\(\sin^{-1}x+\cos^{-1}x\) का मान क्या है जब \(x\in[-1,1]\)?

What is the value of \(\sin^{-1}x+\cos^{-1}x\) when \(x\in[-1,1]\)?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

A. \(\frac{\pi}{2}\)

Step 1

Concept

For every \(x\in[-1,1]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). This identity is often used directly.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{\pi}{2}\). For every \(x\in[-1,1]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). This identity is often used directly.

Step 3

Exam Tip

हर \(x\in[-1,1]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) होता है। यह पहचान अक्सर सीधे उपयोग होती है।

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Mathematics Answer, Explanation and Revision Hints

\(\sin^{-1}x+\cos^{-1}x\) का मान क्या है जब \(x\in[-1,1]\)? / What is the value of \(\sin^{-1}x+\cos^{-1}x\) when \(x\in[-1,1]\)?

Correct Answer: A. \(\frac{\pi}{2}\). Explanation: हर \(x\in[-1,1]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) होता है। यह पहचान अक्सर सीधे उपयोग होती है। / For every \(x\in[-1,1]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). This identity is often used directly.

Which concept should I revise for this Mathematics MCQ?

For every \(x\in[-1,1]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). This identity is often used directly.

What exam hint can help solve this Mathematics question?

हर \(x\in[-1,1]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) होता है। यह पहचान अक्सर सीधे उपयोग होती है।