Solve: $sin ({tan}^{- 1} x), | x | < 1$ is equal to

\frac{x}{\sqrt{1 - x^{2}}}

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\frac{1}{\sqrt{1 - x^{2}}}

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\frac{1}{\sqrt{1 + x^{2}}}

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\frac{x}{\sqrt{1 + x^{2}}}

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Solution

The correct option is D $\frac{x}{\sqrt{1 + x^{2}}}$
We need to find value of $sin ({tan}^{- 1} x)$
Put $y = {tan}^{- 1} x$

$\Rightarrow tan y = x$

$∴ tan y = \frac{sin y}{cos y}$

$\Rightarrow sin y = \frac{tan y}{sec y}$

$\Rightarrow sin y = \frac{x}{\sqrt{1 + x^{2}}}$

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Similar questions

Solveis equal to

(A) (B) (C) (D)

$s i n (t a n^{- 1} x), | x | < 1$ , 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) $\frac{x}{\sqrt{1 + x^{2}}}$

Q. If

0 \leq x \leq 1

, then

sin {{tan}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}}

is equal to

Q. If

- 1 \leq x \leq 0

, then

sin {{tan}^{- 1} \frac{1 - x^{2}}{2 x} - {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}}

is equal to

Q. Solve for

x

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

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Solve: sin(tan−1x),|x|<1 is equal to

The correct option is D x√1+x2We need to find value of sin(tan−1x)Put y=tan−1x⇒tany=x∴tany=sinycosy⇒siny=tanysecy⇒siny=x√1+x2

Solve: $sin ({tan}^{- 1} x), | x | < 1$ is equal to

The correct option is D $\frac{x}{\sqrt{1 + x^{2}}}$
We need to find value of $sin ({tan}^{- 1} x)$
Put $y = {tan}^{- 1} x$

$\Rightarrow tan y = x$

$∴ tan y = \frac{sin y}{cos y}$

$\Rightarrow sin y = \frac{tan y}{sec y}$

$\Rightarrow sin y = \frac{x}{\sqrt{1 + x^{2}}}$