If Sin-1 x-x2-2-x3-4- -Cos-1 x2-x4-2-x6-4- -2 for 0 - -x- - -2 Then x -

Sin^ 1x +Cos^ 1x=π/2 ∴ x x^2/2+x^3/4.....∞ =x^2 x^4/2+x^6/4.....∞ use infinite G.P. result. x = 1 ...

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Byju's Answer

Standard XI

Mathematics

Properties Derived from Trigonometric Identities

If Sin-1 x-x2...

Question

If $S i n^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . . . \infty) + C o s^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . . \infty) = \frac{π}{2} f o r 0 < | x | < \sqrt{2}$ Then x =___

\frac{1}{2}

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- \frac{1}{2}

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-1

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Solution

The correct option is B 1
$S i n^{- 1} x + C o s^{- 1} x = \frac{π}{2}$
$∴ x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . . . \infty = x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . . . \infty$
use infinite G.P. result. x = 1

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Similar questions

Q. If

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

for

0 < | x | < \sqrt{2}

,then

x

equal

Q. If

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

and

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Q. If

S i n^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . . . \infty) + C o s^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . . \infty) = \frac{π}{2} f o r 0 < | x | < \sqrt{2}

Then x =___

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

Q. If sin

^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} + . . . . . . . . . . . . . . \infty) +

cos

^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . . . . . . \infty)

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If Sin−1(x−x22+x34.....∞)+Cos−1(x2−x42+x64....∞)=π2 for 0<|x|<√2 Then x =___

The correct option is B 1Sin−1x+Cos−1x=π2 ∴x−x22+x34.....∞=x2−x42+x64.....∞ use infinite G.P. result. x = 1

If $S i n^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . . . \infty) + C o s^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . . \infty) = \frac{π}{2} f o r 0 < | x | < \sqrt{2}$ Then x =___

The correct option is B 1
$S i n^{- 1} x + C o s^{- 1} x = \frac{π}{2}$
$∴ x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . . . \infty = x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . . . \infty$
use infinite G.P. result. x = 1