sinh^{-1}left(frac{x}{sqrt{1-x^2}}right) is e | vedclass

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

Question

(D) Let $\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight) = 	heta$. Then, $\sinh 	heta = \frac{x}{\sqrt{1-x^2}}$. We know the identity $\cosh^2 	heta

Vedclass · Accepted Answer

$	anh^{-1} x$

Vedclass · Answer

(D) Let $\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight) = 	heta$. Then, $\sinh 	heta = \frac{x}{\sqrt{1-x^2}}$. We know the identity $\cosh^2 	heta - \sinh^2 	heta = 1$, so $\cosh^2 	heta = 1 + \sinh^2 	heta$. Substituting the value of $\sinh 	heta$: $\cosh^2 	heta = 1 + \left(\frac{x}{\sqrt{1-x^2}}ight)^2 = 1 + \frac{x^2}{1-x^2} = \frac{1-x^2+x^2}{1-x^2} = \frac{1}{1-x^2}$. Thus, $\cosh 	heta = \frac{1}{\sqrt{1-x^2}}$. Now, $	anh 	heta = \frac{\sinh 	heta}{\cosh 	heta} = \frac{x/\sqrt{1-x^2}}{1/\sqrt{1-x^2}} = x$. Therefore, $	heta = 	anh^{-1} x$. Hence, $\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight) = 	anh^{-1} x$.

$\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is equal to

Similar Questions

If ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x,$ then $x = $

$\cos ^{-1}\left\{\cot \left(\sum_{i=1}^3 \cot ^{-1} i\right)\right\}=$ . . . . . . .

If $\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1$,then $x$ is equal to

If $\tan^{-1}(1) + \tan^{-1}(3) + \tan^{-1}(5) + \tan^{-1}(1/4) = \pi + \tan^{-1}(\alpha/2)$, then the value of $\alpha$ is... (in $/41$)

In a $\triangle ABC$,if $\angle A = 90^{\circ}$,then $\cos^{-1}\left(\frac{R}{r_2+r_3}\right)$ is equal to (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module