Question

lf $\theta ={\tan }^{-1}a,\varphi ={\tan }^{-1}b$ and $ab=-1$, then $\theta -\varphi$ is equal to

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pm \frac{\pi }{2}$

Answer 1

Answer: (D)

$\pm \frac{\pi }{2}$

Given:

$ab=-1$

$\theta ={\tan }^{-1}a\Rightarrow \tan \theta =a$

$\varphi ={\tan }^{-1}b\Rightarrow \tan \varphi =b$

$\therefore ab=\tan \theta \tan \varphi$
$\therefore \tan (\theta -\varphi )=\frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }=\frac{a-b}{1+ab}$
$\Rightarrow \tan (\theta -\varphi )=\pm \infty$

(If $a>b,$ then $+\infty$, if $a<b,$ then $-\infty$)

$\Rightarrow \theta -\varphi =\pm \frac{\pi }{2}$

Hence, option D.