Tan π4+12cos-1 ab+Tan⁡π4-12Cos-1 ab

put 12Cos-1 ab=θ⇒Cos 2θ=ab; Now Tan π4+12cos-1 ab+Tan⁡π4-12Cos-1 ab =Tan π4+θ+Tan π4-θ=1+tan⁡θ1−tan⁡θ+1−tan⁡θ1+tan⁡θ=(1+tan⁡θ)2+(1−tan⁡θ)21−tan2⁡θ=21+tan2⁡θ1−tan2⁡θ=2cos⁡2θ=2ba

Tan π4+12cos-1 ab+Tan⁡π4-12Cos-1 ab

put 12Cos-1 ab=θ⇒Cos 2θ=ab; Now Tan π4+12cos-1 ab+Tan⁡π4-12Cos-1 ab =Tan π4+θ+Tan π4-θ=1+tan⁡θ1−tan⁡θ+1−tan⁡θ1+tan⁡θ=(1+tan⁡θ)2+(1−tan⁡θ)21−tan2⁡θ=21+tan2⁡θ1−tan2⁡θ=2cos⁡2θ=2ba

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$Tan (\frac{π}{4} + \frac{1}{2} \cos^{- 1} \frac{a}{b}) + Tan (\frac{π}{4} - \frac{1}{2} {Cos}^{- 1} \frac{a}{b})$

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b/a

2a/b

a/b

2 b/a

answer is D.

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Detailed Solution

$p u t \frac{1}{2} {Cos}^{- 1} \frac{a}{b} = θ \Rightarrow Cos 2 θ = \frac{a}{b}; N o w Tan (\frac{π}{4} + \frac{1}{2} \cos^{- 1} \frac{a}{b}) + Tan (\frac{π}{4} - \frac{1}{2} {Cos}^{- 1} \frac{a}{b}) = Tan (\frac{π}{4} + θ) + Tan (\frac{π}{4} - θ)$
$\begin{array}{l} = \frac{1 + \tan θ}{1 - \tan θ} + \frac{1 - \tan θ}{1 + \tan θ} \\ = \frac{(1 + \tan θ)^{2} + (1 - \tan θ)^{2}}{1 - \tan^{2} θ} \\ = \frac{2 (1 + \tan^{2} θ)}{1 - \tan^{2} θ} \\ = \frac{2}{\cos 2 θ} \\ = \frac{2 b}{a} \end{array}$