Question

The value of the angle ${\tan }^{-1}(\tan {65}^{\circ }-2\tan {40}^{\circ })$ in degrees is equal to

• A. $-{20}^{\circ }$
• B. ${20}^{\circ }$
• C. ${25}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is C ${25}^{\circ }$

${\tan }^{-1}\left(\tan \right({65}^0)-2\tan ({40}^0\left)\right)$
$={\tan }^{-1}\left(\tan \right({65}^0)-\tan ({40}^0)-\tan ({40}^0\left)\right)$
$={\tan }^{-1}\left(\tan \right({65}^0-{40}^0\left)\right(1+\tan \left({65}^0\right)\tan \left({40}^0\right))-\tan ({40}^0\left)\right)$

Note ...($\tan (A-B)(1+\tan A\tan B)=\tan A-\tan B$)
$={\tan }^{-1}\left(\tan \right({25}^0)+\tan ({25}^0\left)\tan \right({65}^0\left)\tan \right({40}^0)-\tan ({40}^0\left)\right)$
$={\tan }^{-1}\left(\tan \right({25}^0)+\tan ({25}^0\left)\cot \right({25}^0\left)\tan \right({40}^0)-\tan ({40}^0\left)\right)$
$={\tan }^{-1}\left(\tan \right({25}^0)+\tan ({40}^0)-\tan ({40}^0\left)\right)$
$={\tan }^{-1}\left(\tan \right({25}^0\left)\right)$
$={25}^0$