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

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

$\cos \left(\cos ^{-1} \frac{1}{3}+\cos ^{-1} \frac{1}{5}\right)+\cos \left(\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{1}{5}\right) =$ . . . . . . .

If $2 \sin^{-1} x - 3 \cos^{-1} x = 4$, then $2 \sin^{-1} x + 3 \cos^{-1} x =$

Assertion $(A): \operatorname{cosech}^{-1}(3) = \log \left(\frac{1+\sqrt{10}}{3}\right)$
Reason $(R): e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2 - 2p - x = 0$
The correct option among the following is

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x) = \tan^{-1}(\sin x + \cos x)$ in the interval $[0, \frac{\pi}{2}]$. Then the value of $\tan(M - m)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module