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