If $sin(sin^{-1}\frac{1}{5}+cos^{-1}x)=1$, then x is equal to:

Question

Accepted Answer

To solve the problem, we need to evaluate the expression $ \sin(\sin^{-1}\frac{1}{5} + \cos^{-1}x) = 1 $.

### Explanation for Option 2 being correct:

1. **Understanding the Expression:**
   - The expression involves the sum of two angles: $ \sin^{-1}\frac{1}{5} $ and $ \cos^{-1}x $.
   - We know that $ \sin^{-1}a + \cos^{-1}a = \frac{\pi}{2} $ for any $ a $ in the domain of the inverse trigonometric functions.

2. **Applying the Identity:**
   - Given $ \sin(\sin^{-1}\frac{1}{5} + \cos^{-1}x) = 1 $, we recognize that the sine of an angle is 1 when the angle is $ \frac{\pi}{2} $.
   - Therefore, $ \sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2} $.

3. **Solving for $ x $:**
   - Using the identity $ \sin^{-1}a + \cos^{-1}a = \frac{\pi}{2} $, we can equate:
     $$
     \sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2}
     $$
   - This implies that $ \cos^{-1}x = \cos^{-1}\frac{1}{5} $.
   - Therefore, $ x = \frac{1}{5} $.

Thus, Option 2 is correct because $ x = \frac{1}{5} $ satisfies the equation.

### Explanation for why other options are incorrect:

1. **Option 1 (x = 0):**
   - If $ x = 0 $, then $ \cos^{-1}0 = \frac{\pi}{2} $.
   - This would make $ \sin^{-1}\frac{1}{5} + \frac{\pi}{2} > \frac{\pi}{2} $, which cannot equal $ \frac{\pi}{2} $.
   - Therefore, $ x = 0 $ does not satisfy the equation.

2. **Option 3 (x = 4/5):**
   - If $ x = \frac{4}{5} $, then $ \cos^{-1}\frac{4}{5} $ is not equal to $ \cos^{-1}\frac{1}{5} $.
   - This would not satisfy the identity $ \sin^{-1}a + \cos^{-1}a = \frac{\pi}{2} $.
   - Therefore, $ x = \frac{4}{5} $ does not satisfy the equation.

3. **Option 4 (x = 1):**
   - If $ x = 1 $, then $ \cos^{-1}1 = 0 $.
   - This would make $ \sin^{-1}\frac{1}{5} + 0 = \sin^{-1}\frac{1}{5} $, which is not equal to $ \frac{\pi}{2} $.
   - Therefore, $ x = 1 $ does not satisfy the equation.

In conclusion, Option 2 is the only value of $ x $ that satisfies the given equation, making it the correct answer.