Prove that: `tan{pi/4+1/2\ cos^(-1)(a/b)}+tan{pi/4-1/2\ cos^(-1)(a/b)}=(2b)/a`

Explanation

`L.H.S.`

`=\tan (\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b})+\tan (\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b})`

`=\frac{\tan \frac{\pi}{4}+\tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}{1-\tan \frac{\pi}{4} \tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}+\frac{\tan \frac{\pi}{4}-\tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}{1+\tan \frac{\pi}{4} \tan (\frac{1}{2} \cos ^{-1} \frac{1}{b})}`

`=\frac{1+\tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}{1-\tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}+\frac{1-\tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}{1+\tan (\frac{1}{2} \cos ^{-1} \frac{a}{b})}`

`=\frac{[1+\tan (\frac{1}{2} \cos ^{-1}(\frac{a}{b}))]^{2}+[1-\tan (\frac{1}{2} \cos ^{-1}(\frac{a}{b}))]^{2}}{1-\tan ^{2}(\frac{1}{2} \cos ^{-1} \frac{a}{b})}`


`=\frac{2 \sec ^{2}(\frac{1}{2} \cos ^{-1} \frac{a}{b})}{1-\tan ^{2}(\frac{1}{2} \cos ^{-1} \frac{a}{b})}=\frac{2 \sec ^{2} \theta}{1-\tan ^{2} \theta}=\frac{2(1+\tan ^{2} \theta)}{1-\tan ^{2} \theta}`


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