Question

$2\sin h^{-1}\left(\theta \right)=$

• A. $\sin h^{-1}\left(2\theta \sqrt{1+{\theta }^2}\right)$
• B. $\sin h^{-1}\left(2\theta \sqrt{1-{\theta }^2}\right)$
• C. $\sin h^{-1}\left(\theta \sqrt{1+{\theta }^2}\right)$
• D. none of these

Answer 1

The correct option is A

$\sin h^{-1}\left(2\theta \sqrt{1+{\theta }^2}\right)$

Finding the value:

We know that, $\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(x\right)\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(2x\sqrt{1-x^2}\right)$

Substitute $x=i\theta$

$\Rightarrow$ $2{\sin }^{-1}\left(i\theta \right)={\sin }^{-1}\left(2i\theta \sqrt{1-{\left(i\theta \right)}^2}\right)$

$\Rightarrow$ $2i\sin h^{-1}\left(i\theta \right)=i{\sin }^{-1}\left(2i\theta \sqrt{1+{\left(\theta \right)}^2}\right)\mathbf{}\left(\because i^2=-1\right)$

$\Rightarrow$ $2i\sin h^{-1}\left(\theta \right)=i\sin h^{-1}\left(\left(2\theta \sqrt{1+{\left(\theta \right)}^2}\right)i\right)$

$\Rightarrow$$-2\sin h^{-1}\left(\theta \right)=-\sin h^{-1}\left(2\theta \sqrt{1+{\left(\theta \right)}^2}\right)\mathbf{}\left(\because isin{h}^{-1}\left(i\theta \right)=-2sin{h}^{-1}\left(\theta \right)\right)$

$\Rightarrow$ $2\sin h^{-1}\left(\theta \right)=\sin h^{-1}\left(2\theta \sqrt{1+{\left(\theta \right)}^2}\right)$

Hence, correct answer is option $\left(A\right).$.