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 $\log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)=$

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