Evaluate: tan ^{-1} left( frac{x}{sqrt{a^2 - | vedclass

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(C) To evaluate $	an ^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} ight)$,we use the substitution $x = a \sin 	heta$. Then,$\sin 	heta = \frac{x}{a}$,w

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$\sin ^{-1} \left( \frac{x}{a} \right)$

Vedclass · Answer

(C) To evaluate $	an ^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} ight)$,we use the substitution $x = a \sin 	heta$. Then,$\sin 	heta = \frac{x}{a}$,which implies $	heta = \sin ^{-1} \left( \frac{x}{a} ight)$. Substituting $x = a \sin 	heta$ into the expression: $	an ^{-1} \left( \frac{a \sin 	heta}{\sqrt{a^2 - (a \sin 	heta)^2}} ight) = 	an ^{-1} \left( \frac{a \sin 	heta}{\sqrt{a^2(1 - \sin^2 	heta)}} ight)$ $= 	an ^{-1} \left( \frac{a \sin 	heta}{\sqrt{a^2 \cos^2 	heta}} ight) = 	an ^{-1} \left( \frac{a \sin 	heta}{a \cos 	heta} ight)$ $= 	an ^{-1} (	an 	heta) = 	heta$ Substituting back the value of $	heta$,we get $\sin ^{-1} \left( \frac{x}{a} ight)$.

Evaluate: $\tan ^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)$

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