Question

If \[f\left( x \right) = \frac{2x}{1 + x^2}\] , show that f(tan θ) = sin 2θ.

 

 

Answer 1

Given:

\[f\left( x \right) = \frac{2x}{1 + x^2}\]

Thus,

\[f\left( \tan\theta \right) = \frac{2\left( \tan\theta \right)}{1 + \tan^2 \theta}\]

\[= \frac{2 \times \frac{\sin \theta}{\cos \theta}}{1 + \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)}\]

\[ = \frac{2 \sin \theta}{\cos \theta} \times \frac{\cos^2 \theta}{\cos^2 \theta + \sin^2 \theta}\]

\[ = \frac{2 \sin \theta \cos \theta}{1} \left[ \because \cos^2 \theta + \sin^2 \theta = 1 \right]\]

\[ = \sin 2\theta \left[ \because 2 \sin \theta \cos \theta = \sin 2\theta \right]\]

Hence,  f (tan θ) = sin 2θ.