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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App