tan left[ sec^{-1} sqrt{1 + x^2} right] = | vedclass

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(B) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$.Then,$\sqrt{1 + x^2} = \sqrt{1 + 	an^2 	heta} = \sqrt{\sec^2 	heta} = \sec 	heta$.Substitut

Vedclass · Accepted Answer

$x$

Vedclass · Answer

(B) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$.Then,$\sqrt{1 + x^2} = \sqrt{1 + 	an^2 	heta} = \sqrt{\sec^2 	heta} = \sec 	heta$.Substituting this into the expression:$	an \left[ \sec^{-1} \sqrt{1 + x^2} ight] = 	an \left[ \sec^{-1} (\sec 	heta) ight]$.Since $\sec^{-1} (\sec 	heta) = 	heta$,we have:$	an 	heta = x$.Therefore,the correct option is $B$.

$\tan \left[ \sec^{-1} \sqrt{1 + x^2} \right] = $

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