If tan (u + iv) = i,then the value of v is | vedclass

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(B) Given $	an (u + iv) = i$. Using the identity $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $\frac{	an u + 	an (iv)}{1 -

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$\infty$

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(B) Given $	an (u + iv) = i$. Using the identity $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $\frac{	an u + 	an (iv)}{1 - 	an u 	an (iv)} = i$. Since $	an (iv) = i 	anh v$,the equation becomes $\frac{	an u + i 	anh v}{1 - i 	an u 	anh v} = i$. Multiplying both sides by the denominator: $	an u + i 	anh v = i(1 - i 	an u 	anh v) = i + 	an u 	anh v$. Rearranging terms: $	an u - i = 	anh v (	an u - i)$. This implies $(	an u - i)(1 - 	anh v) = 0$. For this to hold,$1 - 	anh v = 0$,which means $	anh v = 1$. Since $	anh v = \frac{e^v - e^{-v}}{e^v + e^{-v}}$,setting this equal to $1$ leads to $e^v - e^{-v} = e^v + e^{-v}$,which simplifies to $2e^{-v} = 0$. This is only possible as $v 	o \infty$.

If $\tan (u + iv) = i$,then the value of $v$ is

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