If `x=sin^(-1)(a^6+1)+cos^(-1)(a^4+1)-tan^(-1)(a^2+1),a in R ,` then the value of `sec^2x` is_______

Explanation

To solve the problem, we need to evaluate the expression given for \( x \) and then find the value of \( \sec^2 x \).

**Step 1: Write down the expression for \( x \)**
Given:
\[ x = \sin^{-1}(a^6 + 1) + \cos^{-1}(a^4 + 1) - \tan^{-1}(a^2 + 1) \]

**Step 2: Analyze the domains of the inverse trigonometric functions**
- The domain of \( \sin^{-1}(y) \) is \( -1 \leq y \leq 1 \).
- The domain of \( \cos^{-1}(y) \) is \( -1 \leq y \leq 1 \).
- The range of \( \tan^{-1}(y) \) is all real numbers.

For \( \sin^{-1}(a^6 + 1) \) to be valid, we need:
\[ -1 \leq a^6 + 1 \leq 1 \]
This simplifies to:
\[ -2 \leq a^6 \leq 0 \]
Since \( a^6 \) is always non-negative for real \( a \), we conclude:
\[ a^6 = 0 \implies a = 0 \]

**Step 3: Substitute \( a = 0 \) into the expression for \( x \)**
Now, substituting \( a = 0 \):
\[ x = \sin^{-1}(0^6 + 1) + \cos^{-1}(0^4 + 1) - \tan^{-1}(0^2 + 1) \]
This simplifies to:
\[ x = \sin^{-1}(1) + \cos^{-1}(1) - \tan^{-1}(1) \]

**Step 4: Evaluate the inverse trigonometric functions**
- \( \sin^{-1}(1) = \frac{\pi}{2} \)
- \( \cos^{-1}(1) = 0 \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)

Substituting these values back into the expression for \( x \):
\[ x = \frac{\pi}{2} + 0 - \frac{\pi}{4} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

**Step 5: Find \( \sec^2 x \)**
Now, we need to find \( \sec^2 x \):
\[ \sec^2 x = \sec^2\left(\frac{\pi}{4}\right) \]
Since \( \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \), we have:
\[ \sec^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2 \]

Thus, the final answer is:
\[ \sec^2 x = 2 \]