If cos⁡θ=cos⁡αcos⁡β, then cot2⁡β2=

cos⁡θ=cos⁡αcos⁡β⇒cos⁡β=cos⁡θcos⁡α⇒1+cos⁡β1−cos⁡β=cos⁡α+cos⁡θcos⁡α−cos⁡θ⇒cot2⁡β2=2cos⁡θ+α2cos⁡θ−α22sin⁡θ+α2sin⁡θ−α2⇒cot2⁡β2=cot⁡θ+α2cot⁡θ−α2

If cos⁡θ=cos⁡αcos⁡β, then cot2⁡β2=

cos⁡θ=cos⁡αcos⁡β⇒cos⁡β=cos⁡θcos⁡α⇒1+cos⁡β1−cos⁡β=cos⁡α+cos⁡θcos⁡α−cos⁡θ⇒cot2⁡β2=2cos⁡θ+α2cos⁡θ−α22sin⁡θ+α2sin⁡θ−α2⇒cot2⁡β2=cot⁡θ+α2cot⁡θ−α2

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$If \cos θ = \cos αcos β, then \cot^{2} \frac{β}{2} =$

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$\cot (\frac{θ + α}{2}) \cot (\frac{θ - α}{2})$

$\cot (\frac{θ + α}{2}) \cot (\frac{θ + α}{2})$

$\cot (\frac{θ - α}{2}) \cot (\frac{θ - α}{2})$

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Detailed Solution

$\begin{array}{l} \cos θ = \cos αcos β \\ \Rightarrow \cos β = \frac{\cos θ}{\cos α} \\ \Rightarrow \frac{1 + \cos β}{1 - \cos β} = \frac{\cos α + \cos θ}{\cos α - \cos θ} \\ \Rightarrow \cot^{2} \frac{β}{2} = \frac{2 \cos (\frac{θ + α}{2}) \cos (\frac{θ - α}{2})}{2 \sin (\frac{θ + α}{2}) \sin (\frac{θ - α}{2})} \\ \Rightarrow \cot^{2} \frac{β}{2} = \cot (\frac{θ + α}{2}) \cot (\frac{θ - α}{2}) \end{array}$