If $sin^{-1}x=\frac{\pi}{5}$ for some $x\in(-1,1),$ then the value of $cos^{-1}x$ is:

Q: If $sin^{-1}x=\frac{\pi}{5}$ for some $x\in(-1,1),$ then the value of $cos^{-1}x$ is:

To solve the problem, we need to find the value of $ \cos^{-1}x $ given that $ \sin^{-1}x = \frac{\pi}{5} $. ### Explanation: 1. **Understanding the Relationship:** - We know that for any angle $ \theta $, $ \sin^{-1}x = \theta $ implies $ \sin \theta = x $. - Given $ \sin^{-1}x = \frac{\pi}{5} $, it follows that $ \theta = \frac{\pi}{5} $ and $ \sin \frac{\pi}{5} = x $. 2. **Using the Identity:** - The identity that relates the inverse sine and inverse cosine functions is: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \] - Substituting the given value $ \sin^{-1}x = \frac{\pi}{5} $ into the identity, we have: \[ \frac{\pi}{5} + \cos^{-1}x = \frac{\pi}{2} \] 3. **Solving for $ \cos^{-1}x $:** - Rearrange the equation to solve for $ \cos^{-1}x $: \[ \cos^{-1}x = \frac{\pi}{2} - \frac{\pi}{5} \] - Simplify the right-hand side: \[ \cos^{-1}x = \frac{5\pi}{10} - \frac{2\pi}{10} = \frac{3\pi}{10} \] 4. **Conclusion:** - Therefore, the value of $ \cos^{-1}x $ is $ \frac{3\pi}{10} $, which corresponds to Option 3. ### Why Other Options are Incorrect: - **Option 1: $ \frac{9\pi}{10} $** - If $ \cos^{-1}x = \frac{9\pi}{10} $, then $ \sin^{-1}x $ would be $ \frac{\pi}{2} - \frac{9\pi}{10} = \frac{\pi}{10} $, which does not match the given $ \sin^{-1}x = \frac{\pi}{5} $. - **Option 2: $ \frac{7\pi}{10} $** - If $ \cos^{-1}x = \frac{7\pi}{10} $, then $ \sin^{-1}x $ would be $ \frac{\pi}{2} - \frac{7\pi}{10} = \frac{3\pi}{10} $, which again does not match the given $ \sin^{-1}x = \frac{\pi}{5} $. - **Option 4: $ \frac{\pi}{10} $** - If $ \cos^{-1}x = \frac{\pi}{10} $, then $ \sin^{-1}x $ would be $ \frac{\pi}{2} - \frac{\pi}{10} = \frac{4\pi}{10} = \frac{2\pi}{5} $, which is not equal to the given $ \sin^{-1}x = \frac{\pi}{5} $. Thus, the correct answer is indeed Option 3, $ \frac{3\pi}{10} $.

Question

Accepted Answer

To solve the problem, we need to find the value of $ \cos^{-1}x $ given that $ \sin^{-1}x = \frac{\pi}{5} $.

### Explanation:

1. **Understanding the Relationship:**
   - We know that for any angle $ \theta $, $ \sin^{-1}x = \theta $ implies $ \sin \theta = x $.
   - Given $ \sin^{-1}x = \frac{\pi}{5} $, it follows that $ \theta = \frac{\pi}{5} $ and $ \sin \frac{\pi}{5} = x $.

2. **Using the Identity:**
   - The identity that relates the inverse sine and inverse cosine functions is:
     $$
     \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}
     $$
   - Substituting the given value $ \sin^{-1}x = \frac{\pi}{5} $ into the identity, we have:
     $$
     \frac{\pi}{5} + \cos^{-1}x = \frac{\pi}{2}
     $$

3. **Solving for $ \cos^{-1}x $:**
   - Rearrange the equation to solve for $ \cos^{-1}x $:
     $$
     \cos^{-1}x = \frac{\pi}{2} - \frac{\pi}{5}
     $$
   - Simplify the right-hand side:
     $$
     \cos^{-1}x = \frac{5\pi}{10} - \frac{2\pi}{10} = \frac{3\pi}{10}
     $$

4. **Conclusion:**
   - Therefore, the value of $ \cos^{-1}x $ is $ \frac{3\pi}{10} $, which corresponds to Option 3.

### Why Other Options are Incorrect:

- **Option 1: $ \frac{9\pi}{10} $**
  - If $ \cos^{-1}x = \frac{9\pi}{10} $, then $ \sin^{-1}x $ would be $ \frac{\pi}{2} - \frac{9\pi}{10} = \frac{\pi}{10} $, which does not match the given $ \sin^{-1}x = \frac{\pi}{5} $.

- **Option 2: $ \frac{7\pi}{10} $**
  - If $ \cos^{-1}x = \frac{7\pi}{10} $, then $ \sin^{-1}x $ would be $ \frac{\pi}{2} - \frac{7\pi}{10} = \frac{3\pi}{10} $, which again does not match the given $ \sin^{-1}x = \frac{\pi}{5} $.

- **Option 4: $ \frac{\pi}{10} $**
  - If $ \cos^{-1}x = \frac{\pi}{10} $, then $ \sin^{-1}x $ would be $ \frac{\pi}{2} - \frac{\pi}{10} = \frac{4\pi}{10} = \frac{2\pi}{5} $, which is not equal to the given $ \sin^{-1}x = \frac{\pi}{5} $.

Thus, the correct answer is indeed Option 3, $ \frac{3\pi}{10} $.