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What is the correct value of \(\tan(\sin^{-1}x)\)?\(\tan(\sin^{-1}x)\) 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}})
Let (\theta=\sin^{-1}x), then (\sin\theta=x) and (\cos\theta=\sqrt{1-x^2}). Hence (\tan\theta=\frac{x}{\sqrt{1-x^2}}).मान लें (\theta=\sin^{-1}x), तब (\sin\theta=x) और (\cos\theta=\sqrt{1-x^2})। इसलिए (\tan\theta=\frac{x}{\sqrt{1-x^2}})।
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