If -frac{pi}{2}

Question

(B) Let $x = \log \left(	an \left(\frac{\pi}{4}+\frac{	heta}{2}ight)ight)$.Then,$e^x = 	an \left(\frac{\pi}{4}+\frac{	heta}{2}ight) = \frac{

Vedclass · Accepted Answer

$2 	anh ^{-1}\left(	an \frac{	heta}{2}ight)$

Vedclass · Answer

(B) Let $x = \log \left(	an \left(\frac{\pi}{4}+\frac{	heta}{2}ight)ight)$.Then,$e^x = 	an \left(\frac{\pi}{4}+\frac{	heta}{2}ight) = \frac{1+	an \frac{	heta}{2}}{1-	an \frac{	heta}{2}}$.Applying the componendo and dividendo rule:$\frac{e^x-1}{e^x+1} = \frac{(1+	an \frac{	heta}{2}) - (1-	an \frac{	heta}{2})}{(1+	an \frac{	heta}{2}) + (1-	an \frac{	heta}{2})} = \frac{2 	an \frac{	heta}{2}}{2} = 	an \frac{	heta}{2}$.We know that $	anh \left(\frac{x}{2}ight) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}} = \frac{e^x - 1}{e^x + 1}$.Therefore,$	an \frac{	heta}{2} = 	anh \left(\frac{x}{2}ight)$.Taking the inverse hyperbolic tangent on both sides:$\frac{x}{2} = 	anh ^{-1}\left(	an \frac{	heta}{2}ight)$.Thus,$x = 2 	anh ^{-1}\left(	an \frac{	heta}{2}ight)$.Hence,option $(B)$ is correct.

If $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$,then $\log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)=$

Similar Questions

$\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \dots \infty = $

If $|a| < 1$ and $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$,then $a$ is equal to

$\frac{4}{1 \times 3} - \frac{6}{2 \times 4} + \frac{12}{5 \times 7} - \frac{14}{6 \times 8} + \dots \infty = $

$\frac{1}{2} - \frac{1}{2 \cdot 2^2} + \frac{1}{3 \cdot 2^3} - \frac{1}{4 \cdot 2^4} + \ldots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module