First slide

Properties of ITF

Question

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

Moderate

A

$\sqrt{\cot θ}$
B

tan θ
C

$\sqrt{\tan θ}$
D

none of these

Solution

Let $\sqrt{\tan θ} = \tan α$
$\begin{matrix} ∴ A = \cot^{- 1} (\tan α) - \tan^{- 1} (\tan α) \\ = \cot^{- 1} (\cot (\frac{π}{2} - α)) - \tan^{- 1} (\tan α) \\ = \cot^{- 1} (\cot (\frac{π}{2} - α)) - \tan^{- 1} (\tan α) = \frac{π}{2} - α - α \end{matrix}$
$\begin{array}{l} \Rightarrow 2 α = \frac{π}{2} - 2 α or α = \frac{π}{2} - A \\ ∴ \sqrt{\tan θ} = \tan α = \tan (\frac{π}{4} - \frac{A}{2}) . \end{array}$

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