sin^{-1} sqrt{frac{x}{x+a}} is equal to | vedclass

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(C) Let $y = \sin^{-1} \sqrt{\frac{x}{x+a}}$.Substitute $x = a 	an^2 	heta$,which implies $	an 	heta = \sqrt{\frac{x}{a}}$ or $	heta = 	an^{-1}

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$	an^{-1} \sqrt{\frac{x}{a}}$

Vedclass · Answer

(C) Let $y = \sin^{-1} \sqrt{\frac{x}{x+a}}$.Substitute $x = a 	an^2 	heta$,which implies $	an 	heta = \sqrt{\frac{x}{a}}$ or $	heta = 	an^{-1} \sqrt{\frac{x}{a}}$.Then,$\sqrt{\frac{x}{x+a}} = \sqrt{\frac{a 	an^2 	heta}{a 	an^2 	heta + a}} = \sqrt{\frac{a 	an^2 	heta}{a(1 + 	an^2 	heta)}} = \sqrt{\frac{	an^2 	heta}{\sec^2 	heta}} = \sqrt{\sin^2 	heta} = \sin 	heta$.Thus,$y = \sin^{-1}(\sin 	heta) = 	heta$.Substituting back,$y = 	an^{-1} \sqrt{\frac{x}{a}}$.

$\sin^{-1} \sqrt{\frac{x}{x+a}}$ is equal to

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