2tan−1⁡a−ba+btan⁡θ2 is

We know that, 2tan−1⁡t=cos−1⁡1−t21+t2 2tan−1⁡a−ba+b⋅tan⁡θ2=cos−1⁡1−a−ba+btan2⁡θ21+a−ba+btan2⁡θ2=cos−1⁡acos⁡θ+ba+bcos⁡θ (on simplification)

2tan−1⁡a−ba+btan⁡θ2 is

We know that, 2tan−1⁡t=cos−1⁡1−t21+t2 2tan−1⁡a−ba+b⋅tan⁡θ2=cos−1⁡1−a−ba+btan2⁡θ21+a−ba+btan2⁡θ2=cos−1⁡acos⁡θ+ba+bcos⁡θ (on simplification)

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$2 \tan^{- 1} \sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}$ is

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$\cos^{- 1} (\frac{a \cos θ + b}{a})$

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

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

$\cos^{- 1} (\frac{b \cos θ + a}{b})$

answer is B.

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Detailed Solution

We know that, $2 \tan^{- 1} t = \cos^{- 1} \frac{1 - t^{2}}{1 + t^{2}}$

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