Question

$\cos \left({\tan }^{-1}x\right)=$

• A. $\sqrt{1+x^2}$
• B. $\frac{1}{\sqrt{1+x^2}}$
• C. $1+x^2$
• D. None of these

Answer 1

The correct option is B $\frac{1}{\sqrt{1+x^2}}$

Let $\theta ={\tan }^{-1}x\Rightarrow x=\tan \theta$
$\therefore \cos \theta =\frac{1}{\sqrt{1+{\tan }^2\theta }}=\frac{1}{\sqrt{1+x^2}}$
Hence $\cos \theta =\cos \left(ta{n}^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}$