2tan−1a−ba+btanθ2=

2tan−1[a−ba+btanθ2]=cos−11−a−ba+btan2θ21+a−ba+btan2θ2 ∵2tan−1x=cos−11−x21+x2 =cos−1(a+b)−(a−b)tan2θ2(a+b)+(a−b)tan2θ2=cos−1a(1−tan2θ2)+b(1+tan2θ2)a(1+tan2θ2)+b(1−tan2θ2)=cos−1a(1−tan2θ2)1+tan2θ2+ba+b(1−tan2θ21+tan2θ2)=cos−1acosθ+ba+bcosθ

2tan−1a−ba+btanθ2=

2tan−1[a−ba+btanθ2]=cos−11−a−ba+btan2θ21+a−ba+btan2θ2 ∵2tan−1x=cos−11−x21+x2 =cos−1(a+b)−(a−b)tan2θ2(a+b)+(a−b)tan2θ2=cos−1a(1−tan2θ2)+b(1+tan2θ2)a(1+tan2θ2)+b(1−tan2θ2)=cos−1a(1−tan2θ2)1+tan2θ2+ba+b(1−tan2θ21+tan2θ2)=cos−1acosθ+ba+bcosθ

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$2 \tan^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}] =$

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$\cos^{- 1} (\frac{a \cos θ + b}{a + b \cos θ})$

$\cos^{- 1} (\frac{a + b \cos θ}{a \cos θ + b})$

$\cos^{- 1} (\frac{a \cos θ}{a + b \cos θ})$

$\cos^{- 1} (\frac{b \cos θ}{a \cos θ + b})$

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Detailed Solution

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

$(∵ 2 \tan^{- 1} x = \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}})$

$\begin{array}{l} = \cos^{- 1} [\frac{(a + b) - (a - b) \tan^{2} \frac{θ}{2}}{(a + b) + (a - b) \tan^{2} \frac{θ}{2}}] \\ = \cos^{- 1} [\frac{a (1 - \tan^{2} \frac{θ}{2}) + b (1 + \tan^{2} \frac{θ}{2})}{a (1 + \tan^{2} \frac{θ}{2}) + b (1 - \tan^{2} \frac{θ}{2})}] \\ = \cos^{- 1} [\frac{\frac{a (1 - \tan^{2} \frac{θ}{2})}{1 + \tan^{2} \frac{θ}{2}} + b}{a + b (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}})}] \\ = \cos^{- 1} [\frac{a \cos θ + b}{a + b \cos θ}] \end{array}$