First slide

Properties of ITF

Question

If $x < 0,$ then $\tan^{- 1} (\frac{1}{x})$ equals

Moderate

A

$\cot^{- 1} x$
B

$- \cot^{- 1} x$
C

$- π + \cot^{- 1} x$
D

$- π - \cot^{- 1} x$

Solution

Let $\cot^{- 1} x = θ .$ Then, $x = \cot θ$
Also, $x < 0 \Rightarrow \cot θ < 0 \Rightarrow \frac{π}{2} < θ < π$
Now,
$\begin{array}{l} \tan^{- 1} (\frac{1}{x}) \\ \Rightarrow \tan^{- 1} (\frac{1}{x}) = \tan^{- 1} (\tan θ) \\ \Rightarrow \tan^{- 1} (\frac{1}{x}) = \tan^{- 1} (- \tan (π - θ)) \end{array}$
$\begin{array}{l} \Rightarrow \tan^{- 1} (\frac{1}{x}) = \tan^{- 1} (\tan (θ - π)) \\ \Rightarrow \tan^{- 1} (\frac{1}{x}) = θ - π [\frac{π}{2} < θ < π \Rightarrow - \frac{π}{2} < θ - π < 0] \\ \Rightarrow \tan^{- 1} (\frac{1}{x}) = \cot^{- 1} x - π . \end{array}$

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