If tanh^{-1}(x+iy) = frac{1}{2} tanh^{-1}left | vedclass

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

Question

(C) We know the relationship between the inverse hyperbolic tangent and the inverse trigonometric tangent function: $	anh^{-1}(z) = \frac{1}{i} 	an^

Vedclass · Accepted Answer

$i 	an^{-1}(y)$

Vedclass · Answer

(C) We know the relationship between the inverse hyperbolic tangent and the inverse trigonometric tangent function: $	anh^{-1}(z) = \frac{1}{i} 	an^{-1}(iz)$.To find $	anh^{-1}(iy)$,we substitute $z = iy$ into the formula:$	anh^{-1}(iy) = \frac{1}{i} 	an^{-1}(i(iy))$Since $i^2 = -1$,this simplifies to:$	anh^{-1}(iy) = \frac{1}{i} 	an^{-1}(-y)$Using the property $	an^{-1}(-y) = -	an^{-1}(y)$,we get:$	anh^{-1}(iy) = \frac{1}{i} (-	an^{-1}(y)) = -\frac{1}{i} 	an^{-1}(y)$Since $\frac{1}{i} = -i$,we have $-\frac{1}{i} = i$.Therefore,$	anh^{-1}(iy) = i 	an^{-1}(y)$.Thus,option $C$ is correct.

If $\tanh^{-1}(x+iy) = \frac{1}{2} \tanh^{-1}\left(\frac{2x}{1+x^2+y^2}\right) + \frac{i}{2} \tan^{-1}\left(\frac{2y}{1-x^2-y^2}\right)$,where $x, y \in \mathbb{R}$,then $\tanh^{-1}(iy) =$

Similar Questions

$\sinh^{-1}\left(2^{3/2}\right)$ is equal to :

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

If $2 \cosh 2x + 10 \sinh 2x = 5$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module