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प्रश्न
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
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उत्तर
Put \[x^2 = \cos2\theta\], we have
\[\tan^{- 1} \left( \frac{\sqrt{1 + \cos2\theta} + \sqrt{1 - \cos2\theta}}{\sqrt{1 + \cos2\theta} - \sqrt{1 - \cos2\theta}} \right)\]
\[ = \tan^{- 1} \left( \frac{\sqrt{2c {os}^2 \theta} + \sqrt{2 \sin^2 \theta}}{\sqrt{2 \cos^2 \theta} - \sqrt{2 \sin^2 \theta}} \right)\]
\[ = \tan^{- 1} \left( \frac{cos\theta + \sin\theta}{cos\theta - \sin\theta} \right)\]
\[ = \tan^{- 1} \left( \frac{1 + \tan\theta}{1 - \tan\theta} \right)\]
\[= \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} + \theta \right) \right]\]
\[ = \frac{\pi}{4} + \theta \left[ \because - 1 < x < 1 \Rightarrow 0 < x^2 < 1 \Rightarrow 0 < 2\theta < \frac{\pi}{2} \Rightarrow 0 < \theta < \frac{\pi}{4} \right]\]
\[ = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 \left[ \because x^2 = \cos2\theta \Rightarrow 2\theta = \cos^{- 1} x^2 \right]\]
Hence proved.
