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What is the correct value of \(\cos(\tan^{-1}x)\)?\(\cos(\tan^{-1}x)\) का सही मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\frac{1}{\sqrt{1+x^2}})(\frac{1}{\sqrt{1+x^2}})
Taking \(\theta=\tan^{-1}x\), the adjacent side is (1) and the hypotenuse is \(\sqrt{1+x^2}\). Thus \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\).\(\theta=\tan^{-1}x\) लेने पर adjacent side (1) और hypotenuse \(\sqrt{1+x^2}\) मिलती है। इसलिए \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\)।
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