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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $\sin ^{-1} x < \cos ^{-1} x$, then

$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$

Considering only the principal values of inverse trigonometric functions,the number of positive real values of $x$ satisfying $\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$ is:

If $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,where $a \in [0, 1]$,then which of the following is true?

If for $x \in \left(0, \frac{1}{4}\right)$,the derivative of $\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$ is $\sqrt{x} \cdot g(x)$,then $g(x)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module