sin (tan-1x), where |x| < 1, is equal to:

(a) \( \dfrac{x}{\sqrt{1-x^2}}\)

(b) \( \dfrac{1}{\sqrt{1-x^2}}\)

(c) \( \dfrac{1}{\sqrt{1+x^2}}\)

(d) \( \dfrac{x}{\sqrt{1+x^2}}\)

Asked by Abhisek | 1 year ago |  64

1 Answer

Solution :-

Right answer is (d) \( \dfrac{x}{\sqrt{1+x^2}}\)

Explanation:-

\( sin(tan^{−1}x)\)

=\(( sin^{-1}(\dfrac{x}{\sqrt{1+x^2}}))\)

 \(=\dfrac{x}{\sqrt{1+x^2}}\)

Answered by Pragya Singh | 1 year ago

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