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

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(B) Let $	heta = \frac{1}{2}{\cos ^{ - 1}}\frac{a}{b}$.Then,$2	heta = {\cos ^{ - 1}}\frac{a}{b}$,which implies $\cos 2	heta = \frac{a}{b}$.The expr

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$\frac{2b}{a}$

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(B) Let $	heta = \frac{1}{2}{\cos ^{ - 1}}\frac{a}{b}$.Then,$2	heta = {\cos ^{ - 1}}\frac{a}{b}$,which implies $\cos 2	heta = \frac{a}{b}$.The expression becomes $	an \left( {\frac{\pi }{4} + 	heta } ight) + 	an \left( {\frac{\pi }{4} - 	heta } ight)$.Using the formula $	an (A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an (A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$ with $	an \frac{\pi }{4} = 1$:$= \frac{1 + 	an 	heta }{1 - 	an 	heta } + \frac{1 - 	an 	heta }{1 + 	an 	heta }$$= \frac{{(1 + 	an 	heta )^2 + (1 - 	an 	heta )^2}}{{(1 - 	an 	heta )(1 + 	an 	heta )}}$$= \frac{{1 + {{	an }^2}	heta + 2	an 	heta + 1 + {{	an }^2}	heta - 2	an 	heta }}{{1 - {{	an }^2}	heta }}$$= \frac{{2(1 + {{	an }^2}	heta )}}{{1 - {{	an }^2}	heta }}$$= \frac{2}{{\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}}}$$= \frac{2}{{\cos 2	heta }}$Since $\cos 2	heta = \frac{a}{b}$,the expression equals $\frac{2}{{a/b}} = \frac{{2b}}{a}$.

$\tan \left[ {\frac{\pi }{4} + \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] + \tan \left[ {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] = $

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