If solution of `cot(sin^(-1)sqrt(1-x^(2)))=sin(tan^(-1)(xsqrt(6))),x!=0` is `1/psqrt(q/r)` where `p,q,r` are prime then value of `(p+q-r)/p` is

Explanation

To solve the equation \( \cot(\sin^{-1}(\sqrt{1-x^2})) = \sin(\tan^{-1}(x\sqrt{6})) \), we will follow these steps:

### Step 1: Rewrite the left-hand side
We know that:
\[ \cot(\sin^{-1}(y)) = \frac{\sqrt{1-y^2}}{y} \]
Thus, substituting \( y = \sqrt{1-x^2} \):
\[ \cot(\sin^{-1}(\sqrt{1-x^2})) = \frac{\sqrt{1 - (1-x^2)}}{\sqrt{1-x^2}} = \frac{\sqrt{x^2}}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}} \]

### Step 2: Rewrite the right-hand side
Using the identity for sine in terms of tangent:
\[ \sin(\tan^{-1}(z)) = \frac{z}{\sqrt{1+z^2}} \]
Substituting \( z = x\sqrt{6} \):
\[ \sin(\tan^{-1}(x\sqrt{6})) = \frac{x\sqrt{6}}{\sqrt{1+(x\sqrt{6})^2}} = \frac{x\sqrt{6}}{\sqrt{1 + 6x^2}} \]

### Step 3: Set the two sides equal
Now we set the two expressions equal to each other:
\[ \frac{x}{\sqrt{1-x^2}} = \frac{x\sqrt{6}}{\sqrt{1 + 6x^2}} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ x \sqrt{1 + 6x^2} = x\sqrt{6} \sqrt{1 - x^2} \]
Assuming \( x \neq 0 \), we can divide both sides by \( x \):
\[ \sqrt{1 + 6x^2} = \sqrt{6} \sqrt{1 - x^2} \]

### Step 5: Square both sides
Squaring both sides results in:
\[ 1 + 6x^2 = 6(1 - x^2) \]
Expanding the right side:
\[ 1 + 6x^2 = 6 - 6x^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 6x^2 + 6x^2 = 6 - 1 \]
\[ 12x^2 = 5 \]
Thus:
\[ x^2 = \frac{5}{12} \]

### Step 7: Solve for \( x \)
Taking the square root:
\[ x = \pm \sqrt{\frac{5}{12}} = \pm \frac{\sqrt{5}}{2\sqrt{3}} = \pm \frac{\sqrt{15}}{6} \]
Since \( x \neq 0 \), we consider the positive root:
\[ x = \frac{\sqrt{15}}{6} \]

### Step 8: Express in the required form
We need to express \( x \) in the form \( \frac{1}{p}\sqrt{\frac{q}{r}} \):
\[ x = \frac{1}{2} \sqrt{\frac{5}{3}} \]
Thus, we have \( p = 2 \), \( q = 5 \), and \( r = 3 \).

### Step 9: Calculate \( \frac{p + q - r}{p} \)
Now, we compute:
\[ \frac{p + q - r}{p} = \frac{2 + 5 - 3}{2} = \frac{4}{2} = 2 \]

### Final Answer
Thus, the value of \( \frac{p + q - r}{p} \) is \( 2 \).