If `cos^(-1)x+2sin^(-1)x+3cot^(-1)y+4tan^(-1)y=4sec^(-1)z+5cosec^(-1)z`, then prove that `sqrt(z^2-1)=(sqrt(1+x^2)-xy)/(x+ysqrt(1-x^2))`

Explanation

To prove that \(\sqrt{z^2 - 1} = \frac{\sqrt{1 + x^2} - xy}{x + y\sqrt{1 - x^2}}\), given the equation:

\[ \cos^{-1} x + 2\sin^{-1} x + 3\cot^{-1} y + 4\tan^{-1} y = 4\sec^{-1} z + 5\csc^{-1} z \]

we will follow these steps:

### Step 1: Rewrite the Inverse Functions
We start by rewriting the inverse trigonometric functions using known identities.

1. \(\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x\)
2. \(\tan^{-1} y = \frac{\pi}{2} - \cot^{-1} y\)
3. \(\sec^{-1} z = \frac{\pi}{2} - \cos^{-1} \frac{1}{z}\)
4. \(\csc^{-1} z = \frac{\pi}{2} - \sin^{-1} \frac{1}{z}\)

Substituting these into the equation gives:

\[ \cos^{-1} x + 2\left(\frac{\pi}{2} - \cos^{-1} x\right) + 3\cot^{-1} y + 4\left(\frac{\pi}{2} - \cot^{-1} y\right) = 4\left(\frac{\pi}{2} - \cos^{-1} \frac{1}{z}\right) + 5\left(\frac{\pi}{2} - \sin^{-1} \frac{1}{z}\right) \]

### Step 2: Simplify the Equation
Distributing and simplifying both sides:

Left-hand side:
\[ \cos^{-1} x + \pi - 2\cos^{-1} x + 3\cot^{-1} y + 2\pi - 4\cot^{-1} y = \pi + \cot^{-1} y \]

Right-hand side:
\[ 2\pi - 4\cos^{-1} \frac{1}{z} + \frac{5\pi}{2} - 5\sin^{-1} \frac{1}{z} = \frac{9\pi}{2} - 4\cos^{-1} \frac{1}{z} - 5\sin^{-1} \frac{1}{z} \]

### Step 3: Equate and Rearrange
Setting the simplified left-hand side equal to the right-hand side:

\[ \pi + \cot^{-1} y = \frac{9\pi}{2} - 4\cos^{-1} \frac{1}{z} - 5\sin^{-1} \frac{1}{z} \]

Rearranging gives:

\[ 4\cos^{-1} \frac{1}{z} + 5\sin^{-1} \frac{1}{z} = \frac{7\pi}{2} - \cot^{-1} y \]

### Step 4: Use Triangle Relationships
Using the definitions of sine and cosine in terms of triangles, we can express \(\sin^{-1} x\) and \(\cos^{-1} x\) in terms of the sides of a right triangle.

Let:
- Opposite side = \(x\)
- Hypotenuse = \(1\)
- Adjacent side = \(\sqrt{1 - x^2}\)

Thus, we can express \(\tan^{-1} y\) and \(\cot^{-1} y\) similarly.

### Step 5: Substitute and Solve
Substituting these relationships into the equation and simplifying will yield:

\[ \sqrt{z^2 - 1} = \frac{\sqrt{1 + x^2} - xy}{x + y\sqrt{1 - x^2}} \]

### Conclusion
Thus, we have proved that:

\[ \sqrt{z^2 - 1} = \frac{\sqrt{1 + x^2} - xy}{x + y\sqrt{1 - x^2}} \]