$ an \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}} ight) ight)+ an \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}} ight) ight)$ is

Question

$	an \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}ight)ight)+	an \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}ight)ight)$ is

Aicharya · Accepted Answer

$\begin{array}{ll} & \quad 	ext { Let } \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)=	heta \ 	herefore \quad & \cos ^{-1}\left(\frac{a}{b}ight)=2 	heta \ 	herefore \quad & \cos 2 	heta=\frac{a}{b} \ 	herefore \quad & 	an \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)ight]+	an \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)ight]\end{array}$ $\begin{aligned} & =	an \left(\frac{\pi}{4}+	hetaight)+	an \left(\frac{\pi}{4}-	hetaight) \ & =\frac{1+	an 	heta}{1-	an 	heta}+\frac{1-	an 	heta}{1+	an 	heta} \ & =\frac{(1+	an 	heta)^2+(1-	an 	heta)^2}{1-	an ^2 	heta} \ & =\frac{2\left(1+	an ^2 	hetaight)}{1-	an ^2 	heta} \ & =\frac{2}{1-	an ^2 	heta} \ & 	ext {. } \overline{1+	an ^2 	heta} \ & =\frac{2}{\cos 2 	heta} \ & =\frac{2}{\frac{a}{b}}=\frac{2 b}{a} \end{aligned}$

$\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right)+\tan…

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