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: Use $a^2+b^2$ with $a+b$ fixed
With $a+b=\frac{\pi}{2}$, $a^2+b^2=\frac{(a+b)^2+(a-b)^2}{2}$. This is smallest when $a=b=\frac{\pi}{4}$.
Minimum $=\frac{1}{2}\cdot\frac{\pi^2}{4}=\frac{\pi^2}{8}$. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A)}} \]