The output of \(\sin^{-1}x\) lies in \([-\frac{\pi}{2},\frac{\pi}{2}]\). Therefore \(\theta\) must lie in this range.
Step 2
Why this answer is correct
The correct answer is B. \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\). The output of \(\sin^{-1}x\) lies in \([-\frac{\pi}{2},\frac{\pi}{2}]\). Therefore \(\theta\) must lie in this range.
Step 3
Exam Tip
\(\sin^{-1}x\) का आउटपुट \([-\frac{\pi}{2},\frac{\pi}{2}]\) में होता है। इसलिए \(\theta\) इसी range में होना चाहिए।
A. जब (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))/When (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
Step 1
Concept
The principal range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore that interval is the correct choice.
Step 2
Why this answer is correct
The correct answer is A. जब (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) / When (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The principal range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore that interval is the correct choice.
Step 3
Exam Tip
\(\tan^{-1}x\) की प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है। इसलिए वही अंतराल सही चयन है।
The principal range of \(\cos^{-1}x\) is \([0,\pi]\). Therefore (\cos^{-1}\(\cos x\)=x) applies directly only in this range.
Step 2
Why this answer is correct
The correct answer is A. जब \(x\in[0,\pi]\) / When \(x\in[0,\pi]\). The principal range of \(\cos^{-1}x\) is \([0,\pi]\). Therefore (\cos^{-1}\(\cos x\)=x) applies directly only in this range.
Step 3
Exam Tip
\(\cos^{-1}x\) की प्रधान सीमा \([0,\pi]\) है। इसलिए (\cos^{-1}\(\cos x\)=x) इसी सीमा में सीधे लागू होता है।
A. जब \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)/When \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
Step 1
Concept
The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) so (\sin^{-1}\(\sin x\)=x) is directly true there. Without range it can be a mistake.
Step 2
Why this answer is correct
The correct answer is A. जब \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) / When \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) so (\sin^{-1}\(\sin x\)=x) is directly true there. Without range it can be a mistake.
Step 3
Exam Tip
\(\sin^{-1}x\) की प्रधान सीमा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) है इसलिए इसी में (\sin^{-1}\(\sin x\)=x) सीधे सही है। सीमा के बिना यह गलती बन सकती है।
