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For which condition on \(x\) is the identity \(\sin^{-1}x=\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) correct?\(\sin^{-1}x=\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) यह समता किस शर्त (condition) पर \(x\) के लिए सही है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(x\in\left[-1,1\right]\)\(x\in\left[-1,1\right]\)
The expression \(\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) is real only when \(1-x^2\ge 0\), so \(\sqrt{1-x^2}\) is defined. That requires \(x\in\left[-1,1\right]\). On this domain, the inverse-sine and the transformed inverse-tangent represent the same angle, so the identity holds.\(\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) वाला पद वास्तविक (real) तभी होता है जब \(1-x^2\ge 0\), ताकि \(\sqrt{1-x^2}\) परिभाषित रहे। इससे \(x\in\left[-1,1\right]\) आवश्यक है। इसी domain पर \(\sin^{-1}\) और बदले हुए \(\tan^{-1}\) समान कोण को दर्शाते हैं, इसलिए यह समता सही होती है।
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