Inverse trigonometric functions are functions that "undo" the effects of trigonometric functions. They provide a way to find the angle (or value) associated with a given trigonometric ratio. Inverse trigonometric functions are denoted by \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), \(\tan^{-1}(x)\), \(\cot^{-1}(x)\), \(\sec^{-1}(x)\), and \(\csc^{-1}(x)\), representing arcsine, arccosine, arctangent, arccotangent, arcsecant, and arccosecant, respectively.

Here's a brief explanation of each inverse trigonometric function:

1. **arcsin (or \(\sin^{-1}(x)\))**: Gives the angle whose sine is \(x\), where \(x\) is between -1 and 1.

2. **arccos (or \(\cos^{-1}(x)\))**: Gives the angle whose cosine is \(x\), where \(x\) is between -1 and 1.

3. **arctan (or \(\tan^{-1}(x)\))**: Gives the angle whose tangent is \(x\).

4. **arccot (or \(\cot^{-1}(x)\))**: Gives the angle whose cotangent is \(x\).

5. **arcsec (or \(\sec^{-1}(x)\))**: Gives the angle whose secant is \(x\), where \(x \geq 1\) or \(x \leq -1\).

6. **arccsc (or \(\csc^{-1}(x)\))**: Gives the angle whose cosecant is \(x\), where \(x \geq 1\) or \(x \leq -1\).

It's important to note that the range of inverse trigonometric functions is restricted to ensure that they are single-valued and have unique inverses. The specific range depends on the convention used, but commonly accepted ranges are as follows:

- For arcsin and arccos: \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) (or \(-90^\circ \leq \theta \leq 90^\circ\)).
- For arctan: \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\) (or \(-90^\circ < \theta < 90^\circ\)).
- For arccot: \(0 < \theta < \pi\) (or \(0^\circ < \theta < 180^\circ\)).
- For arcsec and arccsc: \(0 \leq \theta < \frac{\pi}{2}\) and \(\frac{\pi}{2} < \theta \leq \pi\) (or \(0^\circ \leq \theta < 90^\circ\) and \(90^\circ < \theta \leq 180^\circ\)).

These functions are essential in solving trigonometric equations, modeling periodic phenomena, and various applications in science, engineering, and mathematics.

Let sin-1(3/5) = x and sin-1(8/17) = y

Therefore  sinx = 3/5 and siny = 8/17

Now, cosx = √(1 - sin2x) = √(1 - (3/5)2) = √(1 - 9/25) = 4/5 and cosy = √(1 - sin2y) = √(1 - (8/17)2) = √(1 - 64/289) = 15/17

We have cos(x - y) = cosx cosy + sinx siny  = 4/5 x 15/17 + 3/5 x 8/17 = 60/85 + 24/85 = 84/85 ⇒ x - y = cos-1(84/85) ⇒ sin-1(3/5) - sin-1(8/17) = cos-1(84/85)

Let's denote \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) \) as \( \alpha \) and \( \tan^{-1} \left( \frac{x + 1}{x + 2} \right) \) as \( eta \).

Given that \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) + \tan^{-1} \left( \frac{x + 1}{x + 2} \right) = \frac{\theta}{4} \), we can use the tangent addition formula:

\[ \tan(\alpha + eta) = \frac{\tan \alpha + \tan eta}{1 - \tan \alpha \cdot \tan eta} \]

Substitute \( \tan \alpha = \frac{x - 1}{x - 2} \) and \( \tan eta = \frac{x + 1}{x + 2} \):

\[ \tan(\alpha + eta) = \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \cdot \frac{x + 1}{x + 2}} \]

\[ \tan(\alpha + eta) = \frac{\frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}}{1 - \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}} \]

\[ \tan(\alpha + eta) = \frac{x^2 + x - 2 + x^2 - x - 2}{(x - 2)(x + 2) - (x^2 - 1)} \]

\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{x^2 + 4 - x^2 + 1} \]

\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{5} \]

Given that \( \tan(\alpha + eta) = \frac{\theta}{4} \), we have:

\[ \frac{2x^2 - 4}{5} = \frac{\theta}{4} \]

\[ 8x^2 - 16 = 5\theta \]

\[ 8x^2 = 5\theta + 16 \]

\[ x^2 = \frac{5\theta + 16}{8} \]

\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]

So, the value of \( x \) depends on the value of \( \theta \).

In LaTeX code:
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]