sin (tan-1x), where |x| < 1, is equal to :(a) \(\frac{x}{\sqrt{1-x^2}}

1.	sin (tan-1x), where \|x\| < 1, is equal to :(a) \(\frac{x}{\sqrt{1-x^2}}\)(b) \(\frac{1}{\sqrt{1-x^2}}\)(c) \(\frac{1}{\sqrt{1+x^2}}\)(d) \(\frac{x}{\sqrt{1+x^2}}\)
Answer» Option : (d) sin(tan⁻¹x) = sin(sin⁻¹(\(\frac{x}{\sqrt{1+x^2}}\))) = \(\frac{x}{\sqrt{1+x^2}}\)

sin (tan-1x), where |x| < 1, is equal to :(a) \(\frac{x}{\sqrt{1-x^2}}\)(b) \(\frac{1}{\sqrt{1-x^2}}\)(c) \(\frac{1}{\sqrt{1+x^2}}\)(d) \(\frac{x}{\sqrt{1+x^2}}\)

Answer»

Option : (d)

sin(tan⁻¹x) = sin(sin⁻¹(\(\frac{x}{\sqrt{1+x^2}}\)))

= \(\frac{x}{\sqrt{1+x^2}}\)

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sin (tan-1x), where |x| &lt; 1, is equal to :(a) \(\frac{x}{\sqrt{1-x^2}}\)(b) \(\frac{1}{\sqrt{1-x^2}}\)(c) \(\frac{1}{\sqrt{1+x^2}}\)(d) \(\frac{x}{\sqrt{1+x^2}}\)

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sin (tan-1x), where |x| < 1, is equal to :(a) \(\frac{x}{\sqrt{1-x^2}}\)(b) \(\frac{1}{\sqrt{1-x^2}}\)(c) \(\frac{1}{\sqrt{1+x^2}}\)(d) \(\frac{x}{\sqrt{1+x^2}}\)