If `1/2"sin"^(-1)(2x)/(1+x^2)+1/2"cos"^(-1)(1-y^2)/(1+y^2)+1/3"tan"^(-1)(3z-z^3)/(1-3z^2)=5pi` then x+y+z=

Options

A. xyz
B. xy+yz+zx
C. 1=xyz
D. none of these

Answer

To solve the equation

\[ \frac{1}{2} \sin^{-1}(2x)/(1+x^2) + \frac{1}{2} \cos^{-1}(1-y^2)/(1+y^2) + \frac{1}{3} \tan^{-1}(3z-z^3)/(1-3z^2) = 5\pi, \]

we will use the properties of inverse trigonometric functions.

### Step 1: Apply the Inverse Trigonometric Identities

Using the identities:
- \(\sin^{-1}(2x) = 2 \tan^{-1}(x)\) for \(|x| \leq \frac{1}{\sqrt{2}}\),
- \(\cos^{-1}(1-y^2) = 2 \tan^{-1}(y)\) for \(|y| \leq 1\),
- \(\tan^{-1}(3z - z^3) = 3 \tan^{-1}(z)\) for \(|z| < 1\).

We can rewrite the equation as:

\[ \frac{1}{2} \cdot 2 \tan^{-1}(x) + \frac{1}{2} \cdot 2 \tan^{-1}(y) + \frac{1}{3} \cdot 3 \tan^{-1}(z) = 5\pi. \]

This simplifies to:

\[ \tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(z) = 5\pi. \]

### Step 2: Use the Addition Formula for Inverse Tangents

We know that:

\[ \tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(z) = \tan^{-1}\left(\frac{x + y + z - xyz}{1 - (xy + yz + zx)}\right). \]

Setting this equal to \(5\pi\):

\[ \tan^{-1}\left(\frac{x + y + z - xyz}{1 - (xy + yz + zx)}\right) = 5\pi. \]

### Step 3: Evaluate the Tangent Function

Since \(\tan(n\pi) = 0\) for any integer \(n\), we have:

\[ \frac{x + y + z - xyz}{1 - (xy + yz + zx)} = 0. \]

### Step 4: Solve the Equation

From the equation above, we can deduce:

\[ x + y + z - xyz = 0. \]

Thus, we can rearrange it to:

\[ x + y + z = xyz. \]

### Conclusion

The final result is:

\[ x + y + z = xyz. \]