(B) Let $\theta = \tan ^{ - 1}x$. Then,$x = \tan \theta$. We know that $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1 + \tan^2 \theta}}$. Sub

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) Let $\theta = \tan ^{ - 1}x$. Then,$x = \tan \theta$. We know that $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1 + \tan^2 \theta}}$. Substituting $x = \tan \theta$ into the expression,we get: $\cos \theta = \frac{1}{\sqrt{1 + x^2}}$. Therefore,$\cos (\tan ^{ - 1}x) = \frac{1}{\sqrt{1 + x^2}}$.