'solve tan ( sin − 1 ( cos ( sin − 1 x ) ) ) tan ( cos − 1 ( sin ( cos − 1 x ) ) ) '

Options

A. 0
B. 1
C. `-1`
D. none of these

Explanation

To solve the expression

\[ \tan\left(\sin^{-1}(\cos(\sin^{-1}(x)))\right) \tan\left(\cos^{-1}(\sin(\cos^{-1}(x)))\right), \]

we will break it down step by step.

### Step 1: Simplify \(\sin^{-1}(\cos(\sin^{-1}(x)))\)

Let \(y = \sin^{-1}(x)\). Then, by definition, \(\sin(y) = x\).

Using the Pythagorean identity, we can find \(\cos(y)\):

\[ \cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2}. \]

Thus, we have:

\[ \sin^{-1}(\cos(\sin^{-1}(x))) = \sin^{-1}(\sqrt{1 - x^2}). \]

### Step 2: Simplify \(\tan(\sin^{-1}(\sqrt{1 - x^2}))\)

Using the definition of tangent in terms of sine and cosine, we have:

\[ \tan(\sin^{-1}(a)) = \frac{a}{\sqrt{1 - a^2}}. \]

So,

\[ \tan(\sin^{-1}(\sqrt{1 - x^2})) = \frac{\sqrt{1 - x^2}}{\sqrt{1 - (1 - x^2)}} = \frac{\sqrt{1 - x^2}}{\sqrt{x^2}} = \frac{\sqrt{1 - x^2}}{x}. \]

### Step 3: Simplify \(\cos^{-1}(\sin(\cos^{-1}(x)))\)

Let \(z = \cos^{-1}(x)\). Then, by definition, \(\cos(z) = x\).

Using the Pythagorean identity, we can find \(\sin(z)\):

\[ \sin(z) = \sqrt{1 - \cos^2(z)} = \sqrt{1 - x^2}. \]

Thus, we have:

\[ \cos^{-1}(\sin(\cos^{-1}(x))) = \cos^{-1}(\sqrt{1 - x^2}). \]

### Step 4: Simplify \(\tan(\cos^{-1}(\sqrt{1 - x^2}))\)

Using the definition of tangent in terms of sine and cosine again, we have:

\[ \tan(\cos^{-1}(a)) = \frac{\sqrt{1 - a^2}}{a}. \]

So,

\[ \tan(\cos^{-1}(\sqrt{1 - x^2})) = \frac{\sqrt{1 - (1 - x^2)}}{\sqrt{1 - x^2}} = \frac{x}{\sqrt{1 - x^2}}. \]

### Step 5: Combine the results

Now we can combine the two results:

\[ \tan\left(\sin^{-1}(\cos(\sin^{-1}(x)))\right) \tan\left(\cos^{-1}(\sin(\cos^{-1}(x)))\right) = \left(\frac{\sqrt{1 - x^2}}{x}\right) \left(\frac{x}{\sqrt{1 - x^2}}\right). \]

### Step 6: Simplify the expression

The expression simplifies to:

\[ \frac{\sqrt{1 - x^2}}{x} \cdot \frac{x}{\sqrt{1 - x^2}} = 1. \]

### Conclusion

Thus, the final result is:

\[ \boxed{1}. \]