The value of {tan ^{ - 1}}left[ {frac{{sqrt { | vedclass

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Question

(A) Let ${x^2} = \cos 2	heta$,which implies $2	heta = {\cos ^{ - 1}}{x^2}$ or $	heta = \frac{1}{2}{\cos ^{ - 1}}{x^2}$.Substituting ${x^2} = \cos 2

Vedclass · Accepted Answer

$\frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$

Vedclass · Answer

(A) Let ${x^2} = \cos 2	heta$,which implies $2	heta = {\cos ^{ - 1}}{x^2}$ or $	heta = \frac{1}{2}{\cos ^{ - 1}}{x^2}$.Substituting ${x^2} = \cos 2	heta$ into the expression:${	an ^{ - 1}}\left[ {\frac{{\sqrt {1 + \cos 2	heta }  + \sqrt {1 - \cos 2	heta } }}{{\sqrt {1 + \cos 2	heta }  - \sqrt {1 - \cos 2	heta } }}} ight]$Using the identities $1 + \cos 2	heta = 2{\cos ^2}	heta$ and $1 - \cos 2	heta = 2{\sin ^2}	heta$:${	an ^{ - 1}}\left[ {\frac{{\sqrt {2{{\cos }^2}	heta }  + \sqrt {2{{\sin }^2}	heta } }}{{\sqrt {2{{\cos }^2}	heta }  - \sqrt {2{{\sin }^2}	heta } }}} ight] = {	an ^{ - 1}}\left[ {\frac{{\sqrt 2 \cos 	heta  + \sqrt 2 \sin 	heta }}{{\sqrt 2 \cos 	heta  - \sqrt 2 \sin 	heta }}} ight]$Dividing numerator and denominator by $\cos 	heta$:${	an ^{ - 1}}\left[ {\frac{{1 + 	an 	heta }}{{1 - 	an 	heta }}} ight] = {	an ^{ - 1}}\left[ {	an \left( {\frac{\pi }{4} + 	heta } ight)} ight]$$= \frac{\pi }{4} + 	heta = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$.

The value of ${\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$,where $|x| < 1$ and $x \ne 0$,is equal to

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