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If $A = \cot^{- 1} \sqrt{\tan θ} - \tan^{- 1} \sqrt{\tan θ}, then \tan (\frac{π}{4} - \frac{A}{2})$ is equal to

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Correct option is C

Let tan⁡θ=tan⁡α∴A=cot−1⁡(tan⁡α)−tan−1⁡(tan⁡α)=cot−1⁡cot⁡π2−α−tan−1⁡(tan⁡α)=π2−α−α=π2−2α⇒2α=π2−A or α=π4−A2∴tan⁡θ=tan⁡α=tan⁡π4−A2

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If A=cot−1⁡tan⁡θ−tan−1⁡tan⁡θ, then tan⁡π4−A2 is equal to

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If $A = \cot^{- 1} \sqrt{\tan θ} - \tan^{- 1} \sqrt{\tan θ}, then \tan (\frac{π}{4} - \frac{A}{2})$ is equal to