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Which statement is correct for \(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\)?\(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\) के लिए सही कथन कौन-सा है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: It is true for all \(x\in\mathbb{R}\)यह सभी \(x\in\mathbb{R}\) के लिए सत्य है
\(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).\(\frac{x}{\sqrt{1+x^2}}\) हमेशा ([-1,1]) में रहता है। इसलिए identity सभी real (x) के लिए मान्य है।
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