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Prove the following.
Prove that 1/2cos⁻¹((1-x)/(1+x)) = tan⁻¹√x.
Prove that $\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}\sqrt{x}$.
To show: $\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}\sqrt{x}$
Answer
Answer
Put $x = \tan^2\theta$, so $\sqrt{x} = \tan\theta$ and $\theta = \tan^{-1}\sqrt{x}$.
$\frac{1-x}{1+x} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$
$\frac{1}{2}\cos^{-1}(\cos 2\theta) = \theta = \tan^{-1}\sqrt{x}$
Hence proved.
- Step 1: Put $x = \tan^2\theta$, so $\sqrt{x} = \tan\theta$ and $\theta = \tan^{-1}\sqrt{x}$ (with $0 \le \theta < \frac{\pi}{2}$).
- Step 2: $\frac{1-x}{1+x} = \frac{1-\tan^2\theta}{1+\tan^2\theta}$, by substituting $x = \tan^2\theta$.
- Step 3: $\frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$, by the identity $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.
- Step 4: $\cos^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$, since $0 \le 2\theta < \pi$ lies in the principal range of $\cos^{-1}$.
- Step 5: $\frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) = \theta = \tan^{-1}\sqrt{x}$. Hence proved.