Chapter 7: Q. 99 (page 487)

Show that $\tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$ , if $v > 0$

Short Answer

Expert verified

The equation $\tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$ can be obtained by the formula

$\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2} c o t^{- 1} v = \frac{π}{2} - \tan^{- 1} v \tan^{- 1} v + c o t^{- 1} v = \frac{π}{2} \tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$

Step by step solution

Step 1. Given data

The given equation for verification is

$\tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$

Step 2. Consider

Consider

$c o t^{- 1} v = A v = c o t A \frac{1}{v} = \frac{1}{c o t A} \frac{1}{v} = \tan A \tan^{- 1} (\frac{1}{v}) = A \tan^{- 1} (\frac{1}{v}) = c o t^{- 1} v \dots (1)$

Step 3. proof

Use formula

$\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2} c o t^{- 1} v = \frac{π}{2} - \tan^{- 1} v$

From equation $1$

$c o t^{- 1} v = \frac{π}{2} - \tan^{- 1} v \tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$

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Show that $\tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$ , if $v > 0$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Consider

Step 3. proof

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Show that $\tan^{- 1} (\frac{1}{v}) = \frac{π}{2} - \tan^{- 1} v$ , if $v > 0$