Write the function in the simplest form: tan | vedclass

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(D) Let the given expression be $y = 	an ^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} ight)$.Substitute $x = 	an 	heta$,which implies $	heta = 	an ^

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$\frac{1}{2} 	an ^{-1} x$

Vedclass · Answer

(D) Let the given expression be $y = 	an ^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} ight)$.Substitute $x = 	an 	heta$,which implies $	heta = 	an ^{-1} x$.Then,the expression becomes:$y = 	an ^{-1} \left( \frac{\sqrt{1+	an ^{2} 	heta}-1}{	an 	heta} ight)$Using the identity $1 + 	an^2 	heta = \sec^2 	heta$,we get:$y = 	an ^{-1} \left( \frac{\sec 	heta - 1}{	an 	heta} ight)$Converting to $\sin 	heta$ and $\cos 	heta$:$y = 	an ^{-1} \left( \frac{\frac{1}{\cos 	heta} - 1}{\frac{\sin 	heta}{\cos 	heta}} ight) = 	an ^{-1} \left( \frac{1 - \cos 	heta}{\sin 	heta} ight)$Using half-angle formulas $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2} ight)$ and $\sin 	heta = 2 \sin \left( \frac{	heta}{2} ight) \cos \left( \frac{	heta}{2} ight)$:$y = 	an ^{-1} \left( \frac{2 \sin^2 \left( \frac{	heta}{2} ight)}{2 \sin \left( \frac{	heta}{2} ight) \cos \left( \frac{	heta}{2} ight)} ight)$$y = 	an ^{-1} \left( 	an \left( \frac{	heta}{2} ight) ight) = \frac{	heta}{2}$Substituting back $	heta = 	an ^{-1} x$,we get:$y = \frac{1}{2} 	an ^{-1} x$.

Write the function in the simplest form: $\tan ^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right), x \neq 0$

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