If sin-1x2-cos-1x2 - a - 0-x-1- a 0- then the value of 2x2-1 is-

(sin^ 1x)^2 (cos^ 1x)^2=a ⇒ nbsp; nbsp;(sin^ 1x+cos^ 1x)(sin^ 1x cos^ 1x) = a ⇒π2(π2 2cos^ 1x ) = a ⇒π2 2cos^ 1 x = 2aπ Taking sine on both sides, we ha ...

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Standard X

Mathematics

Trigonometric Identities

If sin-1x2-co...

Question

If ${({sin}^{- 1} x)}^{2} - {({cos}^{- 1} x)}^{2} = a; 0 < x < 1, a \neq 0$ , then the value of $2 x^{2} - 1$ is:

sin (\frac{2 a}{π})

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sin (\frac{4 a}{π})

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cos (\frac{4 a}{π})

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cos (\frac{2 a}{π})

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Solution

The correct option is A $sin (\frac{2 a}{π})$
$({sin}^{- 1} x)^{2} - ({cos}^{- 1} x)^{2} = a$
$\Rightarrow ({sin}^{- 1} x + {cos}^{- 1} x) ({sin}^{- 1} x - {cos}^{- 1} x) = a$
$\Rightarrow \frac{π}{2} (\frac{π}{2} - 2 {cos}^{- 1} x) = a$
$\Rightarrow \frac{π}{2} - 2 {cos}^{- 1} x = \frac{2 a}{π}$
Taking sine on both sides, we have
$sin (\frac{π}{2} - 2 {cos}^{- 1} x) = sin (\frac{2 a}{π})$
$\Rightarrow cos (2 {cos}^{- 1} x) = sin (\frac{2 a}{π})$
$\Rightarrow 2 {cos}^{2} ({cos}^{- 1} x) - 1 = sin (\frac{2 a}{π})$
$∴ 2 x^{2} - 1 = sin (\frac{2 a}{π})$

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If (sin−1x)2−(cos−1x)2=a ; 0<x<1, a≠0, then the value of 2x2−1 is:

If ${({sin}^{- 1} x)}^{2} - {({cos}^{- 1} x)}^{2} = a; 0 < x < 1, a \neq 0$ , then the value of $2 x^{2} - 1$ is: