If 2 sinh^{-1}left(frac{a}{sqrt{1-a^2}}right) | vedclass

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

Question

(A) We know that $\sinh^{-1}(y) = \log(y + \sqrt{1+y^2})$.Let $y = \frac{a}{\sqrt{1-a^2}}$. Then $\sqrt{1+y^2} = \sqrt{1 + \frac{a^2}{1-a^2}} = \sqrt{

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(A) We know that $\sinh^{-1}(y) = \log(y + \sqrt{1+y^2})$.Let $y = \frac{a}{\sqrt{1-a^2}}$. Then $\sqrt{1+y^2} = \sqrt{1 + \frac{a^2}{1-a^2}} = \sqrt{\frac{1-a^2+a^2}{1-a^2}} = \frac{1}{\sqrt{1-a^2}}$.So,$\sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}ight) = \log\left(\frac{a}{\sqrt{1-a^2}} + \frac{1}{\sqrt{1-a^2}}ight) = \log\left(\frac{a+1}{\sqrt{1-a^2}}ight) = \log\left(\frac{1+a}{\sqrt{(1-a)(1+a)}}ight) = \log\left(\sqrt{\frac{1+a}{1-a}}ight) = \frac{1}{2} \log\left(\frac{1+a}{1-a}ight)$.Given equation is $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}ight) = \log \left(\frac{1+x}{1-x}ight)$.Substituting the value,we get $2 	imes \frac{1}{2} \log\left(\frac{1+a}{1-a}ight) = \log \left(\frac{1+x}{1-x}ight)$.$\log\left(\frac{1+a}{1-a}ight) = \log \left(\frac{1+x}{1-x}ight)$.Comparing both sides,we get $x = a$.

If $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x$ is equal to

Similar Questions

Let $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ be two points that lie on the curve $\tan^2(x+y) + \cos^2(x+y) + y^2 + 2y = 0$ in the $XY$-plane. If the distance between $P$ and $Q$ is $d$,then $\cos d =$

If $\sec \theta \cosh y = \operatorname{cosec} x$ and $\operatorname{cosec} \theta \sinh y = \sec x$,then $\sinh ^2 y =$

$\cot 18^{\circ} \cdot \cot 36^{\circ}+1=$

$\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=$

The equation $\sin x \cos x = 2$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module