}{6} \neq \frac{5\pi}{6} \]

### Final Answers:
(i) \( \sin^{-1} \left( \sin \left( \frac{3\pi}{4} \right) \right) = \frac{\pi}{4} \)
(ii) \( \tan^{-1} \left( \tan \left( \frac{5\pi}{6} \right) \right) = -\frac{\pi}{6} \)