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Solve the following.
If tan⁻¹((√(1+x²) - √(1-x²))/(√(1+x²) + √(1-x²))) = α, prove that sin 2α = x².
If $\tan^{-1}\left(\frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right) = \alpha$, prove that $\sin 2\alpha = x^2$.
Given: $\tan^{-1}\left(\frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right) = \alpha$
To show: $\sin 2\alpha = x^2$
Answer
Answer
Put $x^2 = \cos 2\theta$.
Then $\sqrt{1 + x^2} = \sqrt{2}\cos\theta$ and $\sqrt{1 - x^2} = \sqrt{2}\sin\theta$.
$\frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \tan\left(\frac{\pi}{4} - \theta\right)$
$\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta = \alpha$
$\theta = \frac{\pi}{4} - \alpha \implies 2\theta = \frac{\pi}{2} - 2\alpha$
$x^2 = \cos 2\theta = \cos\left(\frac{\pi}{2} - 2\alpha\right) = \sin 2\alpha$
Hence proved.
- Step 1: Put $x^2 = \cos 2\theta$, so $2\theta = \cos^{-1}(x^2)$.
- Step 2: Then $\sqrt{1+x^2} = \sqrt{1+\cos 2\theta} = \sqrt{2}\cos\theta$.
- Step 3: $\sqrt{1-x^2} = \sqrt{1-\cos 2\theta} = \sqrt{2}\sin\theta$.
- Step 4: Substitute into the expression: $\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} = \frac{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta} = \frac{1-\tan\theta}{1+\tan\theta} = \tan\left(\frac{\pi}{4}-\theta\right)$.
- Step 5: Thus $\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right) = \frac{\pi}{4}-\theta = \alpha$.
- Step 6: $\theta = \frac{\pi}{4}-\alpha \implies 2\theta = \frac{\pi}{2}-2\alpha$.
- Step 7: Substituting into $x^2 = \cos 2\theta$, we get $x^2 = \cos\left(\frac{\pi}{2}-2\alpha\right) = \sin 2\alpha$.