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What is the correct value of \(\sin(\tan^{-1}x)\)?\(\sin(\tan^{-1}x)\) का सही मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\frac{x}{\sqrt{1+x^2}})(\frac{x}{\sqrt{1+x^2}})
If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).यदि \(\theta=\tan^{-1}x\), तो \(\tan\theta=x\) और hypotenuse \(\sqrt{1+x^2}\) है। इसलिए \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\)।
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