logo

Ask Questions, Get Answers

X
 
Home  >>  CBSE XII  >>  Math  >>  Model Papers

Write the principal value of $\tan^{-1}(\sqrt 3)-cot^{-1}(-\sqrt 3).$

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

1 Answer

Toolbox:
  • From the properties of inverse circular functions, we know that $cot^{-1}(-x) = \pi - cot^{-1}(x)$ and $tan^{-1}(x)+cot^{-1}(x) = \large \frac{\pi}{2}$
Given $\tan^{-1}(\sqrt 3)-cot^{-1}(-\sqrt 3)$
From the properties of inverse circular functions, we know that $cot^{-1}(-x) = \pi - cot^{-1}(x)$ and $tan^{-1}(x)+cot^{-1}(x) = \large \frac{\pi}{2}$
Let $x = \sqrt 3 \Rightarrow cot^{-1}(-\sqrt 3) = \pi - cot^{-1}(\sqrt 3)$
Therefore $\tan^{-1}(\sqrt 3)-cot^{-1}(-\sqrt 3) = \tan^{-1}(\sqrt 3)-( \pi - cot^{-1}(\sqrt 3)$
$\Rightarrow \tan^{-1}(\sqrt 3)-cot^{-1}(-\sqrt 3)= tan^{-1}(\sqrt 3)+cot^{-1}(\sqrt 3) - \pi$
Substituting $tan^{-1}(\sqrt 3)+cot^{-1}(\sqrt 3) = \large \frac{\pi}{2}$, we get:
$\Rightarrow \tan^{-1}(\sqrt 3)-cot^{-1}(-\sqrt 3)= \large\frac{\pi}{2}$$ - \pi = -\large \frac{\pi}{2}$
answered Mar 21, 2013 by balaji.thirumalai
 

Related questions

...