`5cos^(-1)((1-x^(2))/(1+x^(2)))+7sin^(-1)((2x)/(1+x^(2)))-4tan^(-1)((2x)/(1-x^(2)))-tan^(-1)x=5pi`, then x is equal to

Options

A. 3
B. `-sqrt(3)`
C. `sqrt(2)`
D. `sqrt(3)`

Explanation

To solve the equation \( 5\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 7\sin^{-1}\left(\frac{2x}{1+x^2}\right) - 4\tan^{-1}\left(\frac{2x}{1-x^2}\right) - \tan^{-1}x = 5\pi \), we will use the relationships between inverse trigonometric functions.

### Step 1: Rewrite the equation using known identities

We know from trigonometric identities that:
- \( \cos^{-1}(y) = \tan^{-1}\left(\frac{\sqrt{1-y^2}}{y}\right) \)
- \( \sin^{-1}(y) = \tan^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) \)
- \( \tan^{-1}(y) = \tan^{-1}(x) \)

Using the identity for \( \cos^{-1} \):
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{\sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2}}{\frac{1-x^2}{1+x^2}}\right) \]

### Step 2: Simplifying the terms

Using the identity for \( \sin^{-1} \):
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \tan^{-1}\left(\frac{\frac{2x}{1+x^2}}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right) \]

Using the identity for \( \tan^{-1} \):
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}(x) + \tan^{-1}(x) \]

### Step 3: Substitute back into the equation

Now substituting these back into the original equation, we can simplify:
\[ 5\tan^{-1}\left(\frac{\sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2}}{\frac{1-x^2}{1+x^2}}\right) + 7\tan^{-1}\left(\frac{\frac{2x}{1+x^2}}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right) - 4\tan^{-1}\left(\frac{2x}{1-x^2}\right) - \tan^{-1}(x) = 5\pi \]

### Step 4: Combine like terms

We can combine the terms involving \( \tan^{-1} \):
\[ (5 + 7 - 4 - 1)\tan^{-1}(x) = 5\pi \]
This simplifies to:
\[ 7\tan^{-1}(x) = 5\pi \]

### Step 5: Solve for \( x \)

Dividing both sides by 7 gives:
\[ \tan^{-1}(x) = \frac{5\pi}{7} \]
Taking the tangent of both sides:
\[ x = \tan\left(\frac{5\pi}{7}\right) \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = \tan\left(\frac{5\pi}{7}\right) \]