Show 2tan⁻¹x + sin⁻¹(2x/(1+x²)) is Constant

Show that \( 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) is constant for \( x \ge 1 \), and find its value.

Solution:

Let

\[ \theta = \tan^{-1}(x) \Rightarrow x = \tan \theta \]

Then,

\[ \frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta) \]

Thus,

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

Since \( x \ge 1 \Rightarrow \theta \ge \frac{\pi}{4} \), so \( 2\theta \in \left[\frac{\pi}{2}, \pi\right) \)

Hence,

\[ \sin^{-1}(\sin 2\theta) = \pi – 2\theta \]

Now substitute back:

\[ 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\theta + (\pi – 2\theta) = \pi \]

Thus, the expression is constant.

\[ 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi \quad \text{for } x \ge 1 \]

Spread the love