The value of tan-1-1-x2-1-x2-1-x2-1-x2- -x- - 1-2- x- 0- is equal to-

The correct option is A π/4+1/2cos^ 1x^2 x^2=cos 2θ; θ =1/2cos ^ 1 x^2 tan^ 1[√(1+cos 2θ)+√(1 cos 2θ)/√(1+cos 2θ) √(1 cos 2θ)] = #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

The value of ...

Question

The value of $t a n^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}]$ , $| x | < \frac{1}{2}, x \neq 0$ , is equal to.

\frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2}

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\frac{π}{4} + c o s^{- 1} x^{2}

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\frac{π}{4} - c o s^{- 1} x^{2}

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\frac{π}{4} - \frac{1}{2} c o s^{- 1} x^{2}

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Solution

The correct option is A $\frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2}$
$x^{2} = c o s 2 θ; θ = \frac{1}{2} c o s^{- 1} x^{2}$
$t a n^{- 1} [\frac{\sqrt{1 + c o s 2 θ} + \sqrt{1 - c o s 2 θ}}{\sqrt{1 + c o s 2 θ} - \sqrt{1 - c o s 2 θ}}]$
= $t a n^{- 1} [\frac{c o s θ + s i n θ}{c o s θ - s i n θ}]$
= $t a n^{- 1} [\frac{1 + t a n θ}{1 - t a n θ}]$
= $t a n^{- 1} [t a n (\frac{π}{4} + θ)]$
= $\frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2}$ .

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Similar questions

Prove that $t a n^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = \frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2} .$

Q.
(a) Prove that

sin [{t a n}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}] = 1

(b) If

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . .) = \frac{π}{2}

for

0 < | x | < \sqrt{2}

, then x equals

If $s i n^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + c o s^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for 0 < 1x < $\sqrt{2}$ , then x equals

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

is equal to

(where x \in R - {0})

lim x \to \infty \frac{e^{1 / x^{2}} - 1}{2 {tan}^{- 1} (x^{2}) - π}

is equal to?

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The value of tan−1[√1+x2+√1−x2√1+x2−√1−x2], |x|<12,x≠0, is equal to.

The correct option is A π4+12cos−1x2x2=cos2θ;θ=12cos−1x2tan−1[√1+cos2θ+√1−cos2θ√1+cos2θ−√1−cos2θ]= tan−1[cosθ+sinθcosθ−sinθ]= tan−1[1+tanθ1−tanθ]= tan−1[tan(π4+θ)]= π4+12cos−1x2.

The value of $t a n^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}]$ , $| x | < \frac{1}{2}, x \neq 0$ , is equal to.