Write the function in the simplest form: tan | vedclass

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(A) Given expression: $	an ^{-1} \left(\frac{1}{\sqrt{x^{2}-1}}ight), |x|>1$Let $x = \sec 	heta$,then $	heta = \sec^{-1} x$.Since $|x| > 1$,$	he

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$\frac{\pi}{2}-\sec ^{-1} x$

Vedclass · Answer

(A) Given expression: $	an ^{-1} \left(\frac{1}{\sqrt{x^{2}-1}}ight), |x|>1$Let $x = \sec 	heta$,then $	heta = \sec^{-1} x$.Since $|x| > 1$,$	heta \in (0, \pi/2) \cup (\pi/2, \pi)$.Substituting $x = \sec 	heta$ into the expression:$= 	an ^{-1} \left(\frac{1}{\sqrt{\sec^{2} 	heta - 1}}ight)$$= 	an ^{-1} \left(\frac{1}{\sqrt{	an^{2} 	heta}}ight)$$= 	an ^{-1} \left(\frac{1}{|	an 	heta|}ight)$For $x > 1$,$	heta \in (0, \pi/2)$,so $	an 	heta > 0$,thus $|	an 	heta| = 	an 	heta$.$= 	an ^{-1} \left(\frac{1}{	an 	heta}ight) = 	an ^{-1} (\cot 	heta)$$= 	an ^{-1} \left(	an \left(\frac{\pi}{2} - 	hetaight)ight)$$= \frac{\pi}{2} - 	heta$$= \frac{\pi}{2} - \sec^{-1} x$.

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

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