tan left[frac{pi}{4}+frac{1}{2} cos ^{-1}left | vedclass

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(B) Let $	heta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)$, then $\cos 2 	heta = \frac{a}{b}$.The expression is $	an \left(\frac{\pi}{4} +

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$\frac{2 b}{a}$

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(B) Let $	heta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)$, then $\cos 2 	heta = \frac{a}{b}$.The expression is $	an \left(\frac{\pi}{4} + 	hetaight) + 	an \left(\frac{\pi}{4} - 	hetaight)$.Using the identity $	an(A+B) + 	an(A-B) = \frac{2 	an A}{1 - 	an^2 A 	an^2 B}$ is not standard, so we expand:$= \left( \frac{1 + 	an 	heta}{1 - 	an 	heta} ight) + \left( \frac{1 - 	an 	heta}{1 + 	an 	heta} ight)$$= \frac{(1 + 	an 	heta)^2 + (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$$= \frac{1 + 2 	an 	heta + 	an^2 	heta + 1 - 2 	an 	heta + 	an^2 	heta}{1 - 	an^2 	heta}$$= \frac{2(1 + 	an^2 	heta)}{1 - 	an^2 	heta} = \frac{2}{\cos 2 	heta}$.Substituting $\cos 2 	heta = \frac{a}{b}$, we get $\frac{2}{a/b} = \frac{2b}{a}$.

$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

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