Q. If $a > b > 0$, then the value of $\tan^{- 1}{\left( \frac{a}{b} \right) + \tan^{- 1}\left( \frac{a + b}{a - b} \right)}$ depends on
A) Both $a$ and $b$
B) $b$ and not $a$
C) $a$ and not $b$
D) Neither $a$ nor $b$
Explanation
(d)
$\tan^{- 1}{\left( \frac{a}{b} \right) + \tan^{- 1}\left( \frac{a + b}{a - b} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left\{ \frac{\frac{a}{b} + \frac{a + b}{a - b}}{1 - \frac{a}{b}\left( \frac{a + b}{a - b} \right)} \right\}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {= tan}^{- 1}\left( - \frac{a^{2} + b^{2}}{a^{2} + b^{2}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{( - 1)}$
$\therefore$ The value is neither depends on $a$ nor $b$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.