Statement-1:
`int_(0)^(sin^(2)x) sin^(-1)sqrt(t )dt+int_(0)^(cos^(2)x) cos^(-1)sqrt(t )dt=(pi)/(4)` for all x.
Statement-2:`(d)/(dx) int_(theta(x))overset(psi(x)) f(t)dt=psi'(x)f(psi(x))-theta'(x)f(theta(x))`

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given problem, we need to analyze both statements and prove the correctness of Statement 1 using the Leibniz rule as mentioned in Statement 2.

### Step-by-Step Solution:

1. **Understanding Statement 1:**
We need to evaluate the expression:
\[ \int_{0}^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt = \frac{\pi}{4} \]
for all \( x \).

2. **Applying Leibniz Rule:**
We will differentiate both integrals with respect to \( x \) using the Leibniz rule, which states:
\[ \frac{d}{dx} \int_{\theta(x)}^{\psi(x)} f(t) \, dt = \psi'(x) f(\psi(x)) - \theta'(x) f(\theta(x)) \]

3. **Differentiating the First Integral:**
For the first integral:
\[ I_1 = \int_{0}^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt \]
Using the Leibniz rule:
\[ \frac{dI_1}{dx} = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) - 0 \]
Since \( \sin^{-1}(\sqrt{\sin^2 x}) = \sin x \) and \( \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x \):
\[ \frac{dI_1}{dx} = \sin x \cdot (2 \sin x \cos x) = 2 \sin^2 x \cos x \]

4. **Differentiating the Second Integral:**
For the second integral:
\[ I_2 = \int_{0}^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt \]
Again applying the Leibniz rule:
\[ \frac{dI_2}{dx} = \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x) - 0 \]
Since \( \cos^{-1}(\sqrt{\cos^2 x}) = \cos x \) and \( \frac{d}{dx}(\cos^2 x) = -2 \cos x \sin x \):
\[ \frac{dI_2}{dx} = \cos x \cdot (-2 \cos x \sin x) = -2 \cos^2 x \sin x \]

5. **Combining the Derivatives:**
Now we combine the derivatives of both integrals:
\[ \frac{d}{dx} \left( I_1 + I_2 \right) = 2 \sin^2 x \cos x - 2 \cos^2 x \sin x \]
This simplifies to:
\[ \frac{d}{dx} \left( I_1 + I_2 \right) = 2 \sin x \cos x (\sin x - \cos x) \]

6. **Setting the Derivative Equal to Zero:**
Since the right-hand side of Statement 1 is a constant \( \frac{\pi}{4} \), its derivative is zero:
\[ 2 \sin x \cos x (\sin x - \cos x) = 0 \]
This implies either \( \sin x = 0 \), \( \cos x = 0 \), or \( \sin x = \cos x \).

7. **Conclusion:**
Since the derivative equals zero for all \( x \), the original statement holds true:
\[ \int_{0}^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt = \frac{\pi}{4} \]
Therefore, Statement 1 is correct.