If sin^{-1} x = theta + beta and sin^{-1} y = | vedclass

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(B) Given that $\sin^{-1} x = 	heta + \beta$ and $\sin^{-1} y = 	heta - \beta$.Therefore,$x = \sin(	heta + \beta)$ and $y = \sin(	heta - \beta)$.N

Vedclass · Accepted Answer

$\sin^2 	heta + \cos^2 \beta$

Vedclass · Answer

(B) Given that $\sin^{-1} x = 	heta + \beta$ and $\sin^{-1} y = 	heta - \beta$.Therefore,$x = \sin(	heta + \beta)$ and $y = \sin(	heta - \beta)$.Now,$1 + xy = 1 + \sin(	heta + \beta) \sin(	heta - \beta)$.Using the identity $\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B$,we get:$1 + xy = 1 + \sin^2 	heta - \sin^2 \beta$.Since $1 - \sin^2 \beta = \cos^2 \beta$,we have:$1 + xy = \sin^2 	heta + \cos^2 \beta$.

If $\sin^{-1} x = \theta + \beta$ and $\sin^{-1} y = \theta - \beta$,then $1 + xy = $

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