Question

The minimum value of \((sin^{-1}x)^2+(cos^{-1}x)^2\) is............

• A. \(\frac{π^2}{8}\)
• B. \(\frac{3π^2}{8}\)
• C. \(\frac{5π^2}{8}\)
• D. \(\frac{7π^2}{8}\)

Hint:

Use $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ and complete the square.

Answer 1

The Correct Option is A

Step 1: Set up
Let \(t=\sin^{-1}x\). Then \(\cos^{-1}x=\frac{\pi}{2}-t\).

Step 2: Form the expression
\[ t^2+\left(\frac{\pi}{2}-t\right)^2=2t^2-\pi t+\frac{\pi^2}{4} \]

Step 3: Minimise
The minimum of \(2t^2-\pi t+\frac{\pi^2}{4}\) occurs at \(t=\frac{\pi}{4}\), which lies in \([-\frac{\pi}{2},\frac{\pi}{2}]\).
\[ 2\cdot\frac{\pi^2}{16}-\frac{\pi^2}{4}+\frac{\pi^2}{4}=\frac{\pi^2}{8} \]

Final Answer:
The minimum value is \(\frac{\pi^2}{8}\), option (A). \[ \boxed{\text{(A)}} \]